Question

3.1 lines & angles: to identify relationships between figures in space; to identify angles formed by two lines & a transversal

1. find xy when x(-7, 10) and y(3, 4).
2. find ab when a(-4, -1) and b(-2, 11).
3. find the distance between the points on the graph.

1.7 midpoint & distance

1. find the coordinates of the mid - point of ef if e(-2, -8) and f(-12, 6).
2. find the coordinates of the mid - point between the two points on the graph.
3. find the coordinates of j if k(-5, 8) is the mid - point of jl and l has coordinates (-4, -1).

Explanation:

3.1 Lines & Angles - Distance Formula

1. Find $XY$ when $X(-7,10)$ and $Y(3,4)$

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, $x_1=-7,y_1 = 10,x_2=3,y_2 = 4$.

Step2: Substitute values

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

Step1: Recall distance formula

Use $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, with $x_1=-4,y_1=-1,x_2=-2,y_2 = 11$.

Step2: Substitute values

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

Step1: Recall mid - point formula

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1=-2,y_1=-8,x_2=-12,y_2 = 6$.

Step2: Substitute values

$M=(\frac{-2+( - 12)}{2},\frac{-8 + 6}{2})=(\frac{-2-12}{2},\frac{-2}{2})=(-7,-1)$

Answer:

$2\sqrt{34}$

2. Find $AB$ when $A(-4,-1)$ and $B(-2,11)$