Section Formula in Two Dimensions

Section Formula for Internal Division

The section formula is a fundamental concept in coordinate geometry that allows us to determine the coordinates of a point that lies on a line segment and divides it into two smaller segments in a specific ratio. This division can be either internal (the point lies between the two endpoints) or external (the point lies outside the segment on the line containing it). This section focuses on internal division.

Internal Division

A point P is said to divide the line segment joining two points A and B internally in the ratio $m_1 : m_2$ if point P lies strictly between the points A and B on the line segment AB, such that the ratio of the length of the segment AP to the length of the segment PB is equal to the given ratio $m_1/m_2$. Mathematically, this is expressed as:

Here, $m_1$ and $m_2$ are positive real numbers representing the ratio of the lengths.

Derivation of the Formula

Given:

Two distinct points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the Cartesian plane.

A point $P(x, y)$ that divides the line segment AB internally in the ratio $m_1 : m_2$, meaning $\frac{AP}{PB} = \frac{m_1}{m_2}$.

To Find:

The coordinates of the point P, i.e., $(x, y)$.

Construction:

Draw perpendicular lines from the points A, P, and B to the x-axis. Let these perpendiculars meet the x-axis at points L, M, and N, respectively. Since these are perpendiculars to the x-axis, the coordinates of L, M, and N are $(x_1, 0)$, $(x, 0)$, and $(x_2, 0)$, respectively.

Draw a line through point A parallel to the x-axis. Let this line intersect the perpendicular from P at R and the perpendicular from B at S.

Proof using Similar Triangles:

From the construction, we can see that the line segments AL, PM, and BN are parallel to the y-axis, and the line segments AR and PS are parallel to the x-axis. Thus, $\triangle APR$ and $\triangle PSB$ are right-angled triangles, with right angles at R ($\angle ARP = 90^\circ$) and S ($\angle BPS = 90^\circ$).

Consider lines AR and PS, which are parallel to the x-axis, and the transversal line AB.

Also, we know that $\angle ARP = \angle BPS = 90^\circ$.

By the AA similarity criterion (or since two angles are equal, the third angle must also be equal, i.e., $\angle APR = \angle PBS$), we can conclude that $\triangle APR$ is similar to $\triangle PSB$.

When two triangles are similar, the ratio of their corresponding sides is equal. Therefore:

We are given that $\frac{AP}{PB} = \frac{m_1}{m_2}$. Substituting this into equation (i):

Now, let's express the lengths of the horizontal and vertical segments (AR, PS, PR, SB) in terms of the coordinates:

• The length AR is the horizontal distance between A and R. Since A is $(x_1, y_1)$ and R is $(x, y_1)$ (R has the same x-coordinate as M, and same y-coordinate as A), AR = $x - x_1$. Alternatively, L, M are points on the x-axis, so AR = LM = OM - OL. The x-coordinate of L is $x_1$, M is $x$. So AR = $x - x_1$.
• The length PS is the horizontal distance between P and S. Since P is $(x, y)$ and S is $(x_2, y)$ (S has the same x-coordinate as N, and same y-coordinate as P), PS = $x_2 - x$. Alternatively, M, N are points on the x-axis, so PS = MN = ON - OM. The x-coordinate of M is $x$, N is $x_2$. So PS = $x_2 - x$.
• The length PR is the vertical distance between P and R. Since P is $(x, y)$ and R is $(x, y_1)$, PR = $y - y_1$. Alternatively, PR = PM - RM. PM is the y-coordinate of P ($y$), and RM is the y-coordinate of R ($y_1$, which is same as AL). So PR = $y - y_1$.
• The length SB is the vertical distance between S and B. Since S is $(x_2, y)$ and B is $(x_2, y_2)$, SB = $y_2 - y$. Alternatively, SB = BN - SN. BN is the y-coordinate of B ($y_2$), and SN is the y-coordinate of S ($y$, which is same as PM). So SB = $y_2 - y$.

Note that we assume $x_1 < x < x_2$ and $y_1 < y < y_2$ (or vice versa) for simplicity in taking differences. If $x_1 > x_2$, the difference would be $x_1 - x_2$, but $(x_2 - x_1)^2 = (x_1 - x_2)^2$, so the formula remains consistent regardless of the order of points A and B or their relative positions. The crucial part is that P is *between* A and B internally.

