Question
Let $$PS$$ be the median of the triangle with vertices $$P\left( {2,\,2} \right),\,Q\left( {6,\, - 1} \right)$$ and $$R\left( {7,\,3} \right).$$ The equation of the line passing through $$\left( {1,\, - 1} \right)$$ and parallel to $$PS$$ is :
A.
$$4x+7y+3=0$$
B.
$$2x-9y-11=0$$
C.
$$4x-7y-11=0$$
D.
$$2x+9y+7=0$$
Answer :
$$2x+9y+7=0$$
Solution :
Let $$P, \,O, \,R,$$ be the vertices of $$\Delta PQR$$
Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$
$${\text{So, }}S = \left( {\frac{{7 + 6}}{2},\,\frac{{3 - 1}}{2}} \right) = \left( {\frac{{13}}{2},\,1} \right)$$
Now, slope of $$PS\,\,\,\,\, = \frac{{2 - 1}}{{2 - \frac{{13}}{2}}} = - \frac{2}{9}$$
Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$
Now, equation of line passing through $$\left( {1,\, - 1} \right)$$ and having slope $$ - \frac{2}{9}$$ is
$$\eqalign{ & y - \left( { - 1} \right) = - \frac{2}{9}\left( {x - 1} \right) \cr & \Rightarrow 9y + 9 = - 2x + 2 \cr & \Rightarrow 2x + 9y + 7 = 0 \cr} $$
Let $$P, \,O, \,R,$$ be the vertices of $$\Delta PQR$$
Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$
$${\text{So, }}S = \left( {\frac{{7 + 6}}{2},\,\frac{{3 - 1}}{2}} \right) = \left( {\frac{{13}}{2},\,1} \right)$$
Now, slope of $$PS\,\,\,\,\, = \frac{{2 - 1}}{{2 - \frac{{13}}{2}}} = - \frac{2}{9}$$
Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$
Now, equation of line passing through $$\left( {1,\, - 1} \right)$$ and having slope $$ - \frac{2}{9}$$ is
$$\eqalign{ & y - \left( { - 1} \right) = - \frac{2}{9}\left( {x - 1} \right) \cr & \Rightarrow 9y + 9 = - 2x + 2 \cr & \Rightarrow 2x + 9y + 7 = 0 \cr} $$