Question

If a (5,2) b(2,-2) c(-2,t) are the vertices of a right angled triangle the find the value of t​

Answer 1

Answer:

I can use Distance formula

Step-by-step explanation:

ab=bc=ac

take any 2 sides

ab and bc

(2-5)^2 +( -2-2)^2 =( -2-2)^2 +(t+2)^2

9 + 16 = 16 + t^2 +4t + 4

25 -20 = t^2 +4t

we get the quadratic equation

t^2 + 4t -5

t^2 + 5t -t -5 (splitting the middle term)

t (t+5) -1 (t+5)

(t-1) (t+5)

t = 1

Answer 2

A(5,2), B(2,-2) and C(-2,t) are the vertices of a right angled triangle with $\bf \angle B = 90^\circ$.

⠀⠀⠀

We have to find, value of t.

⠀⠀⠀

$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{C}} \put(8,3){\large\sf{(-2,t)}}\put(7.7,1){\large\sf{B}}\put(7.8, 0.7){\large\sf{(2,-2)}}\put(10.6,1){\large\sf{A}}\put(10.6,0.7){\large\sf{(5,2)}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

⠀⠀⠀

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Using Pythagoras theorem in right triangle ABC,

$\qquad\qquad\qquad\star\;\bf \purple{AC^2 = AB^2 + BC^2}$

$:\implies\sf (5 + 2)^2 + (2 - t)^2 = {(5 - 2)^2 + (2 + 2)^2} + {(2 + 2)^2 + (- 2 - t)^2}$

$:\implies\sf 7^2 + (2 - t)^2 = {3^2 + 4^2} + {4^2 + (-2 - 1)^2}$

$:\implies\sf 49 + (4 - 4t + t^2) = (9 + 16) + (16 + 4 + 4t + t^2)$

$:\implies\sf t^2 - 4t + 53 = t^2 + 4t \times 45$

$:\implies\sf - 8t = - 8$

$:\implies\sf t = \cancel{ \dfrac{-8}{-8}}$

$:\implies{\boxed{\frak{\pink{t = 1}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Value\;of\;t\;is\; \bf{1}.}}}$