Now, substitute these coordinate expressions for the lengths into the proportion from equation (ii):

and

Let's solve equation (iii) for $x$:

$m_1 (x_2 - x) = m_2 (x - x_1)$

$m_1 x_2 - m_1 x = m_2 x - m_2 x_1$

Collect terms involving $x$ on one side and constant terms on the other:

$m_1 x_2 + m_2 x_1 = m_2 x + m_1 x$

$m_1 x_2 + m_2 x_1 = x (m_2 + m_1)$

Divide by $(m_1 + m_2)$ to find $x$:

Now, let's solve equation (iv) for $y$:

$m_1 (y_2 - y) = m_2 (y - y_1)$

$m_1 y_2 - m_1 y = m_2 y - m_2 y_1$

Collect terms involving $y$ on one side and constant terms on the other:

$m_1 y_2 + m_2 y_1 = m_2 y + m_1 y$

$m_1 y_2 + m_2 y_1 = y (m_2 + m_1)$

Divide by $(m_1 + m_2)$ to find $y$:

Section Formula for Internal Division

Combining the expressions for $x$ and $y$, the coordinates of the point $P(x, y)$ that divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are given by:

$\mathbf{P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} , \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)}$

Special Case: Midpoint Formula

The midpoint of a line segment is the point that divides the segment into two equal parts. This corresponds to an internal division in the ratio $1:1$. In this case, $m_1 = 1$ and $m_2 = 1$.

Substituting $m_1 = 1$ and $m_2 = 1$ into the internal division formula:

$x = \frac{1 \cdot x_2 + 1 \cdot x_1}{1 + 1} = \frac{x_1 + x_2}{2}$

$y = \frac{1 \cdot y_2 + 1 \cdot y_1}{1 + 1} = \frac{y_1 + y_2}{2}$

Thus, the coordinates of the midpoint M of the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ are:

$\mathbf{M(x, y) = \left( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \right)}$

This is the Midpoint Formula.

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Mid-point Formula

The concept of a midpoint is central to geometry. The midpoint of a line segment is defined as the point that is exactly halfway between the two endpoints of the segment. In the context of coordinate geometry, the midpoint formula provides a simple way to find the coordinates of this point when the coordinates of the endpoints are known.

The midpoint is a specific case of the section formula for internal division. It divides the line segment joining two points into two equal parts, meaning the ratio of division is $1 : 1$.

Derivation of the Mid-point Formula

Let A$(x_1, y_1)$ and B$(x_2, y_2)$ be the two endpoints of a line segment.

Let M$(x, y)$ be the midpoint of the line segment AB. By definition, M divides AB internally in the ratio $1 : 1$. This means the ratio $\frac{AM}{MB} = \frac{1}{1}$.

In the general section formula for internal division, the coordinates of a point P dividing the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1 : m_2$ are given by:

$P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} , \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$

For the midpoint M, the ratio is $1 : 1$. So, we substitute $m_1 = 1$ and $m_2 = 1$ into the section formula:

The x-coordinate of the midpoint M is:

Simplifying equation (i):

The y-coordinate of the midpoint M is:

Simplifying equation (iii):

Mid-point Formula

From the derivation above, the coordinates of the midpoint $M(x, y)$ of the line segment joining the points $A(x_1, y_1)$ and $B(x_2, y_2)$ are given by:

$\mathbf{M(x, y) = \left( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \right)}$

In simple terms, the coordinates of the midpoint are found by taking the average of the x-coordinates and the average of the y-coordinates of the two endpoints.

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Section Formula for External Division

While internal division deals with a point lying *within* a line segment, external division deals with a point lying *outside* the segment, but on the line containing it. The section formula can be adapted to find the coordinates of such a point.

External Division

A point P is said to divide the line segment joining points A and B externally in the ratio $m_1 : m_2$ if P lies on the line AB extended (beyond A or beyond B), such that the ratio of the distance from A to P (AP) to the distance from B to P (BP) is equal to the given ratio $m_1/m_2$. Mathematically:

Here, $m_1$ and $m_2$ are positive real numbers. The point P lies on the extension of AB such that A, B, P are collinear. There are two possibilities for the location of P:

• If $m_1 > m_2$, P lies on the extension of AB beyond point B. The order of points on the line is A - B - P.
• If $m_1 < m_2$, P lies on the extension of AB beyond point A. The order of points on the line is P - A - B.

The case where $m_1 = m_2$ is not possible for external division of distinct points A and B, as this would imply $AP=BP$, which can only be true for a point on the perpendicular bisector of AB. A point on the line AB (extended) and equidistant from A and B must be the midpoint, which is an internal division.

Derivation of the Formula

Given:

Two distinct points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the Cartesian plane.

A point $P(x, y)$ that divides the line segment AB externally in the ratio $m_1 : m_2$ ($m_1 \neq m_2$), meaning $\frac{AP}{BP} = \frac{m_1}{m_2}$.

