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| Centroid of a Triangle: Definition and Coordinates | Incentre of a Triangle: Definition and Coordinates | Circumcenter and Orthocenter (Coordinates) |
Triangle Centers in Two Dimensions
Centroid of a Triangle: Definition and Coordinates
Definition
The centroid of a triangle is a fundamental point related to its geometric properties. It is defined as the unique point of intersection of the three medians of the triangle. A median of a triangle is a line segment that connects a vertex to the midpoint of the side opposite that vertex.
Every triangle has exactly three medians, one originating from each vertex. These three medians are always concurrent, meaning they intersect at a single point. This point is the centroid.
The centroid is also commonly referred to as the center of mass or center of gravity of a uniform triangular lamina (a triangle with uniform density). If the triangle were cut out of a piece of cardboard, it would perfectly balance if supported at the centroid.
Coordinates of the Centroid
If the coordinates of the three vertices of a triangle are known, the coordinates of its centroid can be easily calculated. Let the vertices of the triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. The coordinates of the centroid, conventionally denoted by $G$, are given by the arithmetic mean (average) of the corresponding coordinates of the vertices:
$\mathbf{G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)}$
Derivation of the Centroid Formula
Given:
A triangle $\triangle ABC$ 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 $G(x, y)$.
Proof using Section Formula:
Let D be the midpoint of the side BC. According to the definition of a median, AD is the median from vertex A to side BC.
Using the midpoint formula, the coordinates of D, the midpoint of the segment joining $B(x_2, y_2)$ and $C(x_3, y_3)$, are:
$D \equiv \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right)$
... (i)
A crucial property of the centroid G is that it divides each median in the ratio 2 : 1, with the vertex part being twice the length of the part connected to the midpoint of the opposite side. So, G divides the median AD in the ratio $AG : GD = 2 : 1$.
Now, we can use the section formula for internal division to find the coordinates of G. The point G$(x, y)$ divides the line 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 $m_1 : m_2 = 2 : 1$.
The section formula for a point dividing a segment joining $(x_a, y_a)$ and $(x_b, y_b)$ in the ratio $m_1 : m_2$ is $\left( \frac{m_1 x_b + m_2 x_a}{m_1 + m_2}, \frac{m_1 y_b + m_2 y_a}{m_1 + m_2} \right)$.
Here, $(x_a, y_a) = A(x_1, y_1)$, $(x_b, y_b) = D\left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right)$, $m_1 = 2$, and $m_2 = 1$.
Calculating the x-coordinate of G:
$x = \frac{(2) \cdot \left( \frac{x_2 + x_3}{2} \right) + (1) \cdot (x_1)}{2 + 1}$
... (ii)
Simplify equation (ii):
$x = \frac{\frac{2(x_2 + x_3)}{2} + x_1}{3}$
$x = \frac{x_2 + x_3 + x_1}{3}$
$x = \frac{x_1 + x_2 + x_3}{3}$
... (iii)
Calculating the y-coordinate of G:
$y = \frac{(2) \cdot \left( \frac{y_2 + y_3}{2} \right) + (1) \cdot (y_1)}{2 + 1}$
... (iv)
Simplify equation (iv):
$y = \frac{\frac{2(y_2 + y_3)}{2} + y_1}{3}$
$y = \frac{y_2 + y_3 + y_1}{3}$
$y = \frac{y_1 + y_2 + y_3}{3}$
... (v)
From (iii) and (v), the coordinates of the centroid G are $\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.
This derivation using median AD yields the formula. We would obtain the same formula if we used the other two medians (BE or CF, where E and F are midpoints of AC and AB respectively) and applied the section formula in the ratio 2:1.
Properties of the Centroid
Key properties associated with the centroid G of a triangle ABC:
- It is the unique point of intersection of the three medians of the triangle.
- It divides each median in the ratio 2:1, measured from the vertex. For median AD (D being the midpoint of BC), $AG : GD = 2 : 1$. Similarly, if BE and CF are the other medians, $\underbrace{BG : GE = 2 : 1}_{\text{Median BE property}}$ and $\underbrace{CG : GF = 2 : 1}_{\text{Median CF property}}$.
