Question

find all possible values of X for which the distance between the points A(X,-1) and B(5,3) is 5 units.

Answer 1

the answer is as . if u use distance formula then a eqn come.
a2-10a+16.which we solve.
then8,2 comes.

Answer 2

Answer:

The values of X can be 2 and 8 .

Step-by-step explanation:

$A=(x_1,y_1)=(X,-1)$

$B=(x_2,y_2)=(5,3)$

Distance formula : $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We are given that the distance between the given points is 5 units

Substitute the values in the formula

$5=\sqrt{(5-X)^2+(3-(-1))^2}$

$5=\sqrt{(5-X)^2+(3+1)^2}$

$5^2=(5-X)^2+(3+1)^2$

$25=(5-X)^2+(4)^2$

$25=25+X^2-10X+16$

$0=X^2-10X+16$

$X^2-8X-2X+16=0$

$X(X-8)-2(X-8)=0$

$(X-2)(X-8)=0$

$X=2,8$

Hence The values of X can be 2 and 8 .