Question

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

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

Answer 1

The correct option is D (-4, -7)

$LetthetrianglebeABCwithA=(x_1,y_1)=(5,-1),B=(x_2,y_2)=(-2,3)andC=(p,q)TheorthocenterisO=(0,0)SlopeofAO=m_{AO}=\frac{y_1-0}{x_1-0}=-\frac{1}{5}SlopeofBC=m_{BC}=\frac{q-y_2}{p{-x}_2}=\frac{q-3}{p+2}SinceAO\perp BC\therefore m_{AO}\times m_{BC}=-1⟹\frac{q-3}{p+2}\times (-\frac{1}{5})=-1⟹q=5p+\mathrm{13.............}\left(i\right)AgainslopeofBO=m_{BO}=\frac{y_2-0}{x_2--0}=-\frac{3}{2}andslopeofAC=m_{AC}=\frac{q-y_2}{p-x_2}=\frac{q+1}{p-5}SinceBO\perp AC\therefore m_{BO}\times m_{AC}=-\frac{3}{2}\times \frac{q+1}{p-5}=-1⟹q=\frac{2p-13}{3}........\left(ii\right)Eliminatingqfromequation\left(i\right)\&\left(ii\right)weget5p+13=\frac{2p-13}{3}⟹p=-4.Puttingp=-4inequation\left(i\right)wegetq=5\times (-4)+13⟹q=-7\therefore Thethirdvertexis(-4,-7)Ans-OptionD$