NTSE Stage 2 SAT 2015 Question Paper

D is midpoint of BC
The coordinate of
$\mathrm{D}=\left(\frac{-15+5}{2},\frac{15+5}{2}\right)$
$=\left(-5,10\right)$
Now,
$=\left(1,-2\right)$
Now area of ∆PBC
$=\frac{1}{2}|\begin{array}{ccc}1& -2& 1\\ -15& 15& 1\\ 5& 5& 1\\ \end{array}|$
$\frac{1}{2}\left[10-40-150\right]$
$\frac{1}{2}\left[-180\right]=|-90|=90$
and area of ∆ABC $=\frac{1}{2}|\begin{array}{ccc}5& -10& 1\\ -15& 15& 1\\ 5& 5& 1\\ \end{array}|$
$=\frac{1}{2}\left[50-200-150\right]=\frac{1}{2}\left[-300\right]$
$=\left(-150\right)=150$
Now ratio, of area of ∆PBC and ∆ABC $=\frac{90}{150}=\frac{3}{5}$
∴ Option (c) is correct.