Question

IfA(5,2), B (2,-2) and C(-2,t) are the vertices of a right angled triangle with ZB=90°, then find the value of t.​

Answer 1

Answer:

$t = 1$

Step-by-step explanation:

$AB^2 = (2-5)^2 + (-2-2)^2\\AB^2 = (3)^2 + (4)^2\\AB^2 = 9 +16\\AB^2 =25\\\\BC^2 = (-2-2)^2 + (t+2)^2\\BC^2 =(4)^2 + (t+2)^2\\BC^2 = 16 + (t+2)^2\\\\AC^2 = (5+2)^2+(2-t)^2\\AC^2 = (7)^2 + (2-t)^2\\AC^2=49 + (2-t)^2$

${\boxed {\boxed { { AB^2=25 { \ {\ { }}} }} }}$  ${\boxed {\boxed { { BC^2 =16 +(t+2)^2 { \ {\ { }}} }} }}$  ${\boxed {\boxed { { AC^2 =49 +(2-t)^2 { \ {\ { }}} }} }}$

SINCE TRIANGLE ABC IS RIGHT ANGLE TRIANGLE

∴ $AC^2 =AB^2 +BC^2$

$\Rightarrow 49+ (2-t)^2 = 25 +16+(t+2)^2\\\\\Rightarrow 49+4 -4t+t^2 =41 +t^2 +4t +4\\\\\Rightarrow 53 - 4t = 45 +4t\\\\\Rightarrow 53 - 45 = 4t + 4t\\\\\Rightarrow 8 = 8t\\\\\Rightarrow t = \frac{8}{8} \\\\\Rightarrow t = 1$

${\boxed {\boxed {\boxed { t = 1 { \ {\ { }}} }} }}$  

____________________________________