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Worked Examples · Example 26

Q.For what value of kk, points P(3,−2)P(3,-2), Q(8,8)Q(8,8) and R(k,2)R(k,2) are collinear.

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Three points are collinear when the triangle they form has zero area; setting that determinant to zero gives a linear equation for kk.

Points collinear  ⟺  12∣ x1(y2−y3)+x2(y3−y1)+x3(y1−y2) ∣=0.\text{Points collinear} \iff \frac{1}{2}\left|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\right| = 0.

With P(x1,y1)=(3,−2)P(x_1,y_1)=(3,-2), Q(x2,y2)=(8,8)Q(x_2,y_2)=(8,8), R(x3,y3)=(k,2)R(x_3,y_3)=(k,2).

  1. Set the area to zero (drop the 12\tfrac12): 3(8−2)+8(2−(−2))+k(−2−8)=0.3(8 - 2) + 8(2 - (-2)) + k(-2 - 8) = 0. …

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