Show that the perpendicular let fall from any point on the straight line 2x + 11y – 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x – 3y – 2 = 0 are equal to each other.
Given:
Lines 24x + 7y = 20 and 4x – 3y – 2 = 0
To prove:
The perpendicular let fall from any point on the straight line 2x + 11y – 5 = 0 upon the two straight lines 24x + 7y = 20 and 4x – 3y – 2 = 0 are equal to each other.
Concept Used:
Distance of a point from a line.
Assuming:
P(a, b) be any point on 2x + 11y − 5 = 0
Explanation:
∴ 2a + 11b − 5 = 0
⇒ b ……..(i)
Let d1 and d2 be the perpendicular distances from point P
on the lines 24x + 7y = 20 and 4x − 3y − 2 = 0, respectively.
d1
⇒ d1
From (1)
⇒ d1
Similarly,
d2
⇒ d2
From (1)
⇒ d2
∴ d1 = d2
Hence proved.