Let A(1, 0), B(6, 2) and C ( 3 \over 2,6 ) be the vertices of a triangle ABC. 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 ( - 7 \over 6, - 1 \over 3 ) , is \underline\hspace2cm .

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2020 · 7 Jan · Shift 1 · Q32 with Solution

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Correct answer: 5

The point $P$ lies inside $\triangle ABC$ such that the triangles $\triangle APC$ , $\triangle APB$ , and $\triangle BPC$ have equal areas.

Since these three triangles together exactly partition $\triangle ABC$ , if their areas are equal, then each must have area $\frac{1}{3}[ABC].$

This means $P$ is the centroid of triangle $ABC$ .

So, $P=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right).$

Given: $A(1,0),\quad B(6,2),\quad C\left(\frac{3}{2},6\right).$

=\frac{7+\frac{3}{2}}{3} =\frac{\frac{14}{2}+\frac{3}{2}}{3} =\frac{\frac{17}{2}}{3} =\frac{17}{6}.$$ $$y_P=\frac{0+2+6}{3}=\frac{8}{3}.$$ Hence, $$P\left(\frac{17}{6},\frac{8}{3}\right).$$ 3. **Coordinates of $Q$** Given: $$Q\left(-\frac{7}{6},-\frac{1}{3}\right).$$ 4. **Find length $PQ$** Using distance formula, $$PQ=\sqrt{\left(\frac{17}{6}-\left(-\frac{7}{6}\right)\right)^2+\left(\frac{8}{3}-\left(-\frac{1}{3}\right)\right)^2}.$$ Simplify differences: $$\frac{17}{6}+\frac{7}{6}=\frac{24}{6}=4,$$ $$\frac{8}{3}+\frac{1}{3}=\frac{9}{3}=3.$$ So, $$PQ=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5.$$ 5. **Final answer** $$\boxed{5}$$ 6. **Comparison with stored answer** Stored correct answer = $5$. Our derived answer matches the stored answer.

View written solutionFree

Correct answer: 5

The point $P$ lies inside $\triangle ABC$ such that the triangles $\triangle APC$ , $\triangle APB$ , and $\triangle BPC$ have equal areas.

Since these three triangles together exactly partition $\triangle ABC$ , if their areas are equal, then each must have area $\frac{1}{3}[ABC].$

This means $P$ is the centroid of triangle $ABC$ .

So, $P=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right).$

Given: $A(1,0),\quad B(6,2),\quad C\left(\frac{3}{2},6\right).$

=\frac{7+\frac{3}{2}}{3} =\frac{\frac{14}{2}+\frac{3}{2}}{3} =\frac{\frac{17}{2}}{3} =\frac{17}{6}.$$ $$y_P=\frac{0+2+6}{3}=\frac{8}{3}.$$ Hence, $$P\left(\frac{17}{6},\frac{8}{3}\right).$$ 3. **Coordinates of $Q$** Given: $$Q\left(-\frac{7}{6},-\frac{1}{3}\right).$$ 4. **Find length $PQ$** Using distance formula, $$PQ=\sqrt{\left(\frac{17}{6}-\left(-\frac{7}{6}\right)\right)^2+\left(\frac{8}{3}-\left(-\frac{1}{3}\right)\right)^2}.$$ Simplify differences: $$\frac{17}{6}+\frac{7}{6}=\frac{24}{6}=4,$$ $$\frac{8}{3}+\frac{1}{3}=\frac{9}{3}=3.$$ So, $$PQ=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5.$$ 5. **Final answer** $$\boxed{5}$$ 6. **Comparison with stored answer** Stored correct answer = $5$. Our derived answer matches the stored answer.