Question

Find the value(s) of $k$ so that $PQ$ will be parallel to $RS$, given:

(iii) $P(5,-1),Q(6,11),R(6,-4k)\text{and}S(7,k^2)$

Answer 1

Given,
$P(5,-1),Q(6,11),R(6,-4k)\text{and}S(7,k^2)$

We know that, slope of line $=\frac{y_2-y_1}{x_2-x_1}$

$\therefore \text{Slope of PQ}=\frac{11+1}{6-5}=12$

And, $\text{Slope of RS}=\frac{k^2+4k}{7-6}=k^2+4k$

As lines PQ and RS are parallel.

So, slopes are equal to each other.

$\therefore$ Slope of PQ = Slope of RS

$\Rightarrow 12=k^2+4k$

$\Rightarrow k^2+4k-12=0$

$\Rightarrow k^2+6k-2k-12=0$

$\Rightarrow k(k+6)-2(k+6)=0$

$\Rightarrow (k+6)(k-2)=0$

$\Rightarrow (k+6)=0\text{or}(k-2)=0$

$\therefore k=-6\text{or}2$

Hence, the value of $k$ is $-6$ or $2$.