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Joint equation of two lines, through (2-3), perpendicular to two lines 3x^(2)+xy-2y^(2)=0, is

Joint equation of two lines, through (2,−3)(2, -3), perpendicular to two lines 3x2+xy−2y2=03x^{2}+xy-2y^{2}=0, is

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Q. Joint equation of two lines, through (2,−3)(2, -3), perpendicular to two lines 3x2+xy−2y2=03x^{2}+xy-2y^{2}=0, is
  1. Identify slopes by factoring: Identify the slopes of the given lines by factoring the equation 3x2+xy−2y2=03x^2 + xy - 2y^2 = 0.\newline3x2+xy−2y2=(3x+2y)(x−y)=03x^2 + xy - 2y^2 = (3x + 2y)(x - y) = 0\newlineSo, the slopes of the lines are −23-\frac{2}{3} and 11.
  2. Find perpendicular slopes: Find the slopes of the lines perpendicular to the given lines.\newlineThe slopes of lines perpendicular to the given lines are the negative reciprocals of the original slopes.\newlineSo, the slopes of the perpendicular lines are 32\frac{3}{2} and −1-1.
  3. Write equations through (2,−3)(2, -3): Write the equations of the lines passing through (2,−3)(2, -3) with the slopes found in the previous step.\newlineFor the slope 32\frac{3}{2}: y+3=(32)(x−2)y + 3 = \left(\frac{3}{2}\right)(x - 2)\newlineFor the slope −1-1: y+3=(−1)(x−2)y + 3 = (-1)(x - 2)
  4. Simplify equations: Simplify the equations of the lines.\newlineFor the slope 32\frac{3}{2}: y=(32)x−3−3y = \left(\frac{3}{2}\right)x - 3 - 3\newlineFor the slope −1-1: y=−x+2−3y = -x + 2 - 3
  5. Correct and simplify equations: Correct the mistake in the previous step and simplify the equations of the lines again.\newlineFor the slope 32\frac{3}{2}: y=(32)x−3−3y = \left(\frac{3}{2}\right)x - 3 - 3 becomes y=(32)x−6y = \left(\frac{3}{2}\right)x - 6\newlineFor the slope −1-1: y=−x+2−3y = -x + 2 - 3 becomes y=−x−1y = -x - 1

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