If in a parallelogram $A B D C$ , the coordinates of $A, B$ and $C$ are respectively $(1, 2), (3, 4)$ and $(2, 5)$ , then the equation of the diagonal $A D$ is :

5 x + 3 y - 11 = 0

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5 x - 3 y + 1 = 0

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3 x - 5 y + 7 = 0

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3 x + 5 y - 13 = 0

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Solution

The correct option is B $5 x - 3 y + 1 = 0$

Diagonals of a parallelogram bisect each other.
$∴ (\frac{x_{1} + 1}{2}, \frac{y_{1} + 2}{2}) = (\frac{3 + 2}{2}, \frac{5 + 4}{2})$
$\Rightarrow \frac{x_{1} + 1}{2} = \frac{5}{2} \Rightarrow x_{1} = 4$
and $\frac{y_{1} + 2}{2} = \frac{9}{2} \Rightarrow y_{1} = 7$

$∴$ Coordinates of $D$ is $(4, 7)$
$∴$ Equation of $A D$ is
$\frac{y - 2}{x - 1} = \frac{7 - 2}{4 - 1}$

$\Rightarrow \frac{y - 2}{x - 1} = \frac{5}{3}$

$\Rightarrow 3 y - 6 = 5 x - 5$
$\Rightarrow 5 x - 3 y + 1 = 0$

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Similar questions

The equation of the line passing through $(1, 5)$ and perpendicular to the line $3 x - 5 y + 7 = 0$ is

3 x + 5 y = 19

(3, 2)

\frac{x}{2} + \frac{y}{3} = 1

\frac{x}{3} + \frac{y}{2} = 1

A

B

A B

(- 2, 4)

y -

(0, 2) .

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