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Find the equation of the line passing through the point (-3, 4) and perpendicular to the line 2x + 6y = 1 |
Answer» EXPLANATION.Equation of the line PASSING though points (-3,4). Perpendicular to the line : 2x + 6y = 1. As we know that, Slope of a perpendicular line = b/a. Slope of a line : 2x + 6y = 1 ⇒ 6/2 = 3. Slope = 3. Equation of line, ⇒ (y - y₁) = m(x - x₁). Put the VALUE in equation, we get. ⇒ ( y -(4)) = 3(x - (-3)). ⇒ y - 4 = 3(x + 3). ⇒ y - 4 = 3x + 9. ⇒ y - 3x = 13. MORE INFORMATION.Some FACTS about the normal.(1) = The slope of the normal drawn at point p(x₁, y₁) to the curve y = f(x) is -(dy/dx). (2) = If normal makes an angle ∅ with POSITIVE DIRECTION of x-axis then dy/dx = -cot∅. (3) = If normal is parallel to x-axis then dy/dx = ∞. (4) = If normal is parallel to y-axis then dy/dx = 0. |
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