- The centroid is always located inside the triangle, regardless of the type of triangle (acute, obtuse, or right-angled).
- It represents the center of mass or center of gravity of a triangular lamina of uniform density.
- The centroid divides the triangle into six smaller triangles of equal area.
Example 1. Find the coordinates of the centroid of a triangle whose vertices are P(0, 6), Q(5, 2), and R(7, 1).
Answer:
Let the vertices of the triangle be $P(x_1, y_1) = (0, 6)$, $Q(x_2, y_2) = (5, 2)$, and $R(x_3, y_3) = (7, 1)$.
Let $G(x, y)$ be the centroid of $\triangle PQR$.
Using the centroid formula:
$x = \frac{x_1 + x_2 + x_3}{3}$
... (i)
Substitute the x-coordinates into equation (i):
$x = \frac{0 + 5 + 7}{3} = \frac{12}{3}$
$x = 4$
... (ii)
$y = \frac{y_1 + y_2 + y_3}{3}$
... (iii)
Substitute the y-coordinates into equation (iii):
$y = \frac{6 + 2 + 1}{3} = \frac{9}{3}$
$y = 3$
... (iv)
From (ii) and (iv), the coordinates of the centroid G are $(4, 3)$.
The coordinates of the centroid of $\triangle PQR$ are $\textbf{(4, 3)}$.
Centroid Summary (For Competitive Exams)
Definition:
The intersection point of the three medians of a triangle.
Coordinates:
For a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, the centroid $G(x, y)$ is:
$G \equiv \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
Key Property:
The centroid divides each median in the ratio 2:1, with the part towards the vertex being larger (e.g., AG : GD = 2:1).
Location:
Always lies inside the triangle.
Significance:
Represents the center of mass/gravity of the triangle.
Incentre of a Triangle: Definition and Coordinates
Definition
The incentre of a triangle is a unique point that holds significant geometric properties. It is defined as the point where the three internal angle bisectors of the triangle intersect. An angle bisector of a triangle is a line segment that divides one of the vertex angles into two equal angles and extends to the opposite side.
The incentre is also the center of the triangle's incircle. The incircle is the largest possible circle that can be inscribed inside the triangle such that it is tangent to all three sides. The radius of the incircle is called the inradius.
A key property of the incentre is that it is equidistant from all three sides of the triangle. This distance is equal to the inradius.
Coordinates of the Incentre
If the coordinates of the three vertices of a triangle are given, we can find the coordinates of its incentre using a specific formula. Let the vertices of the triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. Let the lengths of the sides opposite to these vertices be $a$, $b$, and $c$ respectively.
- The length of the side opposite to vertex A is $a = BC$. Using the distance formula, $a = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}$.
- The length of the side opposite to vertex B is $b = AC$. Using the distance formula, $b = \sqrt{(x_1 - x_3)^2 + (y_1 - y_3)^2}$.
- The length of the side opposite to vertex C is $c = AB$. Using the distance formula, $c = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
The coordinates of the incentre, denoted by $I(x, y)$, are given by a weighted average of the coordinates of the vertices. The weights used are the lengths of the sides opposite to each vertex:
$\mathbf{I(x, y) = \left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right)}$
Here, $a+b+c$ is the perimeter of the triangle.
Derivation of the Incentre Formula
Given:
A triangle $\triangle ABC$ with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.
Let the side lengths opposite to A, B, and C be $a, b, c$ respectively, where $a=BC$, $b=AC$, $c=AB$.
To Find:
The coordinates of the incentre $I(x, y)$.
Proof using Angle Bisector Theorem and Section Formula:
The derivation of the incentre formula relies on the property that the incentre is the intersection of angle bisectors and the application of the Angle Bisector Theorem along with the Section Formula.
