Chapter 10 Straight Lines (Concepts)

Cartesian Plane

Coordinate geometry, also known as analytical geometry, provides a powerful bridge between algebra and geometry. This field was pioneered by the French mathematician René Descartes. Its fundamental tool is the Cartesian plane (or coordinate plane), a system that uses a pair of numbers, called coordinates, to uniquely determine the position of a point on a flat, two-dimensional surface.

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The Coordinate System

The Cartesian plane is formed by two perpendicular number lines that intersect at their zero points. These lines act as a reference grid for locating any point.

The Coordinate Axes

1. The X-axis: This is the horizontal number line. Positive numbers are to the right of the intersection point, and negative numbers are to the left. It is denoted by the line XX'.

2. The Y-axis: This is the vertical number line. Positive numbers are above the intersection point, and negative numbers are below. It is denoted by the line YY'.

3. The Origin: The point where the x-axis and y-axis intersect is called the origin. It is the starting point for all measurements and is denoted by the letter O. Its coordinates are $(0, 0)$.

Coordinates of a Point

Any point P on the plane is represented by an ordered pair of numbers $(x, y)$. The order is important, for instance, the point $(2, 5)$ is different from the point $(5, 2)$.

• The first coordinate, $x$, is the abscissa or the x-coordinate. It is the perpendicular distance of the point from the y-axis. It tells us how far to move right (positive) or left (negative) from the origin.
• The second coordinate, $y$, is the ordinate or the y-coordinate. It is the perpendicular distance of the point from the x-axis. It tells us how far to move up (positive) or down (negative) from the origin.

A point with abscissa $x$ and ordinate $y$ is written as P$(x, y)$. For example, to locate the point A(3, -4), we start at the origin, move 3 units to the right along the x-axis, and then move 4 units down parallel to the y-axis.

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Quadrants

The two coordinate axes divide the entire plane into four infinite regions. These regions are called quadrants. They are numbered using Roman numerals (I, II, III, IV) in a counter-clockwise direction, starting from the upper-right region.

The signs of the x and y coordinates are unique for each quadrant.

Quadrant	X-coordinate (Abscissa)	Y-coordinate (Ordinate)	Coordinates Form
First Quadrant (I)	Positive ($x > 0$)	Positive ($y > 0$)	(+, +)
Second Quadrant (II)	Negative ($x < 0$)	Positive ($y > 0$)	(−, +)
Third Quadrant (III)	Negative ($x < 0$)	Negative ($y < 0$)	(−, −)
Fourth Quadrant (IV)	Positive ($x > 0$)	Negative ($y < 0$)	(+, −)

Points on the Axes

Points that lie exactly on the coordinate axes are special because they do not belong to any quadrant.

• Any point on the x-axis has a y-coordinate of 0. Its form is $(x, 0)$.
• Any point on the y-axis has an x-coordinate of 0. Its form is $(0, y)$.
• The origin $(0, 0)$ is the unique point that lies on both the x-axis and the y-axis.

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Distance Formula and Section Formula

The Cartesian coordinate system allows us to express geometric ideas using algebra.

We can create formulas based on coordinates to find the distance between points or to locate a point that divides a line segment.

These formulas are fundamental tools in the study of coordinate geometry.

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Distance Formula

The Distance Formula is used to calculate the length of a line segment connecting two points in a plane.

This formula is a direct application of the Pythagorean theorem.

Derivation of the Distance Formula

Given: Two points, $P(x_1, y_1)$ and $Q(x_2, y_2)$, in the Cartesian plane.

To Find: The distance between P and Q.

Construction Required:

1. Draw a line through P parallel to the x-axis.

2. Draw a line through Q parallel to the y-axis.

3. Let these two lines intersect at a point R. The coordinates of R will be $(x_2, y_1)$.

This construction forms a right-angled triangle, $\triangle PQR$, with the right angle at R.

Proof:

Now, we find the lengths of the sides PR and QR.

The length of the horizontal side PR is the difference between the x-coordinates:

$PR = |x_2 - x_1|$

The length of the vertical side QR is the difference between the y-coordinates:

$QR = |y_2 - y_1|$

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Substituting the lengths of PR and QR:

To find the length PQ, we take the square root of both sides. Since distance cannot be negative, we only consider the positive root.

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

This is the Distance Formula.

Note that the result is the same if we use $(x_1 - x_2)^2$ and $(y_1 - y_2)^2$, as squaring removes any negative sign.

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Section Formula

The Section Formula is a vital tool in coordinate geometry used to find the coordinates of a point that divides a line segment, joining two given points, into a specific ratio. This division can occur in two ways: Internally or Externally.

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1. Internal Division

When a point $P(x, y)$ lies on the line segment $AB$ and divides it in the ratio $m:n$, it is called internal division.

Derivation of Internal Section Formula

Given: Two points $A(x_1, y_1)$ and $B(x_2, y_2)$. Point $P(x, y)$ divides the segment $AB$ internally such that $\frac{AP}{PB} = \frac{m}{n}$.

To Prove: $P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$

Construction:

Draw $AL, PN,$ and $BM$ perpendicular to the $X$-axis. Further, draw $AK \perp PN$ and $PH \perp BM$.

Proof:

From the figure, we can determine the lengths of the sides of the triangles:

In $\triangle AKP$ and $\triangle PHB$, the angles are equal (corresponding angles and right angles), so:

Since the triangles are similar, their corresponding sides are proportional:

Substituting the ratio $\frac{m}{n}$ and the lengths derived above:

Solving for $x$ from equation (i):

Similarly, using the $y$ coordinates:

Thus, the coordinates of $P$ are $\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$.

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2. Midpoint Formula

The midpoint is a specific case of the section formula where the point $P$ divides $AB$ into two equal halves.

Derivation

In this case, the ratio $m:n$ becomes $1:1$. Substituting $m=1$ and $n=1$ into the internal section formula:

Hence, the midpoint coordinates are: $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.

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3. External Division

Point $P(x, y)$ divides $AB$ externally if it lies on the line passing through $A$ and $B$, but outside the segment $AB$. The ratio is defined as $\frac{AP}{BP} = \frac{m}{n}$.

Derivation of External Section Formula

Construction: Similar to the internal division, we draw perpendiculars from $A, B,$ and $P$ to the $X$-axis. Let $P$ lie beyond $B$.

Proof:

By using similar triangles (The larger $\triangle$ and the smaller $\triangle$ formed by perpendiculars), we have:

Solving for $x$:

Applying the same logic for $y$:

Thus, the external division coordinates are: $\left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n} \right)$.

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Centroid, Incentre and Area of a Triangle

Coordinate geometry provides formulas to find special points inside a triangle, such as the centroid and incentre.

It also allows us to calculate the area of a triangle using only the coordinates of its vertices.

These formulas are derived from fundamental concepts like the distance and section formulas.

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Centroid of a Triangle

The centroid is a specific point inside a triangle, often considered its "center of mass".

It is the point where the three medians of the triangle intersect.

A median is a line segment that connects a vertex to the midpoint of the opposite side.

An important property is that the centroid divides each median in the ratio $2:1$.

Derivation of the Centroid Formula

Given: A triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

To Find: The coordinates of the centroid, G.

Solution:

First, consider the median AD, where D is the midpoint of the side BC.

We find the coordinates of D using the midpoint formula:

$D = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)$

The centroid G lies on the median AD.

It divides AD in the ratio $AG : GD = 2 : 1$.

Now, we use the internal section formula to find the coordinates of G.

Here, the points are $A(x_1, y_1)$ and $D\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)$, and the ratio is $m:n = 2:1$.

The x-coordinate of G is:

$x = \frac{m x_D + n x_A}{m+n} = \frac{2 \left(\frac{x_2+x_3}{2}\right) + 1(x_1)}{2+1}$

$x = \frac{x_2+x_3+x_1}{3}$

The y-coordinate of G is:

$y = \frac{m y_D + n y_A}{m+n} = \frac{2 \left(\frac{y_2+y_3}{2}\right) + 1(y_1)}{2+1}$

$y = \frac{y_2+y_3+y_1}{3}$

So, the coordinates of the centroid G are the average of the coordinates of the vertices.

$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$

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Incentre of a Triangle

The incentre of a triangle is the point of concurrency of the internal bisectors of the angles of the triangle. It is equidistant from all three sides of the triangle. This equidistant point is the center of the incircle, which is the largest circle that can be inscribed within the triangle, touching all three sides.

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Formula for the Incentre

Let the vertices of a triangle $\triangle ABC$ be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

Let $a, b, c$ be the lengths of the sides opposite to vertices A, B, and C, respectively. These lengths can be calculated using the distance formula:

$a = BC = \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2}$

$b = AC = \sqrt{(x_1-x_3)^2 + (y_1-y_3)^2}$

$c = AB = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$

The coordinates of the incentre, denoted by $I$, are given by the formula:

$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$

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Derivation of the Incentre Formula

The derivation uses the Angle Bisector Theorem and the Section Formula.

1. Angle Bisector Theorem: This theorem states that an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.

Let AD be the angle bisector of $\angle A$, where D is a point on BC. According to the Angle Bisector Theorem:

$\frac{BD}{DC} = \frac{\text{Side adjacent to BD}}{\text{Side adjacent to DC}} = \frac{AB}{AC} = \frac{c}{b}$

This means that the point D divides the side BC in the ratio $c:b$.

