JEE MAIN - Mathematics (2014 (Offline) - No. 14)
Let $$PS$$ be the median of the triangle with vertices $$P(2, 2)$$, $$Q(6, -1)$$ and $$R(7, 3)$$. The equation of the line passing through $$(1, -1)$$ band parallel to PS is :
$$4x + 7y + 3 = 0$$
$$2x - 9y - 11 = 0$$
$$4x - 7y - 11 = 0$$
$$2x + 9y + 7 = 0$$
Explanation
Let $$P,Q,R,$$ be the vertices of $$\Delta PQR$$
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Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$
So, $$S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)$$
Now, slope of $$PS$$ $$ = {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}$$
Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$
Now, equation of line passing through $$(1, -1)$$ and having slope $$ - {2 \over 9}$$ is
$$y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)$$
$$9y + 9 = - 2x + 2$$
$$ \Rightarrow 2x + 9y + 7 = 0$$
Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$
So, $$S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)$$
Now, slope of $$PS$$ $$ = {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}$$
Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$
Now, equation of line passing through $$(1, -1)$$ and having slope $$ - {2 \over 9}$$ is
$$y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)$$
$$9y + 9 = - 2x + 2$$
$$ \Rightarrow 2x + 9y + 7 = 0$$
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