The mid-point of the line segment joining the centroid and the orthocentre of the triangle whose vertices…

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
  1. $\left(\frac{5 a+d}{6}, \frac{b+5 c}{6}\right)$
  2. $\left(\frac{a+5 d}{6}, \frac{5 b+c}{6}\right)$
  3. $(a,c)$
  4. $(0,0)$

Solution

The vertices of the triangle are given as $\mathrm{A}(\mathrm{a}, \mathrm{b}), \mathrm{B}(\mathrm{a}, \mathrm{c})$ and $\mathrm{C}(\mathrm{d}, \mathrm{c})$
Centroid of $\triangle \mathrm{ABC}=\left(\frac{a+a+d}{3}, \frac{b+c+c}{3}\right)$ $\equiv\left(\frac{2 a+d}{3}, \frac{2 c+b}{3}\right)$ Since, $x$-coordinate of point A and B are same. Also $y$-coordinate of point $\mathrm{B}$ and $\mathrm{C}$ are same so it forms a right angle triangle. $\therefore \quad$ Orthocentre of $\triangle \mathrm{ABC}$ is $(\mathrm{a}, \mathrm{c})$ Mid-point of centroid and orthocentre is $=\left(\frac{\frac{2 a+d}{3}+a}{2}, \frac{\frac{2 c+b}{3}+c}{2}\right)=\left(\frac{5 a+d}{6}, \frac{5 c+b}{6}\right)$

Asked in: AP EAMCET 2016

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