Let AD be the internal angle bisector of $\angle A$, where D is a point on the side BC. By the Angle Bisector Theorem, the angle bisector of a vertex angle divides the opposite side in the ratio of the lengths of the other two sides of the triangle.
Applying the Angle Bisector Theorem to $\triangle ABC$ and the angle bisector AD:
$\frac{BD}{DC} = \frac{AB}{AC}$
(Angle Bisector Theorem)
$\frac{BD}{DC} = \frac{c}{b}$
... (i)
Equation (i) tells us that the point D divides the side BC internally in the ratio $c : b$.
Now, we find the coordinates of point D using the section formula. D divides the segment joining $B(x_2, y_2)$ and $C(x_3, y_3)$ internally in the ratio $m_1 : m_2 = c : b$.
Coordinates of $D$: $\left( \frac{c \cdot x_3 + b \cdot x_2}{c + b}, \frac{c \cdot y_3 + b \cdot y_2}{c + b} \right)$.
$D \equiv \left( \frac{b x_2 + c x_3}{b + c}, \frac{b y_2 + c y_3}{b + c} \right)$
... (ii)
The incentre I lies on the angle bisector AD. Now consider the triangle $\triangle ABD$. BI is the angle bisector of $\angle B$. Applying the Angle Bisector Theorem to $\triangle ABD$ and the angle bisector BI, I divides the segment AD in the ratio $AB : BD$.
$\frac{AI}{ID} = \frac{AB}{BD}$
(Angle Bisector Theorem applied to $\triangle ABD$)
We know $AB = c$. We need to find the length of BD. From equation (i), $\frac{BD}{DC} = \frac{c}{b}$, so $BD = \frac{c}{b} DC$. Also, $BD + DC = BC = a$. Substituting $DC = \frac{b}{c} BD$ into the second equation: $BD + \frac{b}{c} BD = a \implies BD \left(1 + \frac{b}{c}\right) = a \implies BD \left(\frac{c+b}{c}\right) = a \implies BD = \frac{ac}{b+c}$.
$BD = \frac{ac}{b + c}$
... (iii)
Now we can find the ratio $\frac{AI}{ID}$:
$\frac{AI}{ID} = \frac{AB}{BD} = \frac{c}{\frac{ac}{b+c}} = c \times \frac{b+c}{ac} = \frac{c(b+c)}{ac} = \frac{b+c}{a}$
... (iv)
Equation (iv) shows that the incentre I divides the median AD internally in the ratio $(b+c) : a$.
Finally, we use the section formula to find the coordinates of I, dividing the segment joining $A(x_1, y_1)$ and $D\left( \frac{b x_2 + c x_3}{b + c}, \frac{b y_2 + c y_3}{b + c} \right)$ internally in the ratio $m_1 : m_2 = (b+c) : a$.
Calculating the x-coordinate of I:
$x = \frac{(b+c) \cdot x_D + a \cdot x_A}{(b+c) + a}$
... (v)
Substitute $x_A=x_1$ and $x_D=\frac{b x_2 + c x_3}{b + c}$ into equation (v):
$x = \frac{(b+c) \cdot \left( \frac{b x_2 + c x_3}{b + c} \right) + a \cdot x_1}{a + b + c}$
$x = \frac{(b x_2 + c x_3) + a x_1}{a + b + c}$
$x = \frac{a x_1 + b x_2 + c x_3}{a + b + c}$
... (vi)
Calculating the y-coordinate of I:
$y = \frac{(b+c) \cdot y_D + a \cdot y_A}{(b+c) + a}$
... (vii)
Substitute $y_A=y_1$ and $y_D=\frac{b y_2 + c y_3}{b + c}$ into equation (vii):
$y = \frac{(b+c) \cdot \left( \frac{b y_2 + c y_3}{b + c} \right) + a \cdot y_1}{a + b + c}$
$y = \frac{(b y_2 + c y_3) + a y_1}{a + b + c}$
$y = \frac{a y_1 + b y_2 + c y_3}{a + b + c}$
... (viii)
From (vi) and (viii), the coordinates of the incentre I are $\left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right)$.
