Let $A(1,0), B(6,2)$ and $C\left(\frac{3}{2}, 6\right)$ be the vertices of a triangle $ABC$. If $P$ is a point inside the triangle ABC such that the triangles $\mathrm{APC}, \mathrm{APB}$ and BPC have equal areas, then the length of the line segment PQ, where Q is the point $\left(-\frac{7}{6},-\frac{1}{3}\right)$ is ______

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Straight Lines

Let

A (1, 0), B (6, 2)

and

C (\frac{3}{2}, 6)

be the vertices of a triangle

A B C

. If

P

is a point inside the triangle ABC such that the triangles

APC, APB

and BPC have equal areas, then the length of the line segment PQ, where Q is the point

(- \frac{7}{6}, - \frac{1}{3})

is ______

Answer: 5

Solution

A point that splits a triangle into three equal-area pieces is its centroid, so

P = (\frac{1 + 6 + 3/2}{3}, \frac{0 + 2 + 6}{3}) = (\frac{17}{6}, \frac{8}{3})

. Its distance to

Q (- \frac{7}{6}, - \frac{1}{3})

P Q = (\frac{17}{6} + \frac{7}{6})^{2} + (\frac{8}{3} + \frac{1}{3})^{2} = 4^{2} + 3^{2} = 5

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Integer answer

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