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Example · Example 3

Q.Find the value of kk so that the line through (k,3)(k, 3) and (2,−4)(2, -4) is perpendicular to the line through (5,1)(5, 1) and (−1,7)(-1, 7).

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Slope of the line through (5,1)(5,1) and (−1,7)(-1,7): m2=7−1−1−5=6−6=−1m_2=\dfrac{7-1}{-1-5}=\dfrac{6}{-6}=-1. For the line through (k,3)(k,3) and (2,−4)(2,-4) to be perpendicular to this, its slope m1m_1 must satisfy m1m2=−1m_1m_2=-1, i.e. m1(−1)=−1m_1(-1)=-1, i.e. m1=1m_1=1. But also m1=−4−32−k=−72−km_1=\dfrac{-4-3}{2-k}=\dfrac{-7}{2-k}. Setting these equal: −72−k=1  ⟹  −7=2−k  ⟹  k=9\dfrac{-7}{2-k}=1 \implies -7 = 2-k \implies k=9. Check: with k=9k=9, m1=−72−9=−7−7=1m_1=\dfrac{-7}{2-9}=\dfrac{-7}{-7}=1, and m1m2=1×(−1)=−1m_1m_2=1\times(-1)=-1 ✓, confirming perpendicularity. [!ANSWER] k=9k=9.

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