2. Find Coordinates of D: Using the section formula, the coordinates of D which divides $B(x_2, y_2)$ and $C(x_3, y_3)$ in the ratio $c:b$ are:

$D = \left( \frac{c x_3 + b x_2}{c+b}, \frac{c y_3 + b y_2}{c+b} \right)$

3. Incentre I divides AD: The incentre I lies on the angle bisector AD. Now, consider $\triangle ABD$. The line segment BI is the angle bisector of $\angle B$. Applying the Angle Bisector Theorem to $\triangle ABD$:

$\frac{AI}{ID} = \frac{AB}{BD}$

We know $AB = c$. We need to find the length of BD. Since D divides BC in the ratio $c:b$, we have:

$BD = \left( \frac{c}{c+b} \right) \times \text{length of BC} = \frac{c}{c+b} \times a = \frac{ac}{b+c}$

Now, substitute the value of BD back into the ratio:

$\frac{AI}{ID} = \frac{c}{\frac{ac}{b+c}} = \frac{c(b+c)}{ac} = \frac{b+c}{a}$

So, the incentre I divides the line segment AD in the ratio $(b+c):a$.

4. Find Coordinates of I: Now, we use the section formula again to find the coordinates of I, which divides the line segment joining $A(x_1, y_1)$ and $D\left( \frac{bx_2 + cx_3}{b+c}, \frac{by_2 + cy_3}{b+c} \right)$ in the ratio $(b+c):a$.

x-coordinate of I:

$x = \frac{(b+c) \times (\text{x-coord of D}) + a \times (\text{x-coord of A})}{(b+c)+a}$

$x = \frac{(b+c) \left( \frac{bx_2 + cx_3}{b+c} \right) + a x_1}{a+b+c} = \frac{bx_2 + cx_3 + ax_1}{a+b+c}$

y-coordinate of I:

$y = \frac{(b+c) \times (\text{y-coord of D}) + a \times (\text{y-coord of A})}{(b+c)+a}$

$y = \frac{(b+c) \left( \frac{by_2 + cy_3}{b+c} \right) + a y_1}{a+b+c} = \frac{by_2 + cy_3 + ay_1}{a+b+c}$

Combining and rearranging the terms gives the coordinates of the incentre I:

$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$

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Area of a Triangle

The area of a triangle can be calculated directly from the coordinates of its three vertices without needing to know the side lengths or angles.

Let the vertices of the triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

The formula for the area of $\triangle ABC$ is:

Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

The absolute value is taken because area must always be a positive quantity.

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Derivation of the Area Formula

We can derive this formula by calculating the area of the region under the vertices using trapeziums formed by dropping perpendiculars from the vertices to the x-axis.

Let's drop perpendiculars from A, B, and C to the x-axis, meeting the axis at P, Q, and R respectively. The coordinates of these points are $P(x_1, 0)$, $Q(x_2, 0)$, and $R(x_3, 0)$.

Assuming $x_1 < x_2 < x_3$ for this derivation, we can see three trapeziums are formed: ABQP, BCRQ, and ACRP.

The area of $\triangle ABC$ can be found as:

Area($\triangle ABC$) = Area(Trapezium ABQP) + Area(Trapezium BCRQ) - Area(Trapezium ACRP)

The area of a trapezium is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$.

Area(Trapezium ABQP) = $\frac{1}{2} (AP + BQ) \times PQ = \frac{1}{2} (y_1 + y_2) (x_2 - x_1)$

Area(Trapezium BCRQ) = $\frac{1}{2} (BQ + CR) \times QR = \frac{1}{2} (y_2 + y_3) (x_3 - x_2)$

Area(Trapezium ACRP) = $\frac{1}{2} (AP + CR) \times PR = \frac{1}{2} (y_1 + y_3) (x_3 - x_1)$

Now, substitute these into the main expression:

Area($\triangle ABC$) = $\frac{1}{2} [(y_1 + y_2) (x_2 - x_1) + (y_2 + y_3) (x_3 - x_2) - (y_1 + y_3) (x_3 - x_1)]$

Expanding the terms:

$= \frac{1}{2} [x_2 y_1 - x_1 y_1 + x_2 y_2 - x_1 y_2 + x_3 y_2 \ $$ - x_2 y_2 + x_3 y_3 \ $$ - x_2 y_3 - (x_3 y_1 \ $$ - x_1 y_1 + \ $$ x_3 y_3 - x_1 y_3)]$

$= \frac{1}{2} [x_2 y_1 - x_1 y_1 - x_1 y_2 + x_3 y_2 - x_2 y_3 - x_3 y_1 + x_1 y_1 + x_1 y_3]$

$= \frac{1}{2} [x_2 y_1 - x_1 y_2 + x_3 y_2 - x_2 y_3 - x_3 y_1 + x_1 y_3]$

Now, group the terms by $x_1, x_2, x_3$:

$= \frac{1}{2} [x_1(y_3 - y_2) + x_2(y_1 - y_3) + x_3(y_2 - y_1)]$

Taking a negative sign out from all terms inside the bracket to match the standard formula:

$= \frac{1}{2} [ -x_1(y_2 - y_3) - x_2(y_3 - y_1) - x_3(y_1 - y_2)]$

$= -\frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$

Since area must be positive, we take the absolute value of this expression, which gives the final formula:

Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

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Condition for Collinearity of Three Points

Three or more points are said to be collinear if they lie on the same straight line.

If three points A, B, and C are collinear, they cannot form a triangle. Therefore, the area of the triangle formed by these three points must be zero.

This gives us a direct method to check for collinearity.

The points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear if and only if Area($\triangle ABC$) = 0.

This leads to the condition for collinearity:

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Locus of a Point and an Equation

The concept of locus is fundamental in coordinate geometry as it provides a bridge between geometric shapes and algebraic equations. It allows us to represent the movement of a point under specific restrictions as a mathematical curve.

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Definition of Locus

The locus of a point in a plane is the curve or the path traced out by a point moving under a given geometrical condition (or conditions). In other words, a locus is a set of points whose coordinates satisfy a particular rule.

If a point satisfies the geometric condition, it must lie on the locus. Conversely, if a point lies on the locus, it must satisfy the geometric condition.

Illustrative Examples of Locus in a Plane

1. Distance from the X-axis (Parallel Line)

Let a point $P$ move in a plane such that its distance from the $x$-axis is always equal to a constant $b$ ($b > 0$).

The point $P$ will trace out a straight line $AB$ parallel to the $x$-axis at a distance $b$ units from it. If the point is above the axis, the locus is $y = b$; if below, it is $y = -b$. Therefore, the locus of the moving point $P$ is the straight line $AB$.

2. Equidistant from Two Fixed Points (Perpendicular Bisector)

Let $A$ and $B$ be two fixed points in a plane. Suppose a point $P$ moves in the plane such that its distances from $A$ and $B$ are always equal ($PA = PB$).

All positions of the moving point $P$ will lie on the perpendicular bisector of the segment $AB$. Thus, the locus of a point equidistant from two fixed points is the perpendicular bisector of the segment joining them.

3. Constant Distance from a Fixed Point (Circle)

Let a point $P$ move in a plane such that its distance from a fixed point $C$ (called the centre) is always equal to a constant $r$ ($r > 0$, called the radius).

Obviously, all positions of the moving point $P$ will lie on the circumference of a circle. Therefore, the locus of the point $P$ is the circle with centre $C$ and radius $r$.

Locus in Three-Dimensional Space

While we primarily focus on the coordinate plane, the definition of locus extends to space. When a point moves in three dimensions under a geometric rule, it traces a surface or a 3D path.

Example: The locus of a point in space such that its distance from a fixed point $C$ is always equal to $r$ is a sphere with centre $C$ and radius $r$. If the movement were restricted to a plane, it would be a circle, but in space, it forms a volumetric surface.

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Locus of an Equation

An equation $f(x, y) = 0$ is said to be the equation of a locus if it perfectly describes the relationship between the $x$ and $y$ coordinates of every point on that path. For an equation to be a valid representation of a locus, it must satisfy two fundamental criteria:

(i) Consistency: The coordinates of every single point that lies on the geometric locus must satisfy the algebraic equation.

(ii) Exclusivity: Conversely, if the coordinates of any point $(x_1, y_1)$ satisfy the equation, that point must physically lie on the geometric path.

For example, the equation $x^2 + y^2 = r^2$ represents a circle. Any point whose coordinates satisfy this sum of squares lies on the boundary of the circle, and no point outside or inside the boundary will satisfy it.

Equation of a Locus

The equation of a locus is an algebraic equation.

It describes the path of a moving point that follows a specific geometric rule.

This equation creates a crucial link between the geometric idea of a locus and its algebraic representation in the coordinate plane.

This section details the systematic process for finding this equation.

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Step-by-Step Procedure to Find the Equation of a Locus

Finding the equation of a locus is a process of translating a geometric restriction into an algebraic expression. Here is the detailed professional procedure:

Step I: Assign Coordinates to the Moving Point

Assume the coordinates of any arbitrary point on the locus to be $P(h, k)$. We use $(h, k)$ as temporary variables to represent the "moving" nature of the point and to avoid confusion with fixed points in the problem.

