Question

Two vertices of a triangle are $(5,-1)$and $(-2,3).$ If orthocentre is the origin then coordinates of the third vertex are

• A. $\left(7,4\right)$
• B. $\left(-4,7\right)$
• C. $\left(4,-7\right)$
• D. $\left(-4,-7\right)$

Answer 1

The correct option is D

$\left(-4,-7\right)$

The explanation for the correct answer.

Step 1: Solve the slope of $AD$

Let $A(h,k),B(5,-1),$and $C(-2,3)$ be the vertices of the triangle.

Let $O(0,0)$be the orthocenter.

$AD$ is perpendicular to the side $BC.$

Let the slope of AD be $m_{AD}.$

Slope of $BC,m_{BC}=\frac{\left(3+1\right)}{\left(-2-5\right)}$$=-\frac{4}{7}$

Slope of $AD,m_{AD}=\frac{\left(k-0\right)}{\left(h-0\right)}$

$=\frac{k}{h}$

Step 2:

$AD\perp BC$

So $m_{AD}\times m_{BC}=-1$

$\left(\frac{k}{h}\right)\times -\left(\frac{4}{7}\right)=-1$

$$\begin{gathered} \Rightarrow 4k–7h=0\dots \left(i\right) \\ BE\perp AC \end{gathered}$$

The slope of $BE\times$ slope of $AC=-1$

So $m_{BE}\times m_{AC}=-1$

Slope of $AC=\frac{\left(3-k\right)}{\left(-2-h\right)}$

$\left(-\frac{1}{5}\right)\times \frac{\left(3-k\right)}{\left(-2-h\right)}=-1$

$\Rightarrow 5h–k+13=0\dots .\left(ii\right)$

Solving$\left(i\right)$ and $\left(ii\right),$ we get $h=-4,k=-7$

So the third vertex is $(-4,-7).$

Hence, option (D) is correct.