Question

the distance between the points (x, 25) and (-15, 11) is 7√29. use the distance formula to find the two possible values of x. -203 203 50 -20 -50 20

Explanation:

Step1: Recall distance formula

The distance formula 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, $(x_1,y_1)=(x,25)$ and $(x_2,y_2)=(-15,11)$, and $d = 7\sqrt{29}$.

Step2: Substitute values into formula

$7\sqrt{29}=\sqrt{(- 15 - x)^2+(11 - 25)^2}$. First, simplify $(11 - 25)^2=(-14)^2 = 196$. So we have $7\sqrt{29}=\sqrt{(-15 - x)^2+196}$.

Step3: Square both sides

$(7\sqrt{29})^2=(-15 - x)^2+196$. Calculate $(7\sqrt{29})^2=7^2\times29=49\times29 = 1421$. Then $1421=(-15 - x)^2+196$.

Step4: Isolate the squared - term

$(-15 - x)^2=1421 - 196=1225$.

Step5: Take square root of both sides

$-15 - x=\pm\sqrt{1225}=\pm35$.

Step6: Solve for x

Case 1: When $-15 - x = 35$, then $x=-15 - 35=-50$.
Case 2: When $-15 - x=-35$, then $x=-15 + 35 = 20$.

Answer:

$x = 20$ or $x=-50$