Question

If two vertices of a triangle are (5, -1) and (-2,3) and its orthocentre is at (0,0), then the third vertex is

• A. (4,-7)
• B. (-4,-7)
• C. (-4,7)
• D. (4,7)

Answer 1

Answer: (B)

Let the third vertex of $\Delta$ ABC be $\left(a,b\right)$.
Orthocentre $=H\left(0,0\right)$
Let A $\left(5,-1\right)$ and B $\left(-2,3\right)$ be other two
vertices of $\Delta$ ABC.
Now, (Slope of AH) $\times$ (Slope of BC) $=-1$
$\Rightarrow \left(\frac{-1-0}{5-0}\right)\left(\frac{b-3}{a+2}\right)=-1$
$\Rightarrow b-3=5\left(a+2\right)...\left(1\right)$
Similarly,
(Slope of BH) $\times$ (Slope of AC) $=-1$
$\Rightarrow -\left(\frac{3}{2}\right)\times \left(\frac{b+1}{a-5}\right)=-1$
$\Rightarrow 3h+3=2a-10$
$\Rightarrow 3b-2a+13=0...\left(2\right)$
On solving equation (1) and (2) we get
$a=-4,b=-7$
Hence, third vertex is $\left(-4,-7\right).$