Question

For what value of $k$ the point $(k,2-2k)$, $(1-k,2k)$ and $(-4-k,6-2k)$ are collinear

• A. $1,-1/2$
• B. $1,1/2$
• C. $-1,1/2$
• D. $-1,-1/2$

Answer 1

The correct option is A $1,-1/2$

Condition for 3 points to be collinear is

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

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

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

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

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

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

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

$-8k(k+1)+4(k+1)=0$

$(k+1)(4-8k)=0$

$k+1=0=-1$

$4-8k=0$

$8k=4$

$k=\frac{1}{2}$

So, the value of k is $-1$ and $\frac{1}{2}$