Step II: Express the Geometric Condition Mathematically

Translate the given geometric rule into an algebraic equation using $h$ and $k$. This often involves using the following tools:

• Distance Formula: $d = \sqrt{(h - x_1)^2 + (k - y_1)^2}$

• Section Formula: $h = \frac{mx_2 + nx_1}{m+n}, k = \frac{my_2 + ny_1}{m+n}$

• Area of Triangle: $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Step III: Eliminate Parameters (if any)

If the problem involves auxiliary variables (like an angle $\theta$ or a slope $m$ that changes), eliminate these variables using trigonometric or algebraic identities so that the equation only contains $h, k$ and given constants.

Step IV: Algebraic Simplification

Simplify the resulting expression by removing radicals (squaring both sides), clearing denominators, and grouping like terms.

Step V: Generalization (The Final Locus)

Finally, replace the temporary variables $h$ with $x$ and $k$ with $y$ to obtain the general equation in the Cartesian plane.

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Shifting of Origin

In coordinate geometry, the position of a point is determined relative to two fixed mutually perpendicular lines called the coordinate axes. Sometimes, a mathematical problem can be simplified significantly by changing the reference point of the system. Shifting the origin to a new point while keeping the direction of the axes parallel to the original ones is known as the Translation of Axes or Shifting of Origin.

This transformation affects the coordinates of every point and the equation of every curve in the plane, yet it leaves the geometric properties like distance and area unchanged.

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Transformation of Coordinates

Let $OX$ and $OY$ be the original coordinate axes with origin $O(0, 0)$. Suppose we shift the origin to a new point $O'$ such that its coordinates in the original system are $(h, k)$.

Let $O'X'$ and $O'Y'$ be the new axes drawn through $O'$ such that $O'X'$ is parallel to $OX$ and $O'Y'$ is parallel to $OY$.

Consider any point $P$ in the plane. Let its coordinates referred to the original axes be $(x, y)$ and referred to the new axes be $(X, Y)$.

Proof and Derivation

Given: Original origin $O(0,0)$, new origin $O'(h, k)$. Old coordinates of $P$ are $(x, y)$ and new coordinates are $(X, Y)$.

To Find: Relationship between $(x, y)$ and $(X, Y)$.

Construction: Draw $PM \perp OX$ meeting $O'X'$ at $M'$. Draw $O'N \perp OX$.

From the geometric construction in the diagram:

Since $NM = O'M'$ (horizontal distance in the new system), and $ON = h$ (horizontal shift):

Similarly, for the vertical distance:

Since $MM' = NO' = k$ (vertical shift), and $M'P = Y$ (vertical distance in the new system):

Thus, the transformation formulas are:

Rearranging these to find the new coordinates from the old ones:

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Key Results of Translation

Based on the derivation above, we can conclude the following:

1. Coordinates of a Point: The point whose original coordinates were $(x, y)$ will have new coordinates $(x - h, y - k)$ after the origin is shifted to $(h, k)$.

2. Old Origin in New System: The coordinates of the old origin $O(0, 0)$ referred to the new axes are $(-h, -k)$.

3. Transformation of an Equation: If a curve is represented by the equation $f(x, y) = 0$, its equation in the new coordinate system is obtained by replacing $x$ with $(X + h)$ and $y$ with $(Y + k)$. The new equation becomes $f(X + h, Y + k) = 0$.

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Invariance of Geometric Magnitudes

An important fact in the translation of axes is that it is a rigid transformation. This means it does not affect the physical size or shape of geometric figures. Specifically:

• The distance between two points remains constant.

• The area of a triangle or any polygon remains unchanged.

• The distance of a point from a line remains the same.

Because these magnitudes are invariant, there is usually no need to shift back to the original axes when calculating lengths or areas.

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Slope of a Straight Line

The slope of a straight line is one of the most vital concepts in coordinate geometry as it quantifies the steepness and the direction of a line. In simpler terms, it tells us how much the $y$-coordinate changes for a unit change in the $x$-coordinate.

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Understanding of Inclination

The inclination of a line is the angle that the line makes with the positive direction of the $x$-axis, measured in the anticlockwise (counter-clockwise) direction. It is usually denoted by the Greek letter $\theta$ (theta).

The range of inclination is strictly defined as:

Why do we not include $180^\circ$? This is because a line with an inclination of $180^\circ$ is geometrically identical to a line with an inclination of $0^\circ$ (both are horizontal). To maintain a unique representation for every unique line orientation, we stop just before $180^\circ$.

Visualizing Different Inclinations

• When $\theta = 0^\circ$, the line is perfectly horizontal (parallel to the $x$-axis).
• When $0^\circ < \theta < 90^\circ$, the line is rising as it moves from left to right (acute angle).
• When $\theta = 90^\circ$, the line is perfectly vertical (parallel to the $y$-axis).
• When $90^\circ < \theta < 180^\circ$, the line is falling as it moves from left to right (obtuse angle).

Classification of Lines

Based on their inclination and slope, lines are categorized into three main types:

1. Horizontal Lines: Any line where the $y$-coordinates of all points are the same. These lines have an inclination $\theta = 0^\circ$.

2. Vertical Lines: Any line where the $x$-coordinates of all points are the same. These lines have an inclination $\theta = 90^\circ$.

3. Oblique Lines: Any line that is neither horizontal nor vertical. Their inclination $\theta$ is such that $\theta \neq 0^\circ$ and $\theta \neq 90^\circ$.

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Formal Definition of Slope ($m$)

The Slope or Gradient of a non-vertical line is defined as the tangent of its angle of inclination $\theta$. It is represented by the letter $m$.

The use of the tangent function is logical because $\tan \theta$ represents the ratio of the "Perpendicular" to the "Base" in a right-angled triangle, which corresponds to the "Rise" divided by the "Run".

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Analysis of the Slope Value

The numerical value of the slope $m$ provides immediate information about the line's orientation:

1. Positive Slope ($m > 0$): This occurs when the angle $\theta$ is acute ($0^\circ < \theta < 90^\circ$). In the first quadrant, the tangent function is positive. The line "climbs" upward from left to right.

2. Negative Slope ($m < 0$): This occurs when the angle $\theta$ is obtuse ($90^\circ < \theta < 180^\circ$). In the second quadrant, the tangent function is negative. The line "descends" downward from left to right.

3. Zero Slope ($m = 0$): For a horizontal line, $\theta = 0^\circ$, and since $\tan 0^\circ = 0$, the slope is zero. There is no "rise" regardless of the "run".

4. Undefined Slope: For a vertical line, $\theta = 90^\circ$. Since $\tan 90^\circ$ is undefined (infinite), we say the slope of a vertical line does not exist or is undefined. In this case, there is "rise" but the "run" is zero, and division by zero is not possible.

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Mathematical Summary Table

Type of Line	Inclination ($\theta$)	Slope ($m = \tan \theta$)	Behavior
Horizontal	$0^\circ$	$0$	Parallel to $x$-axis
Oblique (Acute)	$0^\circ < \theta < 90^\circ$	Positive ($> 0$)	Rising left to right
Vertical	$90^\circ$	Undefined	Parallel to $y$-axis
Oblique (Obtuse)	$90^\circ < \theta < 180^\circ$	Negative ($< 0$)	Falling left to right

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Slope of a Straight Line Joining Two Points

While the definition $m = \tan \theta$ is the theoretical foundation of a line's gradient, it is often more practical to calculate the slope using the coordinates of two points situated on the line. This method allows us to find the steepness without knowing the angle of inclination directly.

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Derivation of the Two-Point Formula

Let a non-vertical line $l$ pass through two distinct points $P(x_1, y_1)$ and $Q(x_2, y_2)$. We assume the inclination of the line is $\theta$. To derive the formula, we consider the geometric relationship between the coordinates and the angle.

Case I: When the Angle of Inclination $\theta$ is Acute

In this scenario, the line rises from left to right ($0^\circ < \theta < 90^\circ$).

Construction:

1. Draw perpendiculars $PM$ and $QN$ from points $P$ and $Q$ to the $x$-axis.

2. Draw a perpendicular $PL$ from point $P$ to the line segment $QN$.

From the figure, we can observe the following lengths:

$PL = MN = ON - OM = x_2 - x_1$

$LQ = NQ - NL = NQ - MP = y_2 - y_1$

Since $PL$ is parallel to the $x$-axis, the angle $\angle QPL$ is equal to the angle of inclination $\theta$ (Corresponding Angles).

In the right-angled triangle $QPL$:

Substituting the coordinate values:

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Case II: When the Angle of Inclination $\theta$ is Obtuse

In this scenario, the line falls from left to right ($90^\circ < \theta < 180^\circ$).

