Question

find value of k which point (7;2)(5;1)(3;k) colliner
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Answer 1

According to the Condition for Collinearity of three vertices,

x1(y2-y3)+x2(y3-y1)+x3(y1-y2)=0

7(1-k)+5(k-2)+3(2-1)=0

7-7k+5k-10+3=0

-2k=0

k=0

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Answer 2

Correct Question:

Find value of k which point (7,2)(5,1)(3,k) collinear.

Your Answer:

Points to know before solving this Question.

The area of Triangle on the Straight Line is 0.

So, the Area formed by the coordinates (7,2),(5,1) and (3,k) will be equal to zero.

Area of Triangle

$\tt = \dfrac{1}{2} | x_{1}(y_{2}-y_{3})+ x_{2}(y_{3}-y_{1})+ x_{3}(y_{1}-y_{2})|$

So,

0 = Area of triangle

$\tt Let \: x_{1}=7,x_{2}=5,x_{3}=3\\\\ \tt y_{1}=2,y_{2}=1,y_{3}=k \\\\ \tt So, \\\\ \tt \Rightarrow0=7(1-k)+5(k-2)+3(2-1)\\\\\tt \Rightarrow 0=7-7k+5k-10+3\\\\\tt \Rightarrow 0 = -2k \\\\ \tt \Rightarrow k = 0$

So, value of k = 0