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(A) The vertices of the triangle are $A(a, b), B(a, c)$ and $C(d, c)$.Since the $x$-coordinate of $A$ and $B$ is $a$,the side $AB$ is vertical. Since

Vedclass · Accepted Answer

$\left(\frac{5 a+d}{6}, \frac{b+5 c}{6}\right)$

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(A) The vertices of the triangle are $A(a, b), B(a, c)$ and $C(d, c)$.Since the $x$-coordinate of $A$ and $B$ is $a$,the side $AB$ is vertical. Since the $y$-coordinate of $B$ and $C$ is $c$,the side $BC$ is horizontal. Thus,$	riangle ABC$ is a right-angled triangle at vertex $B(a, c)$.The orthocentre of a right-angled triangle is the vertex at which the right angle is formed. Therefore,the orthocentre is $B(a, c)$.The centroid $G$ of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}ight)$.$G = \left(\frac{a+a+d}{3}, \frac{b+c+c}{3}ight) = \left(\frac{2 a+d}{3}, \frac{b+2 c}{3}ight)$.The mid-point of the line segment joining the centroid $G\left(\frac{2 a+d}{3}, \frac{b+2 c}{3}ight)$ and the orthocentre $H(a, c)$ is:$M = \left(\frac{\frac{2 a+d}{3}+a}{2}, \frac{\frac{b+2 c}{3}+c}{2}ight) = \left(\frac{2 a+d+3 a}{6}, \frac{b+2 c+3 c}{6}ight) = \left(\frac{5 a+d}{6}, \frac{b+5 c}{6}ight)$.

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

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