Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to the line x – 2y = 1.
Given: (x1,y1) = A(3, 5)
To find:
The distance of a point from the line parallel to another line.
Explanation:
It is given that the required line is parallel to x – 2y = 1
⇒ 2y = x – 1
⇒ y
⇒ sin θ and cos θ
So, the equation of the line is
Formula Used:
⇒
⇒ x – 2y + 7 = 0
Let x – 2y + 7 = 0 intersect the line 2x + 3y = 14 at point P.
Let AP = r
Then, the coordinate of P is given by
⇒ x and y
Thus, the coordinate of P is
Clearly, P lies on the line 2x + 3y = 14
⇒
⇒
⇒ r
Hence, the distance of the point (3, 5) from the line 2x + 3y = 14 is