Question

For what value of k are the points A(k, 2 – 2k) ,B(–k + 1, 2k) and C(–4 – k, 6 – 2k) are collinear ? [4 MARKS]

Answer 1

Formula: 1 Mark
Calculations: 2 Marks
Answer: 1 Mark

For points A,B,C to be collinear, area of triangle ABC formed by these points should be zero.

$\Rightarrow$ The area of $\Delta ABC=0$

$\Rightarrow \frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$

$\Rightarrow \frac{1}{2}\left[k\right(2k-6+2k)+(–k+1\left)\right(6–2k-2+2k)+(–4–k\left)\right(2–2k-2k\left)\right]=0$

$\Rightarrow \frac{1}{2}\left[k\right(4k-6)+(–k+1\left)\right(4)+(–4–k\left)\right(2–4k\left)\right]=0$

$\Rightarrow \frac{1}{2}[4k^2-6k-4k+4-8+16k-2k+4k^2]=0$

$\Rightarrow 8k^2+4k-4=0$

$\Rightarrow 8k^2+4k-4=0$

$\Rightarrow 2k^2+k-1=0$

$\Rightarrow (k+1)(2k-1)=0$

$\Rightarrow k=-1ork=\frac{1}{2}$