Question

If kx - 3y - 14 = 0 and 3x + 4y + 10 = 0 are perpendicular to each other, find k.​

Answer 1

Answer:

k= 4

Step-by-step explanation:

Eq 1 is kx - 3y - 14 = 0

Eq 2 is 3x + 4y + 10 = 0

It is known that when two lines are perpendicular, their product is -1.

The gradient of both equations can be found by bringing them in the form

$y = mx + c$.

Eq 1 ⇒ -3y = -kx + 14

           y=  $\frac{k}{3}$ *x - (14/3)

therefore , m= $\frac{k}{3}$

Eq 2 ⇒ 4y = -3x -10

              y=(-3/4)x - (5/2)

therefore, m = (-3/4)

So, (-3/4) *$\frac{k}{3}$ = -1

 (-k/4) = -1

k= 4