Find the value of $k$ so that the area of triangle $ABC$ with $A(k+1, 1), B(4, -3)$ and $C(7, -k)$ is 6 square units.
Given:
The area of triangle $ABC$ with $A(k+1, 1), B(4, -3)$ and $C(7, -k)$ is 6 square units.
To do:
We have to find the value of $k$.
Solution:
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[(k+1)(-3+k)+4(-k-1)+7(1+3)] \)
\( 6=\frac{1}{2}[-3k+k^2-3+k-4k-4+7(4)] \)
\( 6(2)=(k^2-6k+28-7) \)
\( 12=k^2-6k+21 \)
\( k^2-6k+21-12=0 \)
\( k^2-6k+9=0 \)
\( k^2-3k-3k+9=0 \)
\( k(k-3)-3(k-3)=0 \)
\( (k-3)(k-3)=0 \)
\( (k-3)^2=0 \)
\( \Rightarrow k-3=0 \)
\( k=3 \)
The value of $k$ is $3$.
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