Question

The points $(-4,-4),(-1,-2)$ and $(x,-8)$ are the vertices of a right triangle with the right angle at $(-1,-2)$. Find the value of $x$.

• A. $0$
• B. $2$
• C. $3$
• D. $-1$
• E. $-4$

Answer 1

Answer: (C)

$3$

Since the right angle is at the point $B$, $BA$ and $BC$ are the legs of the right triangle and $AC$ is the hypotenuse.

So we find the squares of distances $AC$, $BC$ and $AB$:

${\left(AB\right)}^2={(-1+4)}^2+{(-2+4)}^2=13$

${\left(BC\right)}^2={(x+1)}^2+{(-8+2)}^2={(x+1)}^2+36$

${\left(AC\right)}^2={(x+4)}^2+{(-8+4)}^2={(x+4)}^2+16$

Apply pythagoras theorem:

$\Rightarrow {\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2$

$\Rightarrow 13+{(x+1)}^2+36={(x+4)}^2+16$

$\Rightarrow 13+x^2+1+2x+36=x^2+16+8x+16$

$\Rightarrow -6x=-18$

$\Rightarrow x=3$

Hence, option C is correct.