Maths / Coordinate Geometry / Area of Triangle Using Coordinate

QUESTION

Find the value of k if the points A(2,3), B(4,k) and C(6,-3) are colinear.

EXPLANATION
Explain TypeExplanation Content
Text

Since the given points are collinear, the area of the triangle formed by them must be 0,i.e.,

$\frac {1}{2}\left [ 2(k+3)+4(-3-3)+6(3-k) \right ]=0$

i.e.,                                          $\frac {1}{2}(-4k)=0$

Therefore,                              $k=0$

area of  $\Delta ABC=\frac {1}{2}\left [ 2(0+3)+4(-3-3)+6(3-0) \right ]=0$

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