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Worked Examples · Example 12

Q.Show that the lines 3x+2y=53x + 2y = 5 and 2x−3y=72x - 3y = 7 are perpendicular to each other.

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Read each slope from the general form using m=−abm = -\dfrac{a}{b}.

For 3x+2y=53x + 2y = 5, i.e. 3x+2y−5=03x + 2y - 5 = 0: m1=−32m_1 = -\dfrac{3}{2}.

For 2x−3y=72x - 3y = 7, i.e. 2x−3y−7=02x - 3y - 7 = 0: m2=−2−3=23m_2 = -\dfrac{2}{-3} = \dfrac{2}{3}.

Now test the perpendicularity condition:

m1⋅m2=(−32)(23)=−66=−1m_1 \cdot m_2 = \left(-\dfrac{3}{2}\right)\left(\dfrac{2}{3}\right) = -\dfrac{6}{6} = -1

Since the product of the slopes is −1-1, the lines are perpendicular. …

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