176
$A(6, -3)$, $B(0, 5)$ and $C(-2, 1)$ are vertices of $\Delta ABC$. Points $P(3, 1)$ and $Q(2, -1)$ lie on sides AB and AC respectively. Check whether $\frac{AP}{PB} = \frac{AQ}{QC}$.
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$AP = \sqrt{3^2 + (-4)^2} = 5$ [$\frac{1}{2}$ mark]
$PB = \sqrt{3^2 + (-4)^2} = 5$ [$\frac{1}{2}$ mark]
$AQ = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
$QC = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
So $\frac{AP}{PB} = 1$ and $\frac{AQ}{QC} = 1$
Therefore $\frac{AP}{PB} = \frac{AQ}{QC}$ [$1$ mark]
$PB = \sqrt{3^2 + (-4)^2} = 5$ [$\frac{1}{2}$ mark]
$AQ = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
$QC = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
So $\frac{AP}{PB} = 1$ and $\frac{AQ}{QC} = 1$
Therefore $\frac{AP}{PB} = \frac{AQ}{QC}$ [$1$ mark]