Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.
Straight Line Eq
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Question

Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.

KnowledgeBoat · Accepted Answer

Let point where line touches x-axis be A. So, A = (-3, 0) Given equation of line, ⇒ 3x + 5y = 1 ⇒ 5y = -3x + 1 ⇒ y = Comparing above equation with y = mx + c we get, Slope (m₁) = Let slope of line perpendicular to 3x + 5y = 1 be m₂. ⇒ m₁ × m₂ = -1 ⇒ ⇒ . By point-slope form, Equation of line with slope = and passing through (-3, 0) is : ⇒ y - y₁ = m(x - x₁) ⇒ y - 0 = [x - (-3)] ⇒ 3y = 5[x + 3] ⇒ 3y = 5x + 15 ⇒ 5x - 3y + 15 = 0. Hence, equation of required line is 5x - 3y + 15 = 0.

Mathematics

Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.

Straight Line Eq

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Mathematics

Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.

Straight Line Eq

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(i) the coordinates of A.
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