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Exercise C · Q3

Q.For what value of "kk" the points (k,7)(k,7), (−4,5)(-4,5) and (1,−5)(1,-5) are collinear.

Andaman Nicobar CbseNCERTSubjective· 3mImportance★★★★★est
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✓ Free question

Setting the area of the triangle formed by the three points to zero gives k=−5k=-5.

Three points are collinear iff the triangle they form has zero area:

x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0.x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0.

  1. Take (x1,y1)=(k,7), (x2,y2)=(−4,5), (x3,y3)=(1,−5)(x_1,y_1)=(k,7),\ (x_2,y_2)=(-4,5),\ (x_3,y_3)=(1,-5).
  2. Apply the collinearity condition:

k(5−(−5))+(−4)(−5−7)+1(7−5)=0k(5-(-5))+(-4)(-5-7)+1(7-5)=0

  1. Simplify: k(10)+(−4)(−12)+1(2)=0 ⇒ 10k+48+2=0k(10)+(-4)(-12)+1(2)=0\ \Rightarrow\ 10k+48+2=0.
  2. Solve: 10k+50=0 ⇒ 10k=−50 ⇒ k=−510k+50=0\ \Rightarrow\ 10k=-50\ \Rightarrow\ k=-5.
✓Final answer

k=−5k=-5.

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