In this geometry, the interior angle of the triangle at $P$, denoted as $\angle LPQ$, is supplementary to the inclination $\theta$. Thus:

$\angle LPQ = 180^\circ - \theta$

The lengths of the sides are:

$PL = MN = OM - ON = x_1 - x_2$

$LQ = NQ - NL = y_2 - y_1$

In the right-angled triangle $QPL$:

$\tan(180^\circ - \theta) = \frac{LQ}{PL} = \frac{y_2 - y_1}{x_1 - x_2}$

Using the trigonometric identity $\tan(180^\circ - \theta) = -\tan \theta$:

To eliminate the negative sign, we multiply the denominator by $-1$:

$\tan \theta = \frac{y_2 - y_1}{-(x_1 - x_2)}$

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The Universal Slope Formula

Since the result is identical in both Case I and Case II, we can conclude that the slope $m$ of a non-vertical line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is always:

Key Observations

1. Order of Points: The order in which we subtract the coordinates does not matter, as long as it is consistent for both $x$ and $y$. That is:

$\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_1 - y_2}{x_1 - x_2}$

2. The "Rise over Run" Concept: The numerator $(y_2 - y_1)$ represents the vertical change (Rise), and the denominator $(x_2 - x_1)$ represents the horizontal change (Run).

3. Vertical Lines: For a vertical line, $x_1 = x_2$. This makes the denominator zero. In mathematics, division by zero is undefined, which confirms why the slope of a vertical line is undefined.

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Angle Between Two Lines

In coordinate geometry, when two straight lines intersect in a plane, they form two pairs of vertically opposite angles. If we denote one of these angles as $\theta$, the adjacent angle will be its supplement, i.e., $180^\circ - \theta$ or $\pi - \theta$.

To establish a mathematical relationship, let us consider two non-vertical and non-perpendicular lines $l_1$ and $l_2$ with slopes $m_1$ and $m_2$ respectively. Let $\theta_1$ and $\theta_2$ be the inclinations of these lines with the positive direction of the $x$-axis.

By definition of the slope of a line:

$m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$

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Derivation of the Angle Formula

Let $\alpha$ be the angle between the lines $l_1$ and $l_2$. From the properties of triangles, specifically the Exterior Angle Theorem, we know that the exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Referring to the geometric figure above, we can state:

This can be rearranged to express the angle $\alpha$ as:

$\alpha = \theta_1 - \theta_2$

Taking the tangent on both sides of the equation:

$\tan \alpha = \tan (\theta_1 - \theta_2)$

By applying the trigonometric identity for the tangent of the difference of two angles, $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, we get:

Now, by substituting the values of slopes $m_1$ and $m_2$ into equation provided above, we obtain the fundamental formula:

$\tan \alpha = \frac{m_1 - m_2}{1 + m_1 m_2}$

Since there is another angle between the lines, which is $\pi - \alpha$, its tangent is given by:

$\tan (\pi - \alpha) = -\tan \alpha = - \left( \frac{m_1 - m_2}{1 + m_1 m_2} \right)$

To include both the acute and obtuse angles, we use the $\pm$ sign or take the absolute value for the acute angle.

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Acute and Obtuse Angles

The relationship between the angle $\alpha$ and the slopes $m_1, m_2$ of the lines is derived from the tangent difference formula. Depending on whether we seek the acute or obtuse angle, the sign of the tangent value changes.

Case 1: The Acute Angle

If $\alpha$ is acute, then $0 < \alpha < 90^\circ$. In the first quadrant, the tangent function is always positive. Therefore, we use the absolute value (modulus) to ensure a positive result:

Case 2: The Obtuse Angle

If the angle is obtuse ($90^\circ < \alpha < 180^\circ$), the tangent value lies in the second quadrant and is therefore negative. The formula is expressed as:

There are special cases regarding vertical lines. If both lines are vertical, they are parallel and the angle is $0^\circ$. If only one line is vertical, its slope $m$ is undefined ($\tan 90^\circ$). In such cases, the angle is found using the inclination of the non-vertical line.

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Slopes of Parallel Lines

Two non-vertical lines $l_1$ and $l_2$ are said to be parallel if and only if the angle $\alpha$ between them is $0^\circ$.

Derivation:

We know that $\tan \alpha = 0$ when $\alpha = 0^\circ$. Substituting this into our general formula:

$0 = \pm \frac{m_1 - m_2}{1 + m_1 m_2}$

For the fraction to be zero, the numerator must be zero:

$m_1 - m_2 = 0$

Conclusion: Two non-vertical lines are parallel if and only if their slopes are equal.

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Slopes of Perpendicular Lines

Two non-vertical lines are perpendicular if and only if the angle between them is $90^\circ$.

Proof of the Condition $m_1 m_2 = -1$

Let $l_1$ and $l_2$ be two lines with inclinations $\theta_1$ and $\theta_2$. If the lines are perpendicular, the difference between their inclinations must be $90^\circ$. Thus:

Taking tangent on both sides:

$\tan \theta_2 = \tan (90^\circ + \theta_1)$

Using the trigonometric identity $\tan(90^\circ + \theta) = -\cot \theta$:

$\tan \theta_2 = -\cot \theta_1$

$\tan \theta_2 = -\frac{1}{\tan \theta_1}$

Substituting $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$:

By cross-multiplication, we get the standard condition:

The slope of a line perpendicular to a given line is the negative reciprocal of the original slope.

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Collinearity of Three Points

Three points $A, B$, and $C$ are defined as collinear if they lie on the same straight line. This concept is closely tied to the property of parallel lines.

If segments $AB$ and $BC$ are part of the same line, they must have the same slope. Since they also share a common point ($B$), they cannot be parallel lines at a distance; they must coincide.

Condition: Points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear if:

Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$, the condition becomes:

$\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_2}{x_3 - x_2}$

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Summary Table

Orientation	Mathematical Condition	Angle ($\alpha$)
Parallel Lines	$m_1 = m_2$	$0^\circ$
Perpendicular Lines	$m_1 m_2 = -1$	$90^\circ$
Collinear Points	$\text{Slope}(AB) = \text{Slope}(BC)$	$0^\circ$ or $180^\circ$

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Various Forms of Equations of a Line

A straight line is one of the most fundamental objects in coordinate geometry. To represent a line algebraically, we use an equation in terms of $x$ and $y$. Depending on the information available—such as its slope, the points it passes through, or its intercepts on the axes—we can express the equation of a line in several distinct forms.

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1. Horizontal and Vertical Lines

In the Cartesian coordinate system, a straight line can be oriented in infinitely many directions. However, the most fundamental lines are those that are parallel to the coordinate axes. These lines represent the simplest form of linear equations where one variable remains constant regardless of the value of the other.

Horizontal Lines (Parallel to the x-axis)

A horizontal line is a line that runs from left to right, staying at a constant "height" or distance from the x-axis. Since every point on such a line has the same vertical displacement, their y-coordinates are identical.

Derivation of the Equation

Let $L$ be a straight line parallel to the x-axis. Let the constant distance of this line from the x-axis be $b$ units. If we pick any arbitrary point $P(x, y)$ on this line and draw a perpendicular $MP$ to the x-axis, the length of $MP$ represents the y-coordinate of point $P$.

Since the line is parallel to the x-axis, the distance of every point on the line from the x-axis is always equal to the constant $b$. Thus, the relationship that defines every point on the line is:

Slope of a Horizontal Line

The slope ($m$) of a line passing through any two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

For a horizontal line, since the y-coordinates of all points are the same ($y_1 = y_2 = b$):

Conclusion: The slope of every horizontal line is always zero. This is because there is no "rise" in the line, only "run".

REMARK: If $b > 0$, the line lies above the x-axis. If $b < 0$, the line lies below the x-axis. If $b = 0$, the line coincides with the x-axis; hence, the equation of the x-axis is $y = 0$.

Vertical Lines (Parallel to the y-axis)

A vertical line is a line that runs straight up and down. Every point on a vertical line is located at the same horizontal distance from the y-axis, meaning their x-coordinates remain constant regardless of the value of $y$.

Derivation of the Equation

Let $L$ be a straight line parallel to the y-axis at a constant distance $a$ from it. For any arbitrary point $P(x, y)$ on this line, the distance from the y-axis (known as the abscissa) is always constant and equal to $a$.

Slope of a Vertical Line

For a vertical line, the x-coordinates of any two points are the same ($x_1 = x_2 = a$). Using the slope formula:

Conclusion: Since division by zero is not defined in mathematics, the slope of a vertical line is undefined. Geometrically, this represents a line with infinite steepness.

REMARK: If $a > 0$, the line is to the right of the y-axis. If $a < 0$, the line is to the left. If $a = 0$, the line is the y-axis itself, represented by the equation $x = 0$.

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Comparison of Horizontal and Vertical Lines

The following table summarizes the key properties and differences between these two types of axial lines:

Feature	Horizontal Line	Vertical Line
Orientation	Parallel to x-axis	Parallel to y-axis
Constant Variable	$y$-coordinate ($y = b$)	$x$-coordinate ($x = a$)
Intercept	y-intercept is $b$	x-intercept is $a$
Slope ($m$)	$0$ (Zero)	Undefined ($\infty$)
Inclination ($\theta$)	$0^\circ$ or $180^\circ$	$90^\circ$ or $270^\circ$
Equation of Axis	x-axis is $y = 0$	y-axis is $x = 0$

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2. Point-Slope Form

The Point-Slope Form is one of the most versatile ways to express the equation of a straight line. It is particularly useful when we are given the direction (slope) of the line and one specific location (point) through which it passes. In coordinate geometry, this form directly utilizes the fundamental definition of the slope to establish a linear relationship between $x$ and $y$.