This derivation is based on the angle bisector from vertex A, but the result is symmetric for all vertices, confirming that the three angle bisectors are concurrent at this point.
Properties of the Incentre
Key properties of the incentre I of a triangle ABC:
- It is the unique point of intersection of the three internal angle bisectors.
- It is the center of the incircle, the circle inscribed within the triangle tangent to all three sides.
- It is equidistant from all three sides of the triangle. This distance is the inradius ($r$).
- The incentre always lies inside the triangle.
- For any point P in the plane, the sum of the weighted distances from P to the vertices ($a \cdot PA + b \cdot PB + c \cdot PC$) is minimized when P is the incentre.
Example 1. Find the coordinates of the incentre of the triangle with vertices A(0, 6), B(8, 6), and C(0, 0).
Answer:
Let the vertices be $A(x_1, y_1) = (0, 6)$, $B(x_2, y_2) = (8, 6)$, and $C(x_3, y_3) = (0, 0)$.
First, we need to calculate the lengths of the sides opposite to each vertex using the distance formula:
- Side $a$ (opposite vertex A, segment BC):
$a = BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} = \sqrt{(0 - 8)^2 + (0 - 6)^2} = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100}$
$a = 10$
... (i)
- Side $b$ (opposite vertex B, segment AC):
$b = AC = \sqrt{(x_1 - x_3)^2 + (y_1 - y_3)^2} = \sqrt{(0 - 0)^2 + (6 - 0)^2} = \sqrt{0^2 + 6^2} = \sqrt{36}$
$b = 6$
... (ii)
- Side $c$ (opposite vertex C, segment AB):
$c = AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(8 - 0)^2 + (6 - 6)^2} = \sqrt{8^2 + 0^2} = \sqrt{64}$
$c = 8$
... (iii)
Notice that $a^2 = 10^2 = 100$, $b^2 = 6^2 = 36$, $c^2 = 8^2 = 64$. Since $b^2 + c^2 = 36 + 64 = 100 = a^2$, this is a right-angled triangle with the right angle at C(0,0). This is not strictly necessary for the incentre calculation but is a useful observation.
The sum of the side lengths is $a+b+c = 10 + 6 + 8 = 24$.
Now, use the incentre formula $I(x, y) = \left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right)$.
Calculating the x-coordinate of I:
$x = \frac{ax_1 + bx_2 + cx_3}{a + b + c}$
... (iv)
Substitute the values from (i), (ii), (iii) and vertex coordinates into equation (iv):
$x = \frac{(10)(0) + (6)(8) + (8)(0)}{24} = \frac{0 + 48 + 0}{24} = \frac{48}{24}$
$x = 2$
... (v)
Calculating the y-coordinate of I:
$y = \frac{ay_1 + by_2 + cy_3}{a + b + c}$
... (vi)
Substitute the values from (i), (ii), (iii) and vertex coordinates into equation (vi):
$y = \frac{(10)(6) + (6)(6) + (8)(0)}{24} = \frac{60 + 36 + 0}{24} = \frac{96}{24}$
$y = 4$
... (vii)
From (v) and (vii), the coordinates of the incentre I are $(2, 4)$.
The coordinates of the incentre of the triangle with vertices A(0, 6), B(8, 6), and C(0, 0) are $\textbf{(2, 4)}$.
Incentre Summary (For Competitive Exams)
Definition:
The intersection point of the three internal angle bisectors. It is the center of the incircle.
Coordinates:
For a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$ and opposite side lengths $a=BC$, $b=AC$, $c=AB$, the incentre $I(x, y)$ is:
$I \equiv \left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right)$
where $a = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}$, $b = \sqrt{(x_1 - x_3)^2 + (y_1 - y_3)^2}$, $c = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Key Properties:
- Equidistant from all three sides of the triangle (distance equals the inradius).
- Always lies inside the triangle.
Circumcenter and Orthocenter (Coordinates)
Apart from the centroid and incentre, there are other important points associated with a triangle, such as the circumcenter and the orthocenter. Their coordinates are typically found by utilizing their geometric properties and solving systems of equations.
