PS is the median of the triangle with vertices at $P(2,2), Q(6,-1)$ and $R(7,3)$, then the intercepts on the…
PS is the median of the triangle with vertices at $P(2,2), Q(6,-1)$ and $R(7,3)$, then the intercepts on the co-ordinate axes of the line passing through point $(1,-1)$ and parallel to PS are respectively
$\frac{7}{2}, \frac{-7}{9}$
$\frac{2}{7}, \frac{9}{7}$
$\frac{-7}{2}, \frac{-7}{9}$
$-2,-9$
Solution
$\mathrm{S}=$ midpoint of $\mathrm{QR}=\left(\frac{6+7}{2}, \frac{-1+3}{2}\right)=\left(\frac{13}{2}, 1\right)$
$\therefore \quad$ 'm' of $\mathrm{PS}=\frac{2-1}{2-\frac{13}{2}}=-\frac{2}{9}$
$\therefore \quad$ The required equation is $y+1=\frac{-2}{9}(x-1)$
$\Rightarrow 2 x+9 y+7=0$
Here, intercept on $\mathrm{X}$-axis is $-\frac{7}{2}$ and intercept
on $Y$-axis is $-\frac{7}{9}$