Question

practice: find the possible values for the missing coordinate given: d = √5 d = √((x2 - x1)^2+(y2 - y1)^2) (-3,-6) & (-2,y) (x1,y1) (x2,y2)

Explanation:

Step1: Substitute given values into distance formula

Let $(x_1,y_1)=(-3,-6)$ and $(x_2,y_2)=(-2,y)$, and $d = \sqrt{5}$. Then $\sqrt{5}=\sqrt{(-2 - (-3))^{2}+(y - (-6))^{2}}$.

Step2: Simplify the equation

First, simplify $(-2 - (-3))^{2}$: $(-2+3)^{2}=1^{2} = 1$. The equation becomes $\sqrt{5}=\sqrt{1+(y + 6)^{2}}$. Square both sides to get $5=1+(y + 6)^{2}$.

Step3: Solve for $(y + 6)^{2}$

Subtract 1 from both sides: $(y + 6)^{2}=5 - 1=4$.

Step4: Solve for y

Take the square - root of both sides: $y+6=\pm2$.

• Case 1: When $y + 6 = 2$, then $y=2-6=-4$.
• Case 2: When $y + 6=-2$, then $y=-2 - 6=-8$.

Answer:

$y=-4$ or $y=-8$