Question

The vertices of a$△$ABC are A(5, 5) B(1, 5) and C(9, 1). A line is drawn to intersect sides AB and AC at p and Q respectively, such that $\frac{AP}{AB}=\frac{AQ}{AC}=\frac{3}{4}$. Find the length of the line segment PQ.

Answer 1

We have,$\frac{AP}{AB}=\frac{AQ}{AC}=\frac{3}{4}$

$\Rightarrow \frac{AP}{AP+PB}=\frac{AQ}{AQ+QC}=\frac{3}{4}$

$\Rightarrow \frac{AP}{AP+PB}=\frac{3}{4},\frac{AQ}{AQ+QC}=\frac{3}{4}$

$\Rightarrow 4AP=3AP+3PBand4AQ=3AQ+3QC$

$\Rightarrow AP=3PBandAQ=3QC$

$\Rightarrow \frac{AP}{PB}=\frac{3}{1}and\frac{AQ}{QC}=\frac{3}{1}$

$\Rightarrow PandQdivideABandACrespectivelyinthesameratio3:1$

Thus, the coordinates of P and Q are

$(\frac{3\times 1+1\times 5}{3+1},\frac{3\times 5+1\times 5}{3+1})=(2,5)and(\frac{3\times 9+1\times 5}{3+1},\frac{3\times 1+1\times 5}{3+1})=(8,2)$

$\therefore PQ=\sqrt{{(2-8)}^2+{(5-2)}^2}=\sqrt{45}=3\sqrt{5}units$