Question

The coordinates of the vertices of triangleABC are A (4,1) B (-3,2) C (0,K) given that area of triangle ABC is 12 square units,find the value of 'k'.

Answer 1

however I do I get k as -13/7
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Answer 2

Answer:

The value of k is $\frac{-13}{7}$.

Step-by-step explanation:

The vertices of triangle ABC are A (4,1) B (-3,2) C (0,K).

The area of triangle ABC is 12 square units.

The area of triangle is

$A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

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

$12=\frac{1}{2}[8-4k-3k+3]$

$24=11-7k$

$7k=11-24$

$7k=-13$

$k=\frac{-13}{7}$

Therefore the value of k is $\frac{-13}{7}$.