To Find:

The coordinates of the point P, i.e., $(x, y)$.

Proof using Projections:

Let the points be $A(x_1, y_1)$, $B(x_2, y_2)$, and $P(x, y)$. Since P divides AB externally in the ratio $m_1 : m_2$, A, B, P are collinear and $\frac{AP}{BP} = \frac{m_1}{m_2}$.

Project the points A, B, and P onto the x-axis, obtaining points $L(x_1, 0)$, $M(x_2, 0)$, and $N(x, 0)$ respectively. Since A, B, P are collinear, their projections L, M, N onto the x-axis are also collinear and maintain the same ratio of directed distances along the x-axis as the ratio of segment lengths on the line AB.

Thus, P dividing AB externally in ratio $m_1:m_2$ implies that N divides LM externally in the same ratio $m_1:m_2$ along the x-axis. So, $\frac{LN}{MN} = \frac{m_1}{m_2}$.

The directed distance from L to N is $x - x_1$.

The directed distance from M to N is $x - x_2$.

Therefore, we have:

Cross-multiply equation (i):

$m_2 (x - x_1) = m_1 (x - x_2)$

$m_2 x - m_2 x_1 = m_1 x - m_1 x_2$

Collect terms involving $x$ on one side and constant terms on the other:

$m_1 x_2 - m_2 x_1 = m_1 x - m_2 x$

$m_1 x_2 - m_2 x_1 = x (m_1 - m_2)$

Since $m_1 \neq m_2$, $(m_1 - m_2) \neq 0$. We can divide by $(m_1 - m_2)$ to find $x$:

Similarly, project the points A, B, and P onto the y-axis, obtaining points $L'(0, y_1)$, $M'(0, y_2)$, and $N'(0, y)$ respectively. N' divides L'M' externally in the ratio $m_1:m_2$. So, $\frac{L'N'}{M'N'} = \frac{m_1}{m_2}$.

The directed distance from L' to N' is $y - y_1$.

The directed distance from M' to N' is $y - y_2$.

Therefore, we have:

Cross-multiply equation (iii):

$m_2 (y - y_1) = m_1 (y - y_2)$

$m_2 y - m_2 y_1 = m_1 y - m_1 y_2$

Collect terms involving $y$ on one side and constant terms on the other:

$m_1 y_2 - m_2 y_1 = m_1 y - m_2 y$

$m_1 y_2 - m_2 y_1 = y (m_1 - m_2)$

Since $m_1 \neq m_2$, we can divide by $(m_1 - m_2)$ to find $y$:

Alternative Approach (Using Internal Division with Negative Ratio):

External division of AB by P in ratio $m_1:m_2$ is equivalent to internal division of AB by P in ratio $m_1:(-m_2)$. This is because the directed distance from P to B is in the opposite direction to the directed distance from A to P or A to B. Replacing $m_2$ with $-m_2$ in the internal section formula $\left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} , \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$ yields the external division formula:

$x = \frac{m_1 x_2 + (-m_2) x_1}{m_1 + (-m_2)} = \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}$

$y = \frac{m_1 y_2 + (-m_2) y_1}{m_1 + (-m_2)} = \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}$

This provides a quicker way to remember the formula if the internal formula is known - simply replace $m_2$ with $-m_2$.

Section Formula for External Division

The coordinates of the point $P(x, y)$ which divides the line segment joining the points $A(x_1, y_1)$ and $B(x_2, y_2)$ externally in the ratio $m_1 : m_2$ ($m_1 \neq m_2$) are given by:

$\mathbf{P(x, y) = \left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2} , \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2} \right)}$

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Applications of Section Formula

The section formula (including the internal and external division formulas, as well as the midpoint formula as a special case) is a versatile tool in coordinate geometry. It allows us to solve a variety of geometric problems involving points on a line segment and their relative positions.

Some key applications are:

1. Finding Points of Trisection or N-section

The section formula can be used to find the coordinates of points that divide a line segment into a specific number of equal parts. For instance, to find the points of trisection, we need to find the two points that divide the segment into three equal parts.

If P and Q are the points of trisection of the line segment joining points A and B, such that P is closer to A and Q is closer to B, then the segment AB is divided into three equal parts: AP, PQ, and QB. This implies that $AP = PQ = QB$.

• Point P divides the line segment AB internally in the ratio $AP : PB$. Since $PB = PQ + QB = AP + AP = 2AP$, the ratio is $AP : 2AP = 1 : 2$. So, P divides AB internally in the ratio $\mathbf{1 : 2}$.
• Point Q divides the line segment AB internally in the ratio $AQ : QB$. Since $AQ = AP + PQ = QB + QB = 2QB$, the ratio is $2QB : QB = 2 : 1$. So, Q divides AB internally in the ratio $\mathbf{2 : 1}$.

