Question

question 4
0/1 pt 3 99 details
find two values of x so the distance between the points (x,1) and (-2,3) is 2√10. answer exactly.
x =

Explanation:

Step1: Recall distance formula

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, $(x_1,y_1)=(x,1)$ and $(x_2,y_2)=(-2,3)$ and $d = 2\sqrt{10}$. So, $2\sqrt{10}=\sqrt{(-2 - x)^2+(3 - 1)^2}$.

Step2: Square both sides

$(2\sqrt{10})^2=(-2 - x)^2+(3 - 1)^2$. Since $(2\sqrt{10})^2=4\times10 = 40$ and $(3 - 1)^2=4$, the equation becomes $40=(-2 - x)^2+4$.

Step3: Simplify the equation

Subtract 4 from both sides: $(-2 - x)^2=40 - 4=36$. Then, $(-2 - x)^2=(x + 2)^2=36$.

Step4: Solve for x

Take the square - root of both sides: $x + 2=\pm6$.
Case 1: If $x + 2 = 6$, then $x=6 - 2=4$.
Case 2: If $x + 2=-6$, then $x=-6 - 2=-8$.

Answer:

$x = 4,-8$