Question

Question 19
Find the values of k, if the points A(k+1, 2k), B(3k,2k+3) and C(5k-1,5k) are collinear.

Answer 1

We know that, if three points are collinear, then the area of triangle formed by these points is zero.
Since, the points A(k+1,2k), B(3k,2k+3) and C(5k-1,5k), are collinear.
Then, area of $\Delta$ ABC=0.
$\Rightarrow \frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0Here,x_1=k+1,x_2=3k,x_3=5k-1and{y}_1-2k,y_2-2k+3,y_3=5k\Rightarrow \frac{1}{2}\left[\right(k+1\left)\right(2k+3-5k)+3k(5k-2k)+(5k-1\left)\right(2k-(2k+3)\left)\right]=0\Rightarrow \frac{1}{2}\left[\right(k+1\left)\right(3-3k)+3k(3k)+(5k-1\left)\right(2k-2k-3\left)\right]=0\Rightarrow \frac{1}{2}[-3k^2+3k-3k+3+9k^2-15k+3]=0\Rightarrow \frac{1}{2}(6k^2-15k+6)=0\left[multiplyby2\right]\Rightarrow 6k^2-15k+6=0\left[byfactorizationmethod\right]\Rightarrow 2k^2-5k+2=0\Rightarrow 2k^2-4k-k+2=0\Rightarrow 2k(k-2)-1(k-2)=0\Rightarrow (k-2)(2k-1)=0$
If k – 2 = 0, then k = 2
If 2k – 1, then k = $\frac{1}{2}$
$\therefore k=2,\frac{1}{2}$
Hence, the required values of k are 2 and $\frac{1}{2}$.