Derivation of the Formula

Let $L$ be a non-vertical straight line with a given slope $m$. Suppose this line passes through a fixed point $A(x_1, y_1)$. To find the equation of this line, we consider an arbitrary point $P(x, y)$ that lies anywhere on the line $L$ such that $P$ is distinct from $A$.

By the fundamental definition of the slope of a straight line, the slope $m$ between any two points $A(x_1, y_1)$ and $P(x, y)$ is the ratio of the change in y-coordinates (rise) to the change in x-coordinates (run).

To eliminate the fraction and express the relationship in a linear algebraic form, we multiply both sides of the equation by the term $(x - x_1)$:

This linear equation represents the geometric condition that must be satisfied by the coordinates $(x, y)$ of every point lying on the line $L$.

Special Case: Line Through the Origin

A special and frequent case in mathematics and physics is when a straight line passes through the origin. In this scenario, the fixed point $A(x_1, y_1)$ becomes $(0, 0)$.

Mathematical Simplification

Substituting $x_1 = 0$ and $y_1 = 0$ into the standard point-slope form equation (i):

REMARK: Since the slope of a vertical line is not defined ($m \to \infty$), this form cannot be used to represent vertical lines. The equation of a vertical line passing through $(x_1, y_1)$ is simply $x = x_1$.

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3. Two-Point Form

The Two-Point Form is used to determine the equation of a straight line when the coordinates of two distinct points, $A(x_1, y_1)$ and $B(x_2, y_2)$, lying on the line are known. According to Euclidean geometry, a unique straight line is determined by any two given points in a plane.

Derivation of the Formula

To derive the equation of a line passing through $A(x_1, y_1)$ and $B(x_2, y_2)$, we treat one point as a fixed reference and use the relationship of the slope that exists between any arbitrary point $P(x, y)$ on the line and these fixed points.

Step 1: Calculation of Slope

The slope ($m$) of the line passing through the two fixed points $A$ and $B$ is constant and is given by the ratio of the difference in y-coordinates to the difference in x-coordinates:

Step 2: Substitution into Point-Slope Form

From the point-slope form, we know that the equation of a line passing through $A(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$. By substituting the expression for $m$ from the above equation into this form, we obtain the Two-Point Form:

Alternatively, the equation can be expressed in a symmetrical form by cross-multiplying the denominators:

Note: If $x_1 = x_2$, the denominator in the slope formula becomes zero, implying the line is vertical. In such a case, the equation of the line is simply $x = x_1$.

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4. Slope-Intercept Form

The Slope-Intercept Form is perhaps the most frequently used form of a linear equation. It expresses the relationship between the slope of the line and its intersection with one of the coordinate axes. This form is particularly useful because it allows us to immediately identify the steepness of the line and its starting point on an axis.

Case I: Equation in terms of y-intercept ($c$)

Consider a non-vertical line $L$ with slope $m$ that intersects the y-axis at a specific distance $c$ from the origin. This distance $c$ is known as the y-intercept. The point where the line crosses the y-axis is $C(0, c)$.

Derivation

To derive the equation, we apply the Point-Slope Form using the fixed point $(0, c)$ and slope $m$:

Substituting $(x_1, y_1) = (0, c)$ into the equation:

In this equation, $m = \tan \theta$ is the slope and $c$ is the y-intercept. The sign of $c$ depends on whether the line intersects the y-axis above or below the origin.

Case II: Equation in terms of x-intercept ($d$)

If a line $L$ with slope $m$ makes an intercept $d$ on the x-axis, it crosses the x-axis at the point $D(d, 0)$. This distance $d$ is known as the x-intercept.

Derivation

Using the Point-Slope Form again with the fixed point $(d, 0)$ and slope $m$:

Substituting $(x_1, y_1) = (d, 0)$ into the equation:

Sign Convention for Intercepts

The following table summarizes the signs of the intercepts based on their position relative to the origin:

Intercept	Position	Sign
y-intercept ($c$)	Above the origin	Positive ($+$)
y-intercept ($c$)	Below the origin	Negative ($-$)
x-intercept ($d$)	Right of the origin	Positive ($+$)
x-intercept ($d$)	Left of the origin	Negative ($-$)

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5. Intercept Form

The Intercept Form of a straight line is used when the segments that the line cuts off from the coordinate axes are known. These segments are called the x-intercept (denoted as $a$) and the y-intercept (denoted as $b$). This form is particularly useful for quickly sketching a line on a Cartesian plane.

Derivation of the Formula

Consider a straight line that intersects the x-axis at a distance $a$ from the origin and the y-axis at a distance $b$ from the origin. Consequently, the line passes through the points $A(a, 0)$ and $B(0, b)$.

Step 1: Applying the Two-Point Form

Since we have two distinct points, we use the Two-Point Form of the equation of a line:

Step 2: Substituting the Coordinates

Substituting the values $(x_1, y_1) = (a, 0)$ and $(x_2, y_2) = (0, b)$ into the above equation:

$y - 0 = \frac{b - 0}{0 - a}(x - a)$

$y = \frac{b}{-a}(x - a)$

Step 3: Algebraic Simplification

To remove the fraction, multiply both sides by $a$:

$ay = -b(x - a)$

$ay = -bx + ab$

Rearranging the variables to one side:

$bx + ay = ab$

Step 4: Final Intercept Form

Dividing the entire equation by the product $ab$ (where $a \neq 0$ and $b \neq 0$):

Sign Convention for Intercepts

The sign of the intercepts $a$ and $b$ depends on the quadrant in which the line intersects the axes. This is summarized in the table below:

Intercept Type	Direction from Origin	Mathematical Sign
x-intercept ($a$)	Right side (Positive x-axis)	Positive ($+$)
x-intercept ($a$)	Left side (Negative x-axis)	Negative ($-$)
y-intercept ($b$)	Above (Positive y-axis)	Positive ($+$)
y-intercept ($b$)	Below (Negative y-axis)	Negative ($-$)

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6. Normal (or Perpendicular) Form

The Normal Form, also known as the Perpendicular Form, is a unique way to represent a straight line. Unlike other forms that rely on slopes or intercepts, this form defines a line based on its spatial relationship with the origin. It is characterized by the length of the perpendicular (normal) dropped from the origin to the line and the angle that this normal makes with the positive x-axis.

Geometric Representation

Let $L$ be a straight line. Let $ON$ be the perpendicular drawn from the origin $O(0, 0)$ to the line $L$. The line is uniquely determined by two parameters:

1. $p$: The length of the perpendicular $ON$. Since it represents a length, $p$ is always positive ($p > 0$).

2. $\alpha$: The angle which the normal $ON$ makes with the positive direction of the x-axis, measured in the anti-clockwise direction ($0 \leq \alpha < 360^\circ$).

Derivation of the Formula

To derive the equation, we determine the intercepts made by the line $L$ on the coordinate axes and then apply the Intercept Form.

Step 1: Finding the x-intercept (OA)

In the right-angled triangle $\Delta OAN$, the side $ON$ is the base relative to the angle $\alpha$, and $OA$ is the hypotenuse.

Step 2: Finding the y-intercept (OB)

In the right-angled triangle $\Delta OBN$, the angle $\angle OBN = \alpha$ (by geometry). Here, $ON$ is the perpendicular relative to $\alpha$, and $OB$ is the hypotenuse.

Step 3: Applying Intercept Form

The intercept form of a line with x-intercept $a$ and y-intercept $b$ is $\frac{x}{a} + \frac{y}{b} = 1$. Substituting $a = OA$ and $b = OB$:

$\frac{x}{\frac{p}{\cos \alpha}} + \frac{y}{\frac{p}{\sin \alpha}} = 1$

Multiplying the terms:

$\frac{x \cos \alpha}{p} + \frac{y \sin \alpha}{p} = 1$

Multiplying the entire equation by $p$, we get the Normal Form Equation:

Summary of Angle $\alpha$ in Different Quadrants

The orientation of the line depends on the quadrant in which the normal $ON$ falls. The signs of $\cos \alpha$ and $\sin \alpha$ vary accordingly:

Quadrant	Angle $\alpha$ Range	Sign of $\cos \alpha$	Sign of $\sin \alpha$
I	$0^\circ < \alpha < 90^\circ$	Positive ($+$)	Positive ($+$)
II	$90^\circ < \alpha < 180^\circ$	Negative ($-$)	Positive ($+$)
III	$180^\circ < \alpha < 270^\circ$	Negative ($-$)	Negative ($-$)
IV	$270^\circ < \alpha < 360^\circ$	Positive ($+$)	Negative ($-$)

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7. Symmetrical (or Distance) Form

The Symmetrical Form, also known as the Parametric Form of a straight line, is a specialized representation used primarily to determine the coordinates of points located at a specific distance from a fixed point. It is extremely useful in problems involving distances along a line, such as finding the intersection of a line with a curve or determining the vertices of geometric figures.

Derivation of the Formula

Let a straight line $L$ pass through a fixed point $A(x_1, y_1)$ and have an inclination $\theta$ with the positive direction of the x-axis. Let $P(x, y)$ be any arbitrary point on this line at a distance $r$ from the fixed point $A$.

