Class 10

Math

All topics

Coordinate Geometry

Find the value of y for which the points $A(−3,9),B(2,y)$ and $C(4,−5)$ are collinear.

Points are $A(−3,9),B(2,y)$ and $C(4,−5)$ are collinear.

Which means area of triangle ABC$=0$

Area of a triangle $=21 [x_{1}(y_{2}−y_{3})+x_{2}(y_{3}−y_{1})+x_{3}(y_{1}−y_{2})]$

$=21 [−3(y+5)+2(−5−9)+4(9−y)]$

$=21 [−3y−15+2×(−14)+36−4y]$

$=21 [−7y−15−28+36]$

$=21 [−7y−7]$

Since points are collinear:

$21 (−7y−7)=0$

or $y=−1$.