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​

1 thought on “Q23. If A(5.2). B(2. – 2) and ((-2.t) are the vertices of a right angled triangle with LB = then find<br />the value of t​”

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

   [tex]\sf\:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }[/tex]

   [tex] \sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)[/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   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,

   [tex]\sf\:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }[/tex]

   ↝ Here,

   [tex] \sf \: \: \: \bull \: \: \: x_1 =5 , \: y_1 = 2, \: x_2=2, \: y_2= – 2[/tex]

   On substituting the values, we get

   [tex]\rm :\longmapsto\:AB = \sqrt{ {(2 – 5)}^{2} + {( – 2 – 2)}^{2} } [/tex]

   [tex]\rm :\longmapsto\:AB = \sqrt{ {( – 3)}^{2} + {( – 4)}^{2} } [/tex]

   [tex]\rm :\longmapsto\:AB = \sqrt{9 + 16} [/tex]

   [tex]\rm :\longmapsto\:AB = \sqrt{25} [/tex]

   [tex]\rm :\longmapsto\:AB = 5 – – – (1)[/tex]

   Now,

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

   ↝ We know that,

   [tex]\sf\:BC = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }[/tex]

   ↝ Here,

   [tex] \sf \: \: \: \bull \: \: \: x_1 =2 , \: y_1 = – 2, \: x_2= – 2, \: y_2= t[/tex]

   ↝ On substituting the values, we get

   [tex]\rm :\longmapsto\:BC = \sqrt{ {( – 2 – 2)}^{2} + {(t + 2)}^{2}} [/tex]

   [tex]\rm :\longmapsto\:BC = \sqrt{ {( -4)}^{2} + {(t + 2)}^{2}} [/tex]

   [tex]\rm :\longmapsto\:BC = \sqrt{ 16 + {(t + 2)}^{2}} – – – (2)[/tex]

   Now,

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

   ↝ We know that,

   [tex]\sf\:CA = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }[/tex]

   ↝ Here,

   [tex] \sf \: \: \: \bull \: \: \: x_1 =5 , \: y_1 = 2, \: x_2= – 2, \: y_2= t[/tex]

   ↝ On substituting the values, we get

   [tex]\rm :\longmapsto\: CA= \sqrt{ {( – 2 – 5)}^{2} + {(t – 2)}^{2}} [/tex]

   [tex]\rm :\longmapsto\: CA= \sqrt{ {( – 7)}^{2} + {(t – 2)}^{2}} [/tex]

   [tex]\rm :\longmapsto\:CA = \sqrt{49 + {(t – 2)}^{2}} – – – – (3)[/tex]

   Now,

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

   So,

   ↝ By Pythagoras Theorem, we have

   [tex]\rm :\longmapsto\: {CA}^{2} = {AB}^{2} + {BC}^{2} [/tex]

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

   [tex]\rm :\longmapsto\:49 + {(t – 2)}^{2} = 25 + 16 + {(t + 2)}^{2} [/tex]

   [tex]\rm :\longmapsto\:49 + \cancel{{t}^{2}} + \cancel{4} – 4t = 41 + \cancel{{t}^{2}} + \cancel{4} + 4t [/tex]

   [tex]\rm :\longmapsto\:8 = 8t[/tex]

   [tex]\bf\implies \:t \: = \: 1[/tex]

   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,

   [tex]{\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}}}}[/tex]

   2. Midpoint Formula :-

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

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

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