Find the value of k so that the area of the triangle with vertices A
Question:

Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, −3) and C(7, −k) is 6 square units.

 

Solution:

Let $A\left(x_{1}, y_{1}\right)=A(k+1,1), B\left(x_{2}, y_{2}\right)=B(4,-3)$ and $C\left(x_{3}, y_{3}\right)=C(7,-k)$. Now

Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\Rightarrow 6=\frac{1}{2}[(k+1)(-3+k)+4(-k-1)+7(1+3)]$

$\Rightarrow 6=\frac{1}{2}\left[k^{2}-2 k-3-4 k-4+28\right]$

$\Rightarrow k^{2}-6 k+9=0$

$\Rightarrow(k-3)^{2}=0 \Rightarrow k=3$

Hence, k = 3.

 

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