Question

Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to the line x - 2y = 1.

Answer 1

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0

Since, it passes through (3,5)

⇒3−10+λ=0

⇒λ=7
Therefore, the line is x−2y+7=0.

To find the point of intersection of x−2y+7=0 and 2x+3y−14=0:
Substitute x=2y-7 in equation 2x+3y−14=0, we get
2(2y-7)+3y-14=0
4y-14+3y-14=0
7y=28
y=4
Substitute y=4 in equation x−2y+7=0, we get
x-8+7=0
x=1
Thus, point of intersection is (1, 4)

The distance between (3,5) and (1,4) is $\sqrt{(3-1{)}^2+(5-4{)}^2}=\sqrt{2^2+1^2}$
$=\sqrt{4+1}=\sqrt{5}$ is the required distance.