Let C be the centroid of the triangle with vertices ( 3 , - 1 ) , ( 1 , 3 ) and ( 2 , 4 ) . Let P be the point of intersection of the lines x + 3   y - 1 = 0 and 3   x - y + 1 = 0 . Then the line passing through the points C and P also passes through the point

Question

Let C be the centroid of the triangle with vertices ( 3 , - 1 ) , ( 1 , 3 ) and ( 2 , 4 ) . Let P be the point of intersection of the lines x + 3 &#160; y - 1 = 0 and 3 &#160; x - y + 1 = 0 . Then the line passing through the points C and P also passes through the point

Aicharya · Accepted Answer

The centroid of a triangle having vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}) .$

The centroid of the triangle having vertices $(3, - 1), (1, 3)$ and $(2, 4)$ is

$C (\frac{3 + 1 + 2}{3}, \frac{- 1 + 3 + 4}{3}) = (2, 2) .$

The equation of a line passing through the point on intersection of lines $L_{1} = 0$ and $L_{2} = 0$ is $L_{1} + λ L_{2} = 0 .$

Thus, the equation of line which passes through point of intersection $P$ of lines
$x + 3 y - 1 = 0$ and $3 x - y + 1 = 0$ is

$(x + 3 y - 1) + λ (3 x - y + 1) = 0 . . . (i)$

$∵$ Line $(i)$ passes through point $C (2, 2),$ so

$(2 + 6 - 1) + λ (6 - 2 + 1) = 0$

$\Rightarrow 7 + 5 λ = 0$

$\Rightarrow λ = - \frac{7}{5} .$

$∴$ Equation of line passes through points $C$ and $P$ is

$(x + 3 y - 1) - \frac{7}{5} (3 x - y + 1) = 0$

$\Rightarrow 5 x + 15 y - 5 - 21 x + 7 y - 7 = 0$

$\Rightarrow - 16 x + 22 y - 12 = 0$

$\Rightarrow 8 x - 11 y + 6 = 0 .$

From the options, point $(- 9, - 6)$ satisfy the line.

Hence, option $(B)$ is correct.

Let C be the centroid of the triangle with vertices ( 3 , - 1 ) , ( 1 , 3 ) and ( 2 , 4 ) . Let P be the…

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