First slide

Pair of straight lines

Question

The equations to a pair of opposite sides of parallelogram are $x^{2} - 5 x + 6 = 0$ and $y^{2} - 6 y + 5 = 0$ , the equations to its diagonals are

Moderate

A

x + 4y = 13, y = 4x – 7
B

4x + y = 13, 4y = x – 7
C

4x + y = 13, y = 4x – 7
D

y – 4x = 13, y + 4x = 7

Solution

$x^{2} - 5 x + 6 = 0 \Rightarrow (x - 2) (x - 3) = 0$
$\Rightarrow$ two of the parallel sides are x = 2, x = 3
$y^{2} - 6 y + 5 = 0 \Rightarrow (y - 1) (y - 5) = 0$
$\Rightarrow$ other two parallel sides are y = 1, y = 5
So, the vertices of the parallelogram are (2, 1), (2, 5), (3, 1), (3, 5). Equation of the diagonal are
$y - 5 = \frac{1 - 5}{3 - 2} (x - 2)$
$\Rightarrow$ 4x + y = 13 and $y - 1 = \frac{5 - 1}{3 - 2} (x - 2)$
$\Rightarrow$ y = 4x - 7

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