Circumcenter
Definition
The circumcenter of a triangle is the point in the plane that is equidistant from all three vertices of the triangle. Geometrically, it is defined as the unique point of intersection of the three perpendicular bisectors of the sides of the triangle.
The circumcenter is also the center of the triangle's circumcircle, which is the unique circle that passes through all three vertices of the triangle. The distance from the circumcenter to each vertex is equal to the radius of the circumcircle, known as the circumradius.
Since the circumcenter lies on the perpendicular bisector of each side, it is equidistant from the endpoints of that side. Because it lies on the perpendicular bisectors of all three sides, it is equidistant from all three vertices.
Finding the Coordinates of the Circumcenter
There is no simple direct formula for the coordinates of the circumcenter in terms of the vertex coordinates alone, unlike the centroid or incentre. The coordinates of the circumcenter, usually denoted by $O(x, y)$, are found using geometric properties and algebraic methods.
Method 1: Intersection of Perpendicular Bisectors
This method directly uses the definition of the circumcenter as the intersection of perpendicular bisectors.
-
Find the midpoints of two sides: Calculate the coordinates of the midpoints of any two sides of the triangle (e.g., side AB and side BC) using the midpoint formula. Let $M_{AB}$ be the midpoint of AB and $M_{BC}$ be the midpoint of BC.
-
Find the slopes of those sides: Calculate the slopes of the two chosen sides (AB and BC) using the slope formula. Handle cases where sides are horizontal (slope 0) or vertical (slope undefined) carefully.
-
Find the slopes of the perpendicular bisectors: The slope of a line perpendicular to a line with slope $m$ is $-1/m$ (negative reciprocal), provided $m \neq 0$. If a side is horizontal (slope 0), its perpendicular bisector is a vertical line. If a side is vertical (undefined slope), its perpendicular bisector is a horizontal line.
-
Find the equations of two perpendicular bisectors: Use the point-slope form of the equation of a line, $y - y_0 = m(x - x_0)$, where $(x_0, y_0)$ is the midpoint of the side and $m$ is the slope of the perpendicular bisector. For vertical lines ($x=c$) or horizontal lines ($y=c$), the equation is simpler.
-
Solve the system of equations: The circumcenter is the point $(x, y)$ that satisfies the equations of both perpendicular bisectors simultaneously. Solve the system of two linear equations in $x$ and $y$ to find the coordinates of the circumcenter.
Method 2: Using the Equidistance Property
This method uses the property that the circumcenter is equidistant from all three vertices.
-
Assume coordinates for the circumcenter: Let the coordinates of the circumcenter be $O(x, y)$.
-
Set up distance equations: The distance from O to A must equal the distance from O to B ($OA = OB$), and the distance from O to B must equal the distance from O to C ($OB = OC$). Since distance is always non-negative, we can work with the squared distances to avoid square roots: $OA^2 = OB^2$ and $OB^2 = OC^2$.
-
Expand using the distance formula: Let $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$.
$(x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2$
($OA^2 = OB^2$)
$(x - x_2)^2 + (y - y_2)^2 = (x - x_3)^2 + (y - y_3)^2$
($OB^2 = OC^2$)
-
Simplify the equations: Expand the squared terms in both equations. Notice that the $x^2$ and $y^2$ terms will cancel out on both sides, leaving two linear equations in $x$ and $y$. For example, expanding the first equation:
$(x^2 - 2xx_1 + x_1^2) + (y^2 - 2yy_1 + y_1^2) = (x^2 - 2xx_2 + x_2^2) + (y^2 - 2yy_2 + y_2^2)$
$- 2xx_1 + x_1^2 - 2yy_1 + y_1^2 = - 2xx_2 + x_2^2 - 2yy_2 + y_2^2$
Rearrange to group $x$ and $y$ terms:
$2x(x_2 - x_1) + 2y(y_2 - y_1) = (x_2^2 + y_2^2) - (x_1^2 + y_1^2)$
Repeat similar steps for the second equation ($OB^2 = OC^2$) to get another linear equation in $x$ and $y$.
