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Find the equation of the line passing through the point (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6). |
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Answer» Slope of the line through the points (2, 5) and (-3, 6) \(\frac{y_2 - y_1}{x_2 -x_1} = \frac{6-(5)}{-3-2} = -\frac{1}{5}\) Then the slope of the required line is 5. Hence the equation is y – y1 = m(x – x1) ⇒ y – 5 = 5(x – (-3) ⇒ y – 5 = 5x + 15 ⇒ 5x – y + 20 = 0. |
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