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Worked Examples · Example 5

Q.Find the equation of the line whose y-intercept is −3-3 and which is perpendicular to the line 3x−2y+5=03x - 2y + 5 = 0.

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Find the slope of the given line, take the negative reciprocal for the perpendicular slope, then use slope–intercept form with the given yy-intercept.

Slope–intercept form: y=mx+cy=mx+c. If two lines are perpendicular, m1m2=−1m_1 m_2 = -1.

  1. Find the slope of the given line 3x−2y+5=03x-2y+5=0. Rearrange to y=mx+cy=mx+c form:

2y=3x+5   ⟹   y=32x+522y = 3x+5 \ \implies\ y = \frac{3}{2}x + \frac{5}{2}

So its slope is m1=32m_1 = \dfrac{3}{2}.

  1. Find the perpendicular slope (negative reciprocal of m1m_1):

m2=−1m1=−13/2=−23m_2 = -\frac{1}{m_1} = -\frac{1}{3/2} = -\frac{2}{3}

  1. Use the given yy-intercept. The required line has yy-intercept c=−3c=-3, i.e. it passes through (0,−3)(0,-3).

  2. Write the line in slope–intercept form.

    y=−23x−3y = -\frac{2}{3}x - 3 …

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