-
Solve the system of equations: Solve the two linear equations simultaneously to find the values of $x$ and $y$, which are the coordinates of the circumcenter.
Properties of the Circumcenter
- It is the point of concurrency of the perpendicular bisectors of the sides.
- It is the center of the unique circle passing through the three vertices (circumcircle).
- It is equidistant from all three vertices ($OA = OB = OC = \text{Circumradius}$).
- Its location depends on the type of triangle:
- For an acute triangle, the circumcenter lies inside the triangle.
- For an obtuse triangle, the circumcenter lies outside the triangle.
- For a right-angled triangle, the circumcenter lies exactly at the midpoint of the hypotenuse.
Orthocenter
Definition
The orthocenter of a triangle is the point of intersection of the three altitudes of the triangle. An altitude of a triangle is a perpendicular line segment drawn from a vertex to the opposite side (or the line containing the opposite side). The foot of the altitude lies on the opposite side or its extension.
Finding the Coordinates of the Orthocenter
There is no simple general formula for the orthocenter's coordinates. It is typically found by determining the equations of two altitudes and finding their intersection point.
-
Find the slopes of two sides: Calculate the slopes of any two sides of the triangle (e.g., side BC and side AC) using the slope formula. Handle horizontal/vertical sides.
-
Find the slopes of the corresponding altitudes: An altitude is perpendicular to the side it is drawn to. If a side has slope $m$, the altitude perpendicular to it has slope $-1/m$ (if $m \neq 0$). If a side is horizontal (slope 0), the altitude to it is vertical (undefined slope). If a side is vertical (undefined slope), the altitude to it is horizontal (slope 0).
-
Find the equations of two altitudes: Use the point-slope form $y - y_0 = m(x - x_0)$. For the altitude from vertex A to side BC, use vertex A$(x_1, y_1)$ and the slope perpendicular to BC. For the altitude from vertex B to side AC, use vertex B$(x_2, y_2)$ and the slope perpendicular to AC. (Or choose any other pair of altitudes).
-
Solve the system of equations: The orthocenter is the point $(x, y)$ that satisfies the equations of both altitudes simultaneously. Solve the system of two linear equations in $x$ and $y$ to find the coordinates of the orthocenter.
Properties of the Orthocenter
- It is the point of concurrency of the altitudes.
- Its location depends on the type of triangle:
- For an acute triangle, the orthocenter lies inside the triangle.
- For an obtuse triangle, the orthocenter lies outside the triangle.
- For a right-angled triangle, the orthocenter lies exactly at the vertex where the right angle is formed.
- The vertices A, B, C, and the orthocenter H form an orthocentric system: any one of these four points is the orthocenter of the triangle formed by the other three.
Example 1. Find the circumcenter and orthocenter of the triangle with vertices A(0, 6), B(8, 6), and C(0, 0).
Answer:
Let the vertices of the triangle be $A(x_1, y_1) = (0, 6)$, $B(x_2, y_2) = (8, 6)$, and $C(x_3, y_3) = (0, 0)$.
First, let's analyze the type of triangle by calculating the slopes of its sides:
- Slope of side AB: $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 6}{8 - 0} = \frac{0}{8} = 0$. Side AB is a horizontal line along $y=6$.
- Slope of side AC: $m_{AC} = \frac{y_3 - y_1}{x_3 - x_1} = \frac{0 - 6}{0 - 0}$. The denominator is 0, so the slope is undefined. Side AC is a vertical line along $x=0$ (the y-axis).
- Slope of side BC: $m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{0 - 6}{0 - 8} = \frac{-6}{-8} = \frac{3}{4}$.
Since side AB is horizontal and side AC is vertical, they are perpendicular to each other. Thus, the angle at vertex A is $90^\circ$. $\triangle ABC$ is a right-angled triangle with the right angle at vertex A(0, 6).
