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9.Straight Line
hard
Let $A (-3, 2)$ and $B (-2, 1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x + 4y + 2 = 0$, then the vertex $C$ lies on the line
A
$4x +3y+5= 0$
B
$3x +4y+3=0$
C
$4x +3y+3=0$
D
$3x + 4y + 5 = 0$
(JEE MAIN-2013)
Solution
Let $C\left( {{x_1},{y_1}} \right)$
Centroid, $E = \left( {\frac{{{x_1} – 5}}{3},\frac{{{x_1} – 3}}{3}} \right)$
Since centroid lies on the line
$3x + 4y + 2 = 0$
$\therefore 3\left( {\frac{{{x_1} – 5}}{3}} \right) + 4\left( {\frac{{{x_1} – 3}}{3}} \right) + 2 = 0$
$ \Rightarrow 3{x_1} + 4{y_1} + 3 = 0$
Hence vertex $\left( {{x_1},{y_1}} \right)$ lies on the line
$3x + 4y + 3 = 0$
Standard 11
Mathematics
