Find the distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to a line having slope 3/4.
Given: (x1,y1) = A(2, 5), tanθ ![]()
⇒ sin θ
and cos θ![]()
To find:
The distance of a point from the line parallel to another line.
Explanation:
So, the equation of the line passing through (2, 5) and having a slope
is
Formula Used: ![]()
⇒ ![]()
⇒ 3x – 4y + 14 = 0
Let 3x – 4y + 7 = 0 intersect the line 3x + y + 4 = 0 at point P.
Let AP = r
Then, the coordinate of P are given by

⇒ x
and y![]()
Thus, the coordinate of P is ![]()
Clearly, P lies on the line 3x + y + 4 = 0
![]()
⇒ ![]()
⇒ 3r = – 15
⇒ r = – 5
Hence, the distance of the point (2, 5) from the line 3x + y + 4 = 0 is 5
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.