Find the equation of the line which is perpendicular to the line x/a -y/b =1 at the point where the given line meets y-axis.

Question

KnowledgeBoat · Accepted Answer

Let A be the point where the line meets y-axis. So, x-co-ordinate of point A will be zero. Substituting x = 0 in equation we get, A = (0, -b). The given line equation is, Comparing above equation with y = mx + c we get, Slope (m) = Let slope of perpendicular line be m1. As product of slope of perpendicular lines is -1, ∴ m × m1 = -1 ⇒ × m1 = -1 ⇒ m1 = Equation of line through A (0, -b) and slope = is : ⇒ y - y1 = m(x - x1) ⇒ y - (-b) = (x - 0) ⇒ b(y + b) = -ax ⇒ by + b2 = -ax ⇒ ax + by + b2 = 0 Hence, equation of required line is ax + by + b2 = 0.

Mathematics

Find the equation of the line which is perpendicular to the line x/a -y/b =1 at the point where the given line meets y-axis.

Straight Line Eq

Related Questions

A line through origin meets the line 2x = 3y + 13 at right angles at point Q. Find the co-ordinates of Q.

Find the equation of the perpendicular from the point P(1, –2) on the line 4x – 3y – 5 = 0. Also, find the co-ordinates of the foot of the perpendicular.

The points B(7, 3) and D(0, –4) are two opposite vertices of a rhombus ABCD. Find the equation of diagonal AC.

The points A(1, 3) and C(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal BD.