To find the relationship between the coordinates of $A$ and $P$, we construct a right-angled triangle by drawing perpendiculars from $A$ and $P$ to the axes. In this triangle, the base represents the horizontal displacement and the height represents the vertical displacement:

From the above equations, we can express the parameter $r$ as:

$\frac{x - x_1}{\cos \theta} = r$        and        $\frac{y - y_1}{\sin \theta} = r$

By equating these two expressions, we obtain the Symmetrical Form of the equation of the line:

Parametric Coordinates of a Point

The coordinates of any point $P(x, y)$ on the line can be expressed in terms of the initial point $(x_1, y_1)$, the inclination $\theta$, and the distance $r$. These are called the parametric coordinates:

Nature of the Parameter $r$

In this form, $r$ is treated as a directed distance. This means:

1. If $r$ is positive, the point $P$ lies on one side of the fixed point $A$.

2. If $r$ is negative, the point $P$ lies on the opposite side of the point $A$.

3. For a fixed absolute distance $|r|$, there are always two distinct points on the line, located at $x_1 \pm r \cos \theta$ and $y_1 \pm r \sin \theta$.

Summary of Key Features

Feature	Description
Input Required	One point $(x_1, y_1)$ and inclination $\theta$.
Primary Use	Finding coordinates of a point at a distance $r$.
Inclination ($\theta$)	The angle is measured anti-clockwise from the positive x-axis.
Parameter ($r$)	Represents the actual distance between points.

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General Equation of a Line

A General Equation of a Line is a first-degree polynomial equation in two variables, usually $x$ and $y$. Mathematically, it is expressed as:

In this equation, $A$, $B$, and $C$ are real constants. A critical condition for this equation to represent a straight line is that $A$ and $B$ cannot both be zero simultaneously. If $A = B = 0$, the equation becomes $C = 0$, which is either a statement of fact (if $C=0$) or a contradiction (if $C \neq 0$), but not a line.

The coordinate geometry often emphasizes the conversion of this general form into specific standard forms to find geometric properties like slope, intercepts, and perpendicular distance from the origin.

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Reduction of General Form to Standard Forms

The general equation $Ax + By + C = 0$ can be transformed into various standard forms to easily identify the geometric properties of the line like its slope and intercepts.

1. Slope-Intercept Form ($y = mx + c$)

To reduce the general equation into the slope-intercept form, the primary objective is to isolate the variable $y$ on the left-hand side of the equation.

Detailed Derivation:

Consider the general equation of the line:

To express this in terms of $y$, we first transpose the terms containing $x$ and the constant $C$ to the right-hand side:

Now, assuming $B \neq 0$ (which implies the line is not vertical), we divide the entire equation by the coefficient $B$:

By comparing above equation with the standard slope-intercept form $y = mx + c$, we can directly extract the formulas for the slope and the y-intercept:

Geometric Visualization

The following diagram illustrates how the coefficients $A, B,$ and $C$ define the slope ($m$) and the vertical displacement ($c$) on the Cartesian plane.

Special Observations

1. Vertical Lines: If $B = 0$, the slope becomes undefined ($m = \infty$). The equation reduces to $Ax + C = 0$, or $x = -C/A$, which is a line parallel to the y-axis.

2. Lines through Origin: If $C = 0$, the y-intercept $c$ becomes $0$. The equation reduces to $y = mx$, meaning the line passes through the origin $(0, 0)$.

3. Horizontal Lines: If $A = 0$, the slope $m$ becomes $0$. The equation reduces to $y = -C/B$, which is a line parallel to the x-axis.

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2. Intercept Form ($\frac{x}{a} + \frac{y}{b} = 1$)

To reduce the general equation of a line $Ax + By + C = 0$ into the intercept form, we perform algebraic manipulations to set the constant term on the right-hand side to unity ($1$).

Detailed Derivation:

We begin with the general first-degree equation in $x$ and $y$:

Step 1: Transpose the constant term $C$ to the right-hand side of the equation.

Step 2: Assuming that the line does not pass through the origin (i.e., $C \neq 0$), we divide the entire equation by $-C$ to make the right-hand side equal to $1$.

Simplifying the right-hand side:

Step 3: To match the standard intercept form, we move the coefficients $A$ and $B$ into the denominators of the respective terms.

By comparing above equation with the standard form $\frac{x}{a} + \frac{y}{b} = 1$, we arrive at the following results:

Geometric Significance and Intercept Table

The intercepts represent the signed distance from the origin. The coordinates of the points where the line meets the axes are summarized below:

Axis	Intercept Value	Point of Intersection
X-Axis	$a = -\frac{C}{A}$	$(a, 0)$
Y-Axis	$b = -\frac{C}{B}$	$(0, b)$

Special Conditions

1. When $C = 0$: If the constant term is zero, the line passes through the origin $(0,0)$. In such a case, the intercept form is not defined because we cannot divide by zero. Both $x$ and $y$ intercepts are zero.

2. Horizontal and Vertical Lines:

• If $A = 0$, the line is $By + C = 0$ (Horizontal). There is no x-intercept ($a = \infty$).
• If $B = 0$, the line is $Ax + C = 0$ (Vertical). There is no y-intercept ($b = \infty$).

3. Area of Triangle: The area of the triangle formed by the line $Ax + By + C = 0$ and the coordinate axes is given by:

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3. Normal Form ($x \cos \alpha + y \sin \alpha = p$)

In this representation, $p$ denotes the length of the perpendicular (normal) from the origin $(0, 0)$ to the line, and $\alpha$ (or sometimes $\omega$) is the angle which the normal makes with the positive direction of the x-axis.

Elaborate Derivation:

To reduce the general equation $Ax + By + C = 0$ to the normal form, we assume that both equations represent the same straight line. Therefore, their corresponding coefficients must be in proportion.

Let the general equation be:

And the normal form be:

Comparing the coefficients of (i) and (ii):

From the above proportionality, we can express $\cos \alpha$ and $\sin \alpha$ as:

We know the fundamental trigonometric identity:

Substituting the values of $\cos \alpha$ and $\sin \alpha$:

$(Ak)^2 + (Bk)^2 = 1$

$k^2(A^2 + B^2) = 1$

Now, substituting the value of $k$ back into the expressions for $\cos \alpha$, $\sin \alpha$, and $p$:

Crucial Rule for Sign Convention: Since $p$ represents the length of a perpendicular distance, it must always be positive. To ensure $p > 0$, the sign of the radical $\sqrt{A^2+B^2}$ is chosen to be opposite to the sign of the constant $C$.

Geometric Visualization

Procedure to Reduce General Equation to Normal Form

To convert any equation $Ax + By + C = 0$ into the normal form, follow these steps:

1. Transpose the constant term $C$ to the right-hand side: $Ax + By = -C$.

2. If the RHS is negative, multiply the whole equation by $-1$ to make the constant term positive.

3. Divide both sides of the equation by $\sqrt{A^2 + B^2}$.

4. Compare the resulting equation with $x \cos \alpha + y \sin \alpha = p$ to find $\alpha$ and $p$.

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Summary of Reductions

Standard Form Name	Equation Format	Key Parameters from $Ax + By + C = 0$
General Form	$Ax + By + C = 0$	$A, B, C$ are Real Constants
Slope-Intercept Form	$y = mx + c$	$\text{Slope } m = -\frac{A}{B}$ , $\text{y-intercept } c = -\frac{C}{B}$
Intercept Form	$\frac{x}{a} + \frac{y}{b} = 1$	$\text{x-intercept } a = -\frac{C}{A}$ , $\text{y-intercept } b = -\frac{C}{B}$
Normal Form	$x \cos \alpha + y \sin \alpha = p$	$\text{Distance } p = \frac{|C|}{\sqrt{A^2+B^2}}$ , $\cos \alpha = \frac{A}{\pm\sqrt{A^2+B^2}}$ , $\sin \alpha = \frac{B}{\pm\sqrt{A^2+B^2}}$

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Special Remarks on Coefficients

The coefficients $A, B,$ and $C$ in the general equation $Ax + By + C = 0$ completely determine the orientation and position of the line in the Cartesian plane. By observing whether these constants are zero or non-zero, we can immediately identify specific types of lines.

Analysis of Line Orientation

1. Horizontal Lines: If the coefficient of $x$ is zero (i.e., $A = 0$), the equation reduces to $By + C = 0$. Solving for $y$, we get:

In this case, the line is horizontal and its slope is zero ($m = 0$).

2. Vertical Lines: If the coefficient of $y$ is zero (i.e., $B = 0$), the equation reduces to $Ax + C = 0$. Solving for $x$, we get:

In this case, the line is vertical and its slope is undefined ($m = \infty$).

3. Lines through the Origin: If the constant term is zero (i.e., $C = 0$), the equation becomes $Ax + By = 0$. Since the point $(0, 0)$ satisfies this equation ($A(0) + B(0) = 0$), the line passes through the origin.

