Question

Prove that triangle ABC with vertices A(1, 1), B(4, 5), and C(4, 1) is a right triangle.

Answer 1

$\large\underline{\sf{Solution-}}$

Given coordinates of triangle ABC are

Coordinates of A is (1, 1)

Coordinates of B is (4, 5)

Coordinates of C of (4, 1)

Let first find the length of sides of triangle using distance formula.

We know,

Distance Formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane, then distance between A and B is given by

$\begin{gathered}\boxed{\tt{ \: \: \: AB \: = \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2}} \: \: \: \: }} \\ \end{gathered}$

So, using this formula,

Distance between A(1, 1) and B(4, 5) is

$\rm \: AB = \sqrt{ {(4 - 1)}^{2} + {(5 - 1)}^{2} }$

$\rm \: = \: \sqrt{ {3}^{2} + {4}^{2} }$

$\rm \: = \: \sqrt{9 + 16 }$

$\rm \: = \: \sqrt{25 }$

$\rm \: = \: 5 \:$

Now, Distance between B(4, 5) and C(4, 1) is

$\rm \: BC = \sqrt{ {(4 - 4)}^{2} + {(5 - 1)}^{2}}$

$\rm \: = \: \sqrt{ {0}^{2} + {4}^{2} }$

$\rm \: = \: \sqrt{0 + 16}$

$\rm \: = \: \sqrt{16}$

$\rm \: = \: 4$

Now, Distance between A(1, 1) and C(4, 1) is

$\rm \: AC = \sqrt{ {(4 - 1)}^{2} + {(1 - 1)}^{2} }$

$\rm \: = \sqrt{ {3}^{2} + {0}^{2} }$

$\rm \: = \sqrt{ 9 + 0}$

$\rm \: = \sqrt{ 9}$

$\rm \: = \: 3 \:$

Thus, we have

$\rm \: AB \: = \: 5 \: \: \rm\implies \: {AB}^{2} = 25$

$\rm \: BC \: = \: 4 \: \: \rm\implies \: {BC}^{2} = 16$

$\rm \: AC \: = \: 3 \: \: \rm\implies \: {AC}^{2} = 9$

So, from this we concluded that

$\rm \: {AC}^{2} + {BC}^{2} = {AB}^{2}$

So, by converse of Pythagoras Theorem, triangle ABC is right angle triangle right-angled at C.

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Additional Information

1. Section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

2. Mid-point formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the mid-point of AB, then the coordinates of C is given by

$\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered}$

3. Centroid of a triangle

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

$\begin{gathered}\boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}} \\ \end{gathered}$

4. Area of a triangle

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle, then the area of triangle is given by

$\begin{gathered}\boxed{\tt{ Area =\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}} \\ \end{gathered}$

5. Condition for 3 points to be Collinear

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the coordinates in cartesian plane, then points A, B and C are collinear, then

$\begin{gathered}\boxed{\tt{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0}} \\ \end{gathered}$