Question

Q23. If A(5.2). B(2. - 2) and ((-2.t) are the vertices of a right angled triangle with LB = then find
the value of t​

Answer 1

Given :-

A right-angle triangle ABC right-angled at B having vertices,

• A (5, 2)

• B (2, - 2)

• C (-2, t)

To Find :-

• The value of 't'.

Formula Used :-

Distance Formula :-

Let us consider a line segment joining the points A and B, then distance between A and B is given by

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

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$

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

The vertices of right-angle triangle are

• A (5, 2)

• B (2, - 2)

• C (-2, t)

↝ Distance between A (5, 2) and B (2, - 2)

↝ We know,

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

↝ Here,

$\sf \: \: \: \bull \: \: \: x_1 =5 , \: y_1 = 2, \: x_2=2, \: y_2= - 2$

On substituting the values, we get

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

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

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

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

$\rm :\longmapsto\:AB = 5 - - - (1)$

Now,

↝ Distance between B (2, - 2) and C (-2, t)

↝ We know that,

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

↝ Here,

$\sf \: \: \: \bull \: \: \: x_1 =2 , \: y_1 = - 2, \: x_2= - 2, \: y_2= t$

↝ On substituting the values, we get

$\rm :\longmapsto\:BC = \sqrt{ {( - 2 - 2)}^{2} + {(t + 2)}^{2}}$

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

$\rm :\longmapsto\:BC = \sqrt{ 16 + {(t + 2)}^{2}} - - - (2)$

Now,

↝ Distance between C (- 2, t) and A (5, 2)

↝ We know that,

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

↝ Here,

$\sf \: \: \: \bull \: \: \: x_1 =5 , \: y_1 = 2, \: x_2= - 2, \: y_2= t$

↝ On substituting the values, we get

$\rm :\longmapsto\: CA= \sqrt{ {( - 2 - 5)}^{2} + {(t - 2)}^{2}}$

$\rm :\longmapsto\: CA= \sqrt{ {( - 7)}^{2} + {(t - 2)}^{2}}$

$\rm :\longmapsto\:CA = \sqrt{49 + {(t - 2)}^{2}} - - - - (3)$

Now,

↝ As triangle ABC is right-angle triangle right-angled at B.

So,

↝ By Pythagoras Theorem, we have

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

↝ On substituting the values of CA, AB and BC, we get

$\rm :\longmapsto\:49 + {(t - 2)}^{2} = 25 + 16 + {(t + 2)}^{2}$

$\rm :\longmapsto\:49 + \cancel{{t}^{2}} + \cancel{4} - 4t = 41 + \cancel{{t}^{2}} + \cancel{4} + 4t$

$\rm :\longmapsto\:8 = 8t$

$\bf\implies \:t \: = \: 1$

Additional Information :-

1. Section Formula :-

Section Formula is used to find the coordinates of the line segment joining the points which divides it in the ratio m : n internally,

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n},  \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

2. Midpoint Formula :-

Midpoint Formula is used to find the midpoint of line segment joinjng the two points,

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{x_2 + x_1}{2},  \dfrac{y_2 + y_1}{2}\Bigg) \quad}}}}$