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Line L is perpendicular to the line 3y+2x = 5 and passes through (-11,7), find the equation of L. |
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Answer» Let slope of line L be m. Since line L is perpendicular to 3y+2x=5,therefore, mx(-2/3)= -1 m=3/2 The equation of line L passing through (-11,7) will be y-7=3/2[x-(-11)] 2y-3x=47 Slope of line 3y + 2x = 5 is m1 = \(\frac{-2}3\) ∴ Slope of line perpendicular to it \(m = \frac{-1}{m_1} = \frac32\) It passes through (-11, 7) ∴ \(y - 7 = \frac32 (x - (-11))\) ⇒ \(y - 7 = \frac32 (x +11)\) ⇒ \(2y - 14 = 3x + 33\) ⇒ \(3x - 2y + 47 = 0\) ∴ Equation of line L is 3x - 2y + 47 = 0. |
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