Question

Prove that the points (3,0),(6,4)& (-1,3) are the vertices of a rightangled isosceles triangle ​

Answer 1

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

Let assume that (3,0),(6,4) and (-1,3) represents the vertices of triangle ABC.

So,

Coordinates of A = (3, 0)

Coordinates of B = (6, 4)

Coordinates of C = (- 1, 3)

Since, we have to show that A, B, C forms the vertices of right - angled isosceles triangle.

So, Using 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

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

So, Consider

$\rm :\longmapsto\:AB$

$\rm \:  =  \: \sqrt{ {(6 - 3)}^{2} + {(4 - 0)}^{2} }$

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

$\rm \:  =  \: \sqrt{9 + 16}$

$\rm \:  =  \: \sqrt{25}$

$\rm \:  =  \: 5$

$\rm\implies \:\boxed{\tt{ \: \: AB \: = \: 5 \: units}}$

Now, Consider

$\rm :\longmapsto\:BC$

$\rm \:  =  \: \sqrt{ {( - 1 - 6)}^{2} + {(3 - 4)}^{2} }$

$\rm \:  =  \: \sqrt{ {( -7)}^{2} + {( - 1)}^{2} }$

$\rm \:  =  \: \sqrt{49 + 1}$

$\rm \:  =  \: \sqrt{50}$

$\rm \:  =  \: \sqrt{5 \times 5 \times 2}$

$\rm \:  =  \: 5 \sqrt{2}$

$\rm\implies \:\boxed{\tt{ \: \: BC \: = \: 5 \sqrt{2} \: units}}$

Now, Consider

$\rm :\longmapsto\:AC$

$\rm \:  =  \: \sqrt{ {( - 1 - 3)}^{2} + {(3 - 0)}^{2} }$

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

$\rm \:  =  \: \sqrt{16 + 9 }$

$\rm \:  =  \: \sqrt{25 }$

$\rm \:  =  \: 5$

$\rm\implies \:\boxed{\tt{ \: \: AC \: = \: 5 \: units}}$

From above, we concluded that

$\bf\implies \:AB = AC$

and

$\rm :\longmapsto\: {AB}^{2} + {AC}^{2}$

$\rm \:  =  \: {5}^{2} + {5}^{2}$

$\rm \:  =  \: 25 + 25$

$\rm \:  =  \: 50$

$\rm \:  =  \: 5 \times 5 \times 2$

$\rm \:  =  \: {(5 \sqrt{2}) }^{2}$

$\rm \:  =  \: {BC}^{2}$

Hence,

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

So, By Converse of Pythagoras Theorem, Triangle ABC is right angled triangle, right angled at A.

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LEARN MORE :-

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

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