Finding the Circumcenter (O)
For a right-angled triangle, the circumcenter is always located at the midpoint of the hypotenuse. The hypotenuse is the side opposite the right angle, which is side BC.
Using the midpoint formula for side BC with $B(8, 6)$ and $C(0, 0)$:
x-coordinate of O $= \frac{x_2 + x_3}{2} = \frac{8 + 0}{2} = \frac{8}{2}$
... (i)
x-coordinate of O $= 4$
... (ii)
y-coordinate of O $= \frac{y_2 + y_3}{2} = \frac{6 + 0}{2} = \frac{6}{2}$
... (iii)
y-coordinate of O $= 3$
... (iv)
From (ii) and (iv), the coordinates of the circumcenter are (4, 3).
The circumcenter of $\triangle ABC$ is $\mathbf{O(4, 3)}$.
Finding the Orthocenter (H)
For a right-angled triangle, the orthocenter is always located at the vertex where the right angle is formed. In $\triangle ABC$, the right angle is at vertex A.
The orthocenter of $\triangle ABC$ is $\mathbf{H(0, 6)}$.
Verification using General Methods (Optional)
We can verify these results using the general methods described earlier.
Circumcenter using Perpendicular Bisectors:
Midpoint of AC: $(\frac{0+0}{2}, \frac{6+0}{2}) = (0, 3)$. Side AC is vertical ($x=0$). Its perpendicular bisector is a horizontal line passing through (0, 3), which is the line $y = 3$.
Midpoint of AB: $(\frac{0+8}{2}, \frac{6+6}{2}) = (4, 6)$. Side AB is horizontal ($y=6$). Its perpendicular bisector is a vertical line passing through (4, 6), which is the line $x = 4$.
The intersection of $y=3$ and $x=4$ is the point (4, 3), which matches the result obtained using the right-triangle property.
Orthocenter using Altitudes:
Altitude from A to BC: Slope of BC is $3/4$. Slope of the altitude is $-1/(3/4) = -4/3$. Equation of altitude from A(0, 6) with slope $-4/3$: $y - 6 = -\frac{4}{3}(x - 0) \implies y - 6 = -\frac{4}{3}x \implies 3y - 18 = -4x \implies 4x + 3y = 18$.
Altitude from B to AC: Side AC is on the vertical line $x=0$. The altitude from B to AC must be a horizontal line passing through B(8, 6). The equation is $y = 6$.
To find the orthocenter, solve the system: $4x + 3y = 18$ $y = 6$
Substitute $y=6$ into the first equation: $4x + 3(6) = 18 \implies 4x + 18 = 18 \implies 4x = 0 \implies x = 0$.
The intersection is (0, 6), which matches the coordinates of vertex A and the result obtained using the right-triangle property.
Circumcenter & Orthocenter Summary (For Competitive Exams)
Circumcenter (O):
- Definition: Intersection point of the perpendicular bisectors of the triangle's sides. It is the center of the circumcircle (the circle passing through all vertices).
- Key Property: Equidistant from all three vertices ($OA = OB = OC = \text{Circumradius}$).
- Finding Coordinates: Solve the system of equations for two perpendicular bisectors OR solve $OA^2 = OB^2$ and $OB^2 = OC^2$.
- Location: Lies inside an acute triangle, outside an obtuse triangle, and at the midpoint of the hypotenuse of a right triangle.
Orthocenter (H):
- Definition: Intersection point of the three altitudes of the triangle. (An altitude is perpendicular from a vertex to the opposite side or its extension).
- Finding Coordinates: Solve the system of equations for two altitudes.
- Location: Lies inside an acute triangle, outside an obtuse triangle, and at the vertex containing the right angle of a right triangle.
Euler Line:
For any triangle that is not equilateral, the Orthocenter (H), Centroid (G), and Circumcenter (O) are collinear and lie on a straight line called the Euler line. The centroid G divides the segment OH in the ratio $1 : 2$, i.e., $OG : GH = 1 : 2$.