Question

which lines are perpendicular ?

Answer 1

Answer:

Lines  C and D

Step-by-step explanation:

For a pair of lines to be perpendicular ,

    the product of their slope must be - 1 .

Slope of A:

$2x - 3y = 6\\\\-3y = -2x + 6\\\\y = \frac{-2x}{-3} + \frac{6}{-3}$

$y = \frac{2}{3}x - 2\\\\slope_A = \frac{2}{3}$

Slope of B:

$3x - 2y = - 9\\\\-2y = - 3x - 9\\\\y = \frac{-3x}{-2} - \frac{9}{-2}\\\\y =\frac{3x}{2} + \frac{9}{2}\\\\slope_B = \frac{3}{2}$

Slope of C:

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

Slope of D:

$y = \frac{2}{3}x + 2\\\\slope_D = \frac{2}{3}$

Product of the slopes = - 1

$slope_A \times slope_B = \frac{2}{3} \times \frac{3}{2} = 1 \neq - 1 \\\\Therefore, not\ perpendicular.\\\\Slope_B \times slope_C = \frac{3}{2} \times \frac{-3}{2} = \frac{-9}{4} \neq -1\\\\Therefore , not \ perpendiucalr.\\\\Slope_C \times slope_D = -\frac{3}{2} \times \frac{2}{3} = - 1\\\\Therefore , perpendicular\\\\\\Slope_A \times slope_D = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9} \neq 1\\\\therefore , not \ perpendicular.$