Question

If two vertices of a triangle are $(-2,3)$ and $(5,-1)$, the orthocenter lies at the origin, and the centroid on the line $x+y=7$, then the third vertex lies at

• A. $(-4,7)$
• B. $(8,14)$
• C. $(12,21)$
• D. $(\frac{64}{11},\frac{112}{11})$

Answer 1

Answer: (A)

$(-4,7)$

Slope of perpendicular from $A$ or $BC$

$$m_1=\frac{-1-0}{5-0}=\frac{-1}{5}$

As perpendicular from $A$ and $BC$ are perpendicular

$m_1{m}_2=-1m_2=-1\times -5$

slope of $BC=5$

Equation of $BC=5m=5;BC(-2,3)$

$y=5x+c$

$3=5(-2)+c$

$c=3+10c=13$

$y=5x+13$....(i)

slope of perpendicular from $B$ in $AC$

$m_1=\frac{0-3}{0-(-2)}=\frac{-3}{2}\Rightarrow m_1{m}_2=-1$

$m_2=\frac{2}{3}$

equation of $ACm=\frac{2}{3};A(5,-1)$

$y=\frac{2}{3}x+c$

$-1=\frac{2}{3}\left(5\right)+c\Rightarrow c=\frac{-13}{3}$

$3y=2x-13$.....(ii)

Solving equation (i) and (ii)

$3y=15x+39$

$-3y=2x-13$ $\Rightarrow x=-4$

$\overline{0=13x+42}$

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

$(x,y)=(-4,-7)$.