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Find the equation of the straight line passing through the point (-2, 1) and perpendicular to the line 3x+5y+2=0. |
Answer» EXPLANATION.Equation of STRAIGHT lines. Passing through POINT = (-2,1). Perpendicular to the line : 3X + 5y + 2 = 0. As we know that, SLOPE of the perpendicular line = b/a. Slope of the line : 3x + 5y + 2 = 0 is 5/3. ⇒ Slope = m = 5/3. Equation of straight line. ⇒ (y - y₁) = m(x - x₁). Put the values in the equation, we get. ⇒ (y - 1) = 5/3(x - (-2)). ⇒ (y - 1) = 5/3(x + 2). ⇒ 3(y - 1) = 5(x + 2). ⇒ 3y - 3 = 5X + 10. ⇒ 5x + 10 - 3y + 3 = 0. ⇒ 5x - 3y + 13 = 0. MORE INFORMATION.Slope of line.(1) = m = tanθ, where θ is the acute angle made by a line with the positive direction of x axis in anticlockwise. (2) = The slope of a line joining two points (x₁, y₁) and (x₂, y₂) is given by, m = (y₂ - y₁)/(x₂ - x₁). |
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