Question

Question 15 Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Class X1 - Maths -Straight Lines Page 234

Answer 1

The equation of line AB is 4x + 7y + 5 = 0-----(1)
and equation of line PQ is 2x - y = 0-------(2)

solve the equations (1) and (2) for obtaining point Q .
from equation (2) put y = 2x in equation (1)
4x + 7 × 2x + 5 = 0
4x + 14x + 5 = 0
18x + 5 = 0
x = -5/18 , put it in equation (2)
y = 2x = -5/9

so, co-ordinate of Q = { -5/18, -5/9}

now,
Length of PQ = √{(x2-x1)² + (y2-y1)²}
= √{(1+5/18)²+(2+5/9)²}
= √{(23/18)² + (23/9)²}
= √{23²{1/18² + 1/9²}
= 23√{1/9²(1/4+1)}
=23/9 √(1/4+1)
= 23/9 √(5/4)
= 23√5/18 unit