First slide

Equation of line in Straight lines

Question

Two consecutive sides of a parallelogram are 4x + 5y = 0 and 7x + 2y = 0. If the equation to one diagonal is 11x + 7y = 9, then the equation of the other diagonal is

Easy

A

$x - y = 0$
B

$x + y = 0$
C

$x + y = 1$
D

$x - y = 1$

Solution

$Let the equation of sides A B and A D of the parallelogram ABCD be$
$4 x + 5 y = 0 - - - - (1)$
$and 7 x + 2 y = 0 - - - (2)$
respectively. Solving (l) and (2), we have x = 0, y= 0
A = (0,0) Equation of one diagonal of the parallelogram is
$11 x + 7 y = 9 - - - (3)$
Clearly, I (0, 0) does not lie on diagonal (3), therefore (3) is the equation of diagonal BD.
$Solving (1) and (3), we get B \equiv (\frac{5}{3}, - \frac{4}{3}) and D \equiv (- \frac{2}{3}, \frac{7}{3}) .$
$Since H is the middle point of B D ∴ H \equiv (\frac{1}{2}, \frac{1}{2}) .$
Now the equation of diagonal AC which passes through
$A (0, 0) and H (\frac{1}{2}, \frac{1}{2}) is y - 0 = \frac{0 - \frac{1}{2}}{0 - \frac{1}{2}} (x - 0)$
$or y - x = 0$

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