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.

Answer 1

Here is your answer hope so it is helpful for you

Answer 2

Answer:

The value of k is 3.

Step-by-step explanation:

Since, the area of a triangle having vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$,

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

$=\frac{1}{2}|(k+1)(-3+k)+4(-k-1)+7(1+3)|$

$=\frac{1}{2}|-3k+k^2-3+k-4k-4+28|$

$=\frac{1}{2}|k^2-6k+21|$

According to the question,

A = 6 unit²,

$\implies \frac{1}{2}|k^2-6k+21| = 6$

$k^2-6k+21=12$

$k^2-6k+9=0$

$(k-3)^2=0$

$\implies k =3$

Hence, the value of k is 3.