Class 10

Math

All topics

Coordinate Geometry

For what value of $k(k>0)$ is the area of the triangle with vertices $(−2,5),(k,−4)$ and $(2k+1,10)$ equal to $53$ square units?

Area of triangle $=53$ square units

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

$53=21 [−2(−4−10)+k(10−5)+(2k+1)(5+4)]$

$53=21 [−2×(−14)+k×5+(2k+1)×9]$

$106=23k+37$

$k=2369 =3$

The value of k is $3$.