Question

If vertices of right-angle triangle are (5,2), (2,-2)and (-2,a). It is right angled between points (5,2),(2,-2) and (2,-2),(-2,a), find the value of a.

• A. 1
• B. 2
• C. 3
• D. -1

Answer 1

The correct option is A

1

Let A (5,2), b(2,-2), c(-2,a)

According to question, $\angle B={90}^{\circ }$

$\therefore A{C}^2=A{B}^2+B{C}^2$

Now, $A{C}^2=(5+2{)}^2+(2-a{)}^2$

$=49+4+a^2-4a$

$B{C}^2=(2+2{)}^2+(-2-a^2)$

$=16+4+a^2+4a$

$A{B}^2=(5-2{)}^2+(2+2{)}^2$

=9+16 =25

$\therefore 53+a^2-4a=25+20+a^2+4a$

$\Rightarrow 53-4a=45+4a$

$\Rightarrow 8=8a$

$\therefore a=1$.