We can use the internal section formula with these specific ratios ($m_1=1, m_2=2$ for P and $m_1=2, m_2=1$ for Q) to find the coordinates of the points of trisection.

More generally, to divide a line segment into $n$ equal parts, there will be $n-1$ points. The first point divides the segment in the ratio $1 : (n-1)$, the second point in the ratio $2 : (n-2)$, and the $k$-th point in the ratio $k : (n-k)$.

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2. Finding the Ratio of Division

If the coordinates of the two endpoints of a line segment A and B, and the coordinates of a point P that lies on the line containing the segment are given, we can use the section formula to find the ratio in which P divides the segment AB.

Assume the point P$(x, y)$ divides the line segment joining A$(x_1, y_1)$ and B$(x_2, y_2)$ in the ratio $k : 1$. We use the section formula for internal division and solve for $k$. The formulas become:

$x = \frac{k x_2 + 1 \cdot x_1}{k + 1} \implies x(k+1) = kx_2 + x_1 \implies xk + x = kx_2 + x_1 \implies x - x_1 = kx_2 - xk \implies x - x_1 = k(x_2 - x)$

So, the ratio based on x-coordinates is $k = \frac{x - x_1}{x_2 - x}$ (provided $x_2 - x \neq 0$).

Similarly, from the y-coordinate formula: $y = \frac{k y_2 + 1 \cdot y_1}{k + 1} \implies y(k+1) = ky_2 + y_1 \implies yk + y = ky_2 + y_1 \implies y - y_1 = ky_2 - yk \implies y - y_1 = k(y_2 - y)$

So, the ratio based on y-coordinates is $k = \frac{y - y_1}{y_2 - y}$ (provided $y_2 - y \neq 0$).

We only need to use one of these formulas (either for $x$ or for $y$) to find the value of $k$. If $k$ is positive, the division is internal, and the ratio is $k : 1$. If $k$ is negative, the division is external, and the ratio is $|k| : 1$. If the denominator is zero (i.e., $x = x_2$ or $y = y_2$), it means the point P coincides with B, which is a trivial case of division (ratio would be undefined or considered as division in a limiting sense). If $x = x_1$ and $y = y_1$, P coincides with A (ratio $0:1$).

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3. Finding the Centroid of a Triangle

The centroid of a triangle is a special point of concurrency where the three medians of the triangle intersect. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. A key property of the centroid is that it divides each median in the ratio $\mathbf{2 : 1}$, with the longer segment being from the vertex to the centroid.

Derivation of the Centroid Formula:

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

Consider the median from vertex A to the side BC. Let D be the midpoint of the side BC. Using the midpoint formula, the coordinates of D are:

The centroid G divides the median AD in the ratio $2 : 1$ (A to D). So, G divides the segment joining A$(x_1, y_1)$ and D$\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)$ internally in the ratio $m_1 : m_2 = 2 : 1$.

Using the internal section formula for G$(x, y)$ dividing AD:

Substitute $m_1=2$, $m_2=1$, $x_A=x_1$, and $x_D=\frac{x_2+x_3}{2}$ into equation (ii):

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

$x = \frac{\frac{2(x_2 + x_3)}{2} + x_1}{3}$

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

Similarly, for the y-coordinate:

Substitute $m_1=2$, $m_2=1$, $y_A=y_1$, and $y_D=\frac{y_2+y_3}{2}$ into equation (iv):

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

$y = \frac{\frac{2(y_2 + y_3)}{2} + y_1}{3}$

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

Centroid Formula:

The coordinates of the centroid $G(x, y)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are given by the average of the coordinates of its vertices:

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

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4. Problems related to Parallelograms

A key property of parallelograms is that their diagonals bisect each other. This means that the point of intersection of the two diagonals is the midpoint of both diagonals. This property is extremely useful when working with parallelogram vertices in coordinate geometry.

If ABCD is a parallelogram, then the midpoint of diagonal AC is the same point as the midpoint of diagonal BD.

Let the vertices be $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, and $D(x_4, y_4)$.

Midpoint of AC = $\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right)$

Midpoint of BD = $\left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)$

Since these midpoints are the same point, their corresponding coordinates must be equal:

These relationships ($x_1 + x_3 = x_2 + x_4$ and $y_1 + y_3 = y_2 + y_4$) can be used to find the coordinates of a missing vertex if the other three are known, or to verify if four given points form a parallelogram.

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