Question

If the distance between the points $(5,-2)$, $(1,a)$ is $5$ units, find the value of a.

Answer 1

$\Rightarrow$ The distance between two points $P(x_1,y^1)$ and $Q(x_2,y_2)$ is given by

$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

here $P(x_1,y_2)=P(5,-2)$

& $Q(x_2,y_2)=Q(1,a)$

& $PQ=5$ units

i.e. $5=\sqrt{(1-5{)}^2+(a-(-2){)}^2}$

Taking square on both side we get,

$\therefore 25=(-4{)}^2+(a+2{)}^2$

$\therefore 25-16=(a+2{)}^2$

$\therefore (a+2{)}^2=9$

$\therefore a^2+4a+4=9$

$\therefore a^2+4a-5=0$

$\therefore a^{22}+5a-a-5=0$

$\therefore a(a+5)-1(a+5)=0$

$\therefore a(a+5)-1(a+5)=0$

$\therefore a-1=0$ or $a+5=0$

$a=1$ or $a=-5$

$\to$ The values of a will be $1$ or $-5$