Question

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

• A. $(4,7)$
• B. $(3,7)$
• C. $(-4,-7)$
• D. None of these

Answer 1

Answer: (C)

$(-4,-7)$

Given vertices of triangles are $A(5,1)B(-2,3)$ let $C$ be $(a,b)$

Orthocenter $D(0,0)$

Orthocenter is a point where altitude intersects

Slope of AB $=\frac{3+1}{-2-5}=\frac{-4}{7}$

Therefore slope of CG is $\frac{7}{4}$

Equation of CG $=y-b=\frac{7}{4}(x-a)$

Equation of CG $\Rightarrow 7x-4y=7a-4b$

Slope of AC $=\frac{b+1}{a-5}$

Slope of BF$=-\left(\frac{a-5}{b+1}\right)$

Equation of BF $=-\left(\frac{a-5}{b+1}\right)(x+2)$

$by+y=-3b-3=5x+10-ax-2a$

Equation of BF $\Rightarrow x(5-a)-y(b+1)=2a-3b-13$

Intersection of line BF & CG is $D(0,0)$

$7x-4y=7a-4b$

$(x,y)=(a,0)7a=4b.........\left(1\right)x(5-a)-y(b+1)=2a-3b-13(x,y)=(0,0)2a-3b=\mathrm{13.........}\left(2\right)2a-3\left(\frac{7a}{4}\right)=13-13a=13\times 4a=-4\therefore b=-7$

Therefore third vertex will be $(-4,-7)$