Summary of Coefficient Properties

Condition	Equation Form	Geometric Nature
$A = 0, B \neq 0$	$y = -\frac{C}{B}$	Horizontal Line
$B = 0, A \neq 0$	$x = -\frac{C}{A}$	Vertical Line
$C = 0$	$Ax + By = 0$	Passes through $(0,0)$
$A \neq 0, B \neq 0, C \neq 0$	$Ax + By + C = 0$	Oblique line with intercepts

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Intersection of Two Straight Lines

When two straight lines are present in the same plane, they can either intersect at a unique point, be parallel to each other, or coincide. Finding the Point of Intersection involves solving the two linear equations simultaneously.

Algebraic Solution and Derivation

Consider two lines $L_1$ and $L_2$ given by:

To find the coordinates $(x, y)$ that satisfy both equations, we use the Method of Cross-Multiplication:

From the above relation, the unique point of intersection is derived as:

Conditions for Consistency

The nature of the solution depends on the ratios of the coefficients of the two lines. The following table summarizes the three possible cases:

Ratio Comparison	Geometric Nature	Algebraic interpretation
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$	Intersecting Lines	Consistent; Unique Solution
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$	Parallel Lines	Inconsistent; No Solution
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$	Coincident Lines	Dependent; Infinite Solutions

Visualizing the Intersection

In a graphical representation, the intersection point $(x, y)$ is the only point that lies on both the lines $L_1$ and $L_2$. For parallel lines, since the slopes $m_1 = -a_1/b_1$ and $m_2 = -a_2/b_2$ are equal, the distance between them remains constant, preventing any intersection.

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Families of Lines

In coordinate geometry, every first-degree equation in $x$ and $y$, given by $ax + by + c = 0$ (where at least one of $a$ or $b$ is non-zero), represents a straight line. Although it appears to contain three arbitrary constants ($a$, $b$, and $c$), it actually depends on only two independent constants.

By dividing the equation by one of the non-zero coefficients, say $b$, we get:

This can be rewritten as $x + By + C = 0$. Since two independent constants determine a line, we need two distinct geometric conditions to fix a line uniquely. If only one condition is provided, one constant remains arbitrary. This results in a system of infinite lines satisfying that single condition, known as a one-parameter family of lines. The unknown arbitrary constant is called the parameter.

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Standard One-Parameter Families

1. Family of Lines with a Fixed y-intercept

A one-parameter family of lines is a set of infinite lines where each member satisfies a specific geometric property, while another property remains variable. When we fix the y-intercept, we are looking at all possible lines that cross the y-axis at the same height.

Mathematical Derivation and Interpretation

The slope-intercept form of a straight line is given by:

In this equation, $m$ represents the slope ($\tan \theta$) and $c$ represents the y-intercept. When $c$ is a fixed constant, the only variable remaining is $m$. Since $m$ can take any real value from $-\infty$ to $+\infty$, above equation represents an infinite number of lines.

Every member of this family must pass through the fixed point $(0, c)$ on the y-axis. Geometrically, this looks like a set of lines "pivoting" or rotating around the fixed point $(0, c)$ as the angle of inclination $\theta$ changes.

Case Study: The Family $y = mx + 3$

Consider the family where the y-intercept is fixed at $c = 3$. The equation becomes $y = mx + 3$. Depending on the value of the parameter $m$, we get different members:

1. If $m = 0$, the line is $y = 3$. This is a horizontal line parallel to the x-axis.

2. If $m = 1$, the line is $y = x + 3$. This line is inclined at an angle of $45^\circ$ to the positive x-axis.

3. If $m = -1$, the line is $y = -x + 3$. This line is inclined at $135^\circ$.

4. As $m \to \infty$, the line approaches a vertical orientation, although the vertical line $x = 0$ (y-axis) cannot be strictly represented by a finite value of $m$ in this form.

2. Family of Lines Passing Through a Fixed Point (Pencil of Lines)

The collection of all lines in a plane that pass through a single common point $(x_1, y_1)$ is called a Pencil of Lines. This fixed point is known as the vertex of the pencil.

Mathematical Representation

From the point-slope form, the equation of a line passing through $(x_1, y_1)$ with a slope $m$ is:

Here, $(x_1, y_1)$ are the coordinates of the fixed point. The slope $m$ is the parameter of the family. By substituting different real values for $m$, we obtain different lines, all of which satisfy the condition of passing through $(x_1, y_1)$.

The Limitation of the Point-Slope Form

For competitive exams, it is crucial to note that the equation $y - y_1 = m(x - x_1)$ cannot represent the vertical line passing through $(x_1, y_1)$.

The equation of the vertical line through $(x_1, y_1)$ is:

Since the slope of a vertical line is undefined ($m = \tan 90^\circ$), there is no real value of $m$. Therefore, while the "pencil" technically includes the vertical line, the algebraic parameterization using $m$ misses it.

Example: Pencil of lines passing through $(2, -1)$

The equation of this family is $y - (-1) = m(x - 2)$, which simplifies to:

If we are asked to find a specific member of this family, we need one additional condition (like a second point it passes through, its slope, or its distance from the origin) to calculate the value of the parameter $m$.

Summary Table of Parameters

Fixed Property	Fixed Point in Plane	Variable Parameter	Geometric Effect
y-intercept ($c$)	$(0, c)$	Slope ($m$)	Rotation around a point on the y-axis.
Vertex $(x_1, y_1)$	$(x_1, y_1)$	Slope ($m$)	Pencil of lines "radiating" from the vertex.

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Family of Lines Parallel to a Given Line

A family of parallel lines consists of infinite lines that share the same inclination (slope) but differ in their distance from the origin. In the Cartesian plane, two lines are parallel if and only if their slopes are equal.

General Representation

If we have a fixed line $L$ given by the equation $ax + by + c = 0$, any line parallel to it must have the same coefficients for $x$ and $y$ (or coefficients in the same ratio) to ensure the slope remains constant. The constant term, however, can vary. Thus, the equation of the family is represented as:

By changing the value of $k$, we obtain different members of this family. Each value of $k \in \mathbb{R}$ corresponds to exactly one line parallel to the given line.

Formal Proof and Derivation

To prove that $ax + by + k = 0$ is always parallel to $ax + by + c = 0$, we compare their slopes by converting them into the slope-intercept form ($y = mx + c$).

Step 1: Consider the given line $L_1$:

Rearranging for $y$:

The slope of this line ($m_1$) is $-\frac{a}{b}$.

Step 2: Consider the family of lines $L_2$:

Rearranging for $y$:

The slope of this line ($m_2$) is $-\frac{a}{b}$.

Conclusion: Since $m_1 = m_2 = -\frac{a}{b}$, the lines represented by equation (ii) are parallel to the line represented by equation (i) for all real values of $k$.

Geometric Visualization

Geometrically, as we vary the parameter $k$, the line slides across the plane without rotating. If $b$ is positive, increasing $k$ generally shifts the line downwards, while decreasing $k$ shifts it upwards.

Key Properties and Observations

The following table summarizes the behavior of the family $ax + by + k = 0$:

Condition	Geometric Result
$k = c$	The line is identical (coincident) to the original line.
$k \neq c$	The line is strictly parallel and will never intersect the original line.
$b = 0$	The family represents vertical lines $ax + k = 0$ or $x = -\frac{k}{a}$.
$a = 0$	The family represents horizontal lines $by + k = 0$ or $y = -\frac{k}{b}$.

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Family of Lines Perpendicular to a Given Line

A family of perpendicular lines consists of infinite lines that intersect a given line at a right angle ($90^\circ$). In coordinate geometry, the condition for two lines with slopes $m_1$ and $m_2$ to be perpendicular is that the product of their slopes must be $-1$.

To generate the equation of a line perpendicular to a given line, we swap the coefficients of $x$ and $y$ and change the sign of one of them. The constant term is replaced by an arbitrary parameter $k$.

Mathematical Representation and Derivation

Let the equation of the given fixed line be:

The equation of the family of lines perpendicular to equation (i) is represented as:

By substituting different real values for $k$, we obtain all possible lines in the plane that are perpendicular to the original line.

Formal Proof of Perpendicularity

To prove that any line in family (ii) is perpendicular to line (i), we compare their slopes.

Step 1: Finding the slope of the given line.

Transforming $ax + by + c = 0$ into slope-intercept form ($y = mx + c$):

$by = -ax - c$

Therefore, the slope of the given line ($m_1$) is:

Step 2: Finding the slope of the family of lines.

Transforming $bx - ay + k = 0$ into slope-intercept form:

$-ay = -bx - k$

Therefore, the slope of the family ($m_2$) is:

Step 3: Verifying the product of slopes.

The product of the slopes is calculated as:

Since the product of the slopes is $-1$, the line represented by equation (ii) is always perpendicular to line (i) for any real value of $k$.

Geometric Significance

While a parallel family shifts the line along its normal, a perpendicular family rotates the entire system of lines by $90^\circ$. Changing the parameter $k$ moves the perpendicular line across the original line, allowing it to pass through any specific point in the 2D plane.

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Summary Table: Perpendicular Transformations

In competitive exams, it is useful to quickly identify the perpendicular form. The constant term $k$ is found using an additional condition (e.g., passing through a point).

