Question

If two vertices of a triangle are $(5,-1)$ and $(-2,3)$, and the orthocentre lies at the origin, the coordinates of the third vertex are

• A. $(-3,-2)$
• B. $(2,3)$
• C. $(-4,-7)$
• D. $(7,11)$

Answer 1

Answer: (C)

$(-4,-7)$

Let $A(5,-1)$ and $B(-2,3)$ and orthocentre is $O(0,0)$

$\therefore$ Slope of altitude coming from A

$=\frac{-1-0}{5-0}=\frac{-1}{5}$

Slope of BC $=\left[\frac{1}{\frac{-1}{5}}\right]=5$

$\therefore$ Slope of $BC\times$ Slope of altitude coming from $A=-1$

Equation of line through $BC$ is,

$y=5x+c$

If $B(-2,3)$ lies on above the line

i.e,$3=-5\times +C$ $\Rightarrow C=13$

i.e,$y=5x+13\to$ (1)

Similarly, the slope of AC $=\frac{2}{3}$

Substituting $(5,-1)$ in the equation

$-1=\frac{2}{3}\times 5+C\Rightarrow C=-1\frac{-10}{3}=\frac{-13}{3}$

i.e,$y=\frac{2}{3}x-\frac{13}{3}$

$\Rightarrow 3y=2x-13\to \left(2\right)$

eq Subtract Eqn (1) and (2)

$0=[5x-\frac{2}{3}x]+[13-(\frac{-13}{3}\left)\right]=\frac{13x}{3}+\left[\frac{39+13}{3}\right]=\frac{13x+52}{3}=0⟹13x=-52i.e,x=-4$

Put the value of $x$ in eqn (1)

$y=5(-4)+13=-7$

$\therefore$ The third vertex is $(-4,-7)$