Question

The mid-point of the line segment joining the centroid and the orthocentre of the triangle whose vertices are $(a,b),(a,c)$ and $(d,c)$, is

• A. $\left(\frac{5a+d}{6},\frac{b+5c}{6}\right)$
• B. $\left(\frac{a+5d}{6},\frac{5b+c}{6}\right)$
• C. (a, c)
• D. (0,0)

Answer 1

Answer: (A)

Centroid of $\Delta ABC=\left(\frac{a+a+d}{3},\frac{b+c+c}{3}\right)$
$\equiv \left(\frac{2a+d}{3},\frac{2c+b}{3}\right)$
Since, $\Delta ABC$ is right triangle and $\angle B=90^{\circ }$
$\therefore$ Orthocentre of $\Delta ABC$ is $(a,c)$
Mid-point of centroid and orthocentre is
$=\left(\frac{\frac{2a+d}{3}+a}{2},\frac{\frac{2c+b}{3}+c}{2}\right)=\left(\frac{5a+d}{6},\frac{5c+b}{6}\right)$