9.Straight Line
hard

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

A

$5x - 3y +1 = 0$

B

$5x + 3y -11 = 0$

C

$3x - 5y + 7 = 0$

D

$3x + 5y -13 = 0$

(JEE MAIN-2019)

Solution

$E$ is $\left( {\frac{5}{2},\frac{9}{2}} \right)$

Slope of  $AD = \frac{5}{3}$

Equation of $AD$ is $y – 2 = \frac{5}{3}\left( {x – 1} \right)$

$ \Rightarrow 5x – 3y + 1 = 0$

Standard 11
Mathematics

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