Question

If the acute angle between the lines 4x-y+7=0 kx-5y-9=0is 45 degree then find the value of k

Answer 1

Given:

4x - y + 7 = 0

kx - 5y - 9 = 0

Acute angle between the lines. = 45°

To find:

The value of k

Solution:

y = mx + c

4x - y + 7 = 0

y = 4x + 7

m₁ = 4

kx - 5y - 9 = 0

5y = kx - 9

y = (k/5)x - 9

m₂ = k/5

$tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2} |\\\\tan 45\° = |\frac{4 - k/5}{1+4*k/5} | \\\\1 = |\frac{20 - k}{5 + 4k} |$

5 + 4k = 20 - k

4k + k = 20 - 5

5k = 15

k = 3

Therefore, the value of k is 3.