Question

find two possible locations for the ordered pair with a missing coordinate. 5) (-7, -4) and (6, y); distance = 13 units 6) (x, 2) and (-5, 5); distance = 5 units

Explanation:

5)

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, $(x_1,y_1)=(-7,-4)$ and $(x_2,y_2)=(6,y)$, and $d = 13$.
Substitute the values into the formula: $13=\sqrt{(6-(-7))^2+(y - (-4))^2}$.

Step2: Simplify the equation

First, simplify $(6 - (-7))^2=(6 + 7)^2=169$. The equation becomes $13=\sqrt{169+(y + 4)^2}$. Square both sides: $13^2=169+(y + 4)^2$. So, $169=169+(y + 4)^2$. Then $(y + 4)^2=0$. Solving for $y$, we get $y+4 = 0$, so $y=-4$. There is only one solution in this case.

6)

Step1: Apply distance formula

Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, with $(x_1,y_1)=(x,2)$ and $(x_2,y_2)=(-5,5)$ and $d = 5$. We have $5=\sqrt{(-5 - x)^2+(5 - 2)^2}$.

Step2: Simplify the equation

First, $(5 - 2)^2=9$. The equation is $5=\sqrt{(-5 - x)^2+9}$. Square both sides: $25=(-5 - x)^2+9$. Then $(-5 - x)^2=25 - 9=16$.

Step3: Solve for $x$

Taking the square - root of both sides: $-5 - x=\pm4$.
Case 1: When $-5 - x = 4$, then $x=-9$.
Case 2: When $-5 - x=-4$, then $x=-1$.

Answer:

1. $(6,-4)$
2. $(-9,2)$ and $(-1,2)$