Prove the following

Question:

Let $A(1,0), B(6,2)$ and $\mathrm{C}\left(\frac{3}{2}, 6\right)$ be the vertices of a triangle $A B C$. If $P$ is a point inside the triangle $A B C$ such that the triangles $A P C, A P B$ and $B P C$ have equal areas, then the length of the line segment

$P Q$, where $Q$ is the point $\left(-\frac{7}{6},-\frac{1}{3}\right)$, is_________.

Solution:

P will be centroid of $\triangle A B C$

$P\left(\frac{17}{6}, \frac{8}{3}\right) \Rightarrow P Q=\sqrt{(4)^{2}+(3)^{2}}=5$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now