Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.

Given, ⇒ 4x + 5y = 6 ⇒ 5y = -4x + 6 ⇒ y = Comparing above equations with y = mx + c we get, Slope (m 1 ) = Let slope of line perpendicular to 4x + 5y = 6 be m 2 . Since, product of slopes of perpendicular lines be -1, By point-slope form, ⇒ y - y 1 = m(x - x 1 ) ⇒ y - 1 = [x - (-2)] ⇒ 4(y - 1) = 5(x + 2) ⇒ 4y - 4 = 5x + 10 ⇒ 5x - 4y + 10 + 4 = 0 ⇒ 5x - 4y + 14 = 0. Hence, equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6 is 5x - 4y + 14 = 0.

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Mathematics

Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.

Straight Line Eq

Related Questions

Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4.

Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).

Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).

B(-5, 6) and D(1, 4) are the vertices of rhombus ABCD. Find the equations of diagonals BD and AC.