Question

distance formula

1. find xy when x(-7, 10) and y(3, 4).

d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}
d = \sqrt{(3 - - 7)^2+(4 - 10)^2}
10^2+ - 6^2=100 + 36 = \sqrt{136}

1. find ab when a(-4, -1) and b(-2, 11).

d = \sqrt{(-2 - - 4)^2+(11 - - 1)^2}
+10^2+12^2=4 + 144 = \sqrt{148}

1. find the distance between the points on the graph.

(3, 11)
x_1 y_1
(-2, 3)
x_2 y_2
d = \sqrt{(-2 - 3)^2+(3 - 11)}

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Solve for $XY$

Given $X(-7,10)$ and $Y(3,4)$, substitute $x_1=-7,y_1 = 10,x_2 = 3,y_2=4$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(3-(-7))^2+(4 - 10)^2}\\ &=\sqrt{(3 + 7)^2+(4-10)^2}\\ &=\sqrt{10^2+(-6)^2}\\ &=\sqrt{100 + 36}\\ &=\sqrt{136}\\ &=2\sqrt{34} \end{align*}$$

\]

Step3: Solve for $AB$

Given $A(-4,-1)$ and $B(-2,11)$, substitute $x_1=-4,y_1=-1,x_2=-2,y_2 = 11$ into the formula:
\[

$$\begin{align*} d&=\sqrt{(-2-(-4))^2+(11-(-1))^2}\\ &=\sqrt{(-2 + 4)^2+(11 + 1)^2}\\ &=\sqrt{2^2+12^2}\\ &=\sqrt{4+144}\\ &=\sqrt{148}\\ &=2\sqrt{37} \end{align*}$$

\]

Step4: Solve for distance between points $(3,11)$ and $(-2,3)$

Substitute $x_1 = 3,y_1=11,x_2=-2,y_2 = 3$ into the formula:
\[

$$\begin{align*} D&=\sqrt{(-2 - 3)^2+(3-11)^2}\\ &=\sqrt{(-5)^2+(-8)^2}\\ &=\sqrt{25 + 64}\\ &=\sqrt{89} \end{align*}$$

\]

Answer:

1. $XY = 2\sqrt{34}$
2. $AB=2\sqrt{37}$
3. Distance between points is $\sqrt{89}$