We have to find angle between the lines. Now, let the angle between the given lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is θ First we need to find the slope of both the lines. 7x - 4y = 0 ⇒ y = (7/4) x Therefore, the slope of the line 7x - 4y = 0 is m1=7∕4 Again, 3x - 11y + 5 = 0 ⇒ y = (3/11) x + (5/11) Therefore, the slope of the line 3x - 11y + 5 = 0 is m2=3∕11 We know that tan‌θ=|
m2−m1
1+m1m2
| ⇒tan‌θ=|
3
11
−
7
4
1+
3
11
×
7
4
|=|
12−77
44+21
|=|
−65
65
|=1 ⇒ tan θ = 1 = tan 45° ∴ θ = 45° Therefore, the required acute angle between the given lines is 45°