Given Line Form	Perpendicular Family Form
$ax + by + c = 0$	$bx - ay + k = 0$ OR $ay - bx + k = 0$
$x = \alpha$ (Vertical)	$y = k$ (Horizontal)
$y = \beta$ (Horizontal)	$x = k$ (Vertical)
$y = mx + c$	$y = -\frac{1}{m}x + k$

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Family of Lines Through the Intersection of Two Lines

When two lines $L_1 = 0$ and $L_2 = 0$ intersect at a specific point, there are infinite lines that can pass through that same point. This collection of lines is known as a pencil of lines passing through the point of intersection. Mathematically, instead of solving for the point of intersection first, we can represent this entire family using a single equation with one parameter.

Mathematical Representation

If $L_1 \equiv a_1x + b_1y + c_1 = 0$ and $L_2 \equiv a_2x + b_2y + c_2 = 0$ are two intersecting lines, then the equation of any line passing through their point of intersection is:

Expanding this, we get:

Elaborate Derivation and Proof

To establish this formula, we must prove two things: first, that the equation represents a straight line, and second, that it passes through the point of intersection of the given lines.

Step 1: Proof of Linearity

By rearranging the terms of the combined equation, we collect the coefficients of $x$ and $y$:

Since this is a first-degree equation in $x$ and $y$ (of the form $Ax + By + C = 0$), it represents a straight line in the Cartesian plane.

Step 2: Proof of Intersection

Let $P(\alpha, \beta)$ be the unique point of intersection of $L_1$ and $L_2$. Since $P$ lies on both lines, its coordinates must satisfy both equations:

Now, let us test if $P(\alpha, \beta)$ satisfies the family equation $L_1 + \lambda L_2 = 0$:

Substituting values from equations (iii) and (iv):

The result $0 = 0$ is true for any real value of $\lambda$. This proves that every line represented by equation (i) must pass through the point of intersection $P$.

Important Geometric Observations

1. The Missing Line: The form $L_1 + \lambda L_2 = 0$ can represent every line passing through the intersection except $L_2 = 0$ itself. This is because there is no finite value of $\lambda$ that can eliminate the $L_1$ part of the equation. If we specifically need to include $L_2$ as a possibility, we use the parameter $\mu$ as $L_2 + \mu L_1 = 0$.

2. Concurrence: This concept is the basis for proving the concurrence of three lines. If a third line $L_3 = 0$ can be written as a linear combination of $L_1$ and $L_2$ (i.e., $L_3 = L_1 + \lambda L_2$), then all three lines are concurrent (pass through the same point).

Summary Table: Family Properties

Equation Form	Variable Component	Geometric Invariant
$L_1 + \lambda L_2 = 0$	The slope ($m$) changes with $\lambda$.	Passes through point of intersection of $L_1, L_2$.
$L_1 + \lambda L_2 = 0$	Intercepts change with $\lambda$.	The point $P(x, y)$ remains common to all.

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Comprehensive Summary of One-Parameter Families

The table below outlines the relationship between the fixed geometric constraint and the resulting mathematical form of the family. In each case, the symbol $k$ or $m$ represents the arbitrary real parameter.

Geometric Property	Standard Equation Form	Variable Parameter
Fixed y-intercept ($c$)	$y = mx + c$	Slope ($m$)
Fixed x-intercept ($d$)	$y = m(x - d)$	Slope ($m$)
Parallel to $ax + by + c = 0$	$ax + by + k = 0$	Constant ($k$)
Perpendicular to $ax + by + c = 0$	$bx - ay + k = 0$	Constant ($k$)
Passing through Fixed Point $(x_1, y_1)$	$y - y_1 = m(x - x_1)$	Slope ($m$)
Through Intersection of $L_1$ and $L_2$	$L_1 + kL_2 = 0$	Ratio ($k$)
Fixed Distance from Origin ($p$)	$x \cos \alpha + y \sin \alpha = p$	Angle ($\alpha$)

Key Points for Problem Solving

1. The Two-Constant Rule

The general equation of a line $ax + by + c = 0$ effectively has only two independent constants. This is why two conditions are required to find a unique line. If only one condition is given (e.g., "the line is parallel to $2x+3y=0$"), the resulting equation will always contain one unknown parameter ($k$).

2. Identifying the Parameter

In a family equation, the parameter is the part of the expression that changes to produce different members of the family. For example, in $y = 3x + k$, the slope is fixed at $3$, while the intercept $k$ is the parameter that "slides" the line across the plane.

3. The Intersection Advantage

Using the family $L_1 + kL_2 = 0$ is often much faster than finding the actual point of intersection $(\alpha, \beta)$ and then using the point-slope form. By using the family form, we avoid dealing with potentially messy fractional coordinates of the intersection point.

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Distance of a Point from a Line

The distance of a point from a line is defined as the length of the perpendicular segment drawn from the given point to the line. In geometry, this perpendicular distance represents the shortest distance between the point and the straight line.

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Derivation of the Distance Formula

To derive the formula for the perpendicular distance $d$ from a point $P(x_1, y_1)$ to the line $ax + by + c = 0$, we analyze the geometry of the triangle formed by the point and the line's intercepts on the coordinate axes.

1. Finding Intercepts on Axes

Let the given line be:

This line intersects the x-axis at point $A$ and the y-axis at point $B$. By setting $y = 0$, we find the x-coordinate of $A$, and by setting $x = 0$, we find the y-coordinate of $B$.

2. Calculating the Length of Base AB

Using the distance formula between points $A$ and $B$:

$AB = \sqrt{\left(0 - \left(-\frac{c}{a}\right)\right)^2 + \left(-\frac{c}{b} - 0\right)^2}$

$AB = \sqrt{\frac{c^2}{a^2} + \frac{c^2}{b^2}} = \sqrt{\frac{c^2(b^2 + a^2)}{a^2b^2}}$

3. Area of Triangle PAB

Draw a perpendicular $PM$ from point $P(x_1, y_1)$ to the line $AB$. Let the length of this perpendicular be $d$. We can now calculate the area of $\Delta PAB$ in two different ways to solve for $d$.

Method A: Using Base and Height

$\text{Area } (\Delta PAB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times d$

Substituting the value of $AB$ from equation (ii):

Method B: Using Coordinate Formula

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substituting the coordinates $P(x_1, y_1)$, $A\left(-\frac{c}{a}, 0\right)$, and $B\left(0, -\frac{c}{b}\right)$:

$\text{Area} = \frac{1}{2} \left|x_1\left(0 - \left(-\frac{c}{b}\right)\right) + \left(-\frac{c}{a}\right)\left(-\frac{c}{b} - y_1\right) + 0(y_1 - 0)\right|$

$\text{Area} = \frac{1}{2} \left|x_1\frac{c}{b} + \frac{c^2}{ab} + \frac{cy_1}{a}\right|$

Taking $\frac{c}{ab}$ as a common factor inside the absolute value:

4. Final Equation for Distance

By equating the results from Method A (iii) and Method B (iv):

$\frac{d \cdot |c|}{2ab} \sqrt{a^2 + b^2} = \frac{|c|}{2ab} |ax_1 + by_1 + c|$

Cancelling the common factor $\frac{|c|}{2ab}$ from both sides, we obtain the standard formula for the perpendicular distance:

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Distance Between Two Parallel Lines

Two lines are parallel if they have the same slope but different intercepts. Geometrically, the perpendicular distance between two parallel lines remains constant at every point along the lines. To find this distance, we can pick any arbitrary point on one line and calculate its perpendicular distance to the second line.

Mathematical Derivation

Let the equations of the two parallel lines be:

Since the lines are parallel, the coefficients of $x$ and $y$ are the same (or can be made the same by multiplication). To find a point on $L_1$, let us set $x = 0$:

Thus, a point $P$ on line $L_1$ is $\left(0, -\frac{c_1}{b}\right)$. Now, we apply the perpendicular distance formula from point $P$ to line $L_2$:

Substituting the coordinates of $P\left(0, -\frac{c_1}{b}\right)$:

$d = \frac{|a(0) + b\left(-\frac{c_1}{b}\right) + c_2|}{\sqrt{a^2 + b^2}}$

$d = \frac{|-c_1 + c_2|}{\sqrt{a^2 + b^2}}$

Note: While using this formula, ensure that the coefficients of $x$ and $y$ (i.e., $a$ and $b$) are identical in both equations. For example, if the lines are $2x + 3y + 5 = 0$ and $4x + 6y + 12 = 0$, divide the second equation by $2$ before calculating the distance.

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Special Case: Distance from the Origin

The origin is the point $(0, 0)$ in the Cartesian plane. Finding the distance from the origin is a frequent requirement in physics and advanced coordinate geometry problems (such as finding the radius of an incircle or checking for tangency).

Derivation

Starting from the general perpendicular distance formula from point $(x_1, y_1)$ to the line $ax + by + c = 0$:

Substituting $x_1 = 0$ and $y_1 = 0$:

$d = \frac{|a(0) + b(0) + c|}{\sqrt{a^2 + b^2}}$

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Summary Table of Distance Formulas

Scenario	Points / Lines Involved	Formula
Point to Line	$P(x_1, y_1)$ and $ax + by + c = 0$	$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$
Between Parallel Lines	$ax + by + c_1 = 0$ and $ax + by + c_2 = 0$	$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$
Origin to Line	$O(0, 0)$ and $ax + by + c = 0$	$d = \frac{|c|}{\sqrt{a^2 + b^2}}$

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