Question

Find the value of $k$, if the point $P(-2,2)$ lies on the locus $x^2-7x+Ky=0$. If the point $Q(3,a)$ also lies on the locus. Find the values of ${}^{\prime }{a}^{\prime }$.

Answer 1

Given that,

the point $P(-2,2)$ lies on the locus

$x^2-7x+ky=0$

then put $P(-2,2)$ and we get,

$x^2-7x+ky=0$

$(-2{)}^2-7(-2)+k(2)=0$

$\Rightarrow 4+14+2k=0$

$\Rightarrow 18+2k=0$

$\Rightarrow 2k=-18$

$\Rightarrow k=-9$

again point $Q(3,a)$ lies on the locus

the, put $Q(3,a)$ and $k=-9$. We get

$x^2-7x+ky=0$

$(3{)}^2-7(3)+(-9)\times a=0$

$9-21-9a=0$

$-9a=+12$

$a=\frac{-12}{9}$

$a=\frac{-4}{3}$

Hence, this is the answer.