Question

1. s(-4, 6) and t(-7, -3): find rt.
2. b is the mid - point of ac and e is the mid - point of bd. if a(-9, -4), c(-1, 6), and e(-4, -3), find the coordinates of d.

directions: suppose q is the mid - point of pr. use the information to find the missing value.

1. pq = 3x + 14 and qr = 7x - 10; find x.
2. pq = 2x + 1 and qr = 5x - 44; find pq.
3. pq = 6x + 25 and qr = 16 - 3x; find pr.
4. pr = 9x - 31 and qr = 43; find x.

Explanation:

11.

Step1: Use 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, $x_1=-4,y_1 = 6,x_2=-7,y_2=-3$.
$ST=\sqrt{(-7+4)^2+(-3 - 6)^2}$

Step2: Simplify the expression

$ST=\sqrt{(-3)^2+(-9)^2}=\sqrt{9 + 81}=\sqrt{90}=3\sqrt{10}$

Step1: Use mid - point formula for $B$

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. If $A(-9,-4)$ and $C(-1,6)$, then the coordinates of $B$ (mid - point of $AC$) are $(\frac{-9-1}{2},\frac{-4 + 6}{2})=(-5,1)$.

Step2: Use mid - point formula for $D$

Let the coordinates of $D$ be $(x,y)$. Since $E(-4,-3)$ is the mid - point of $BD$ and $B(-5,1)$, we have $\frac{-5+x}{2}=-4$ and $\frac{1 + y}{2}=-3$.
From $\frac{-5+x}{2}=-4$, we get $-5+x=-8$, so $x=-3$. From $\frac{1 + y}{2}=-3$, we get $1 + y=-6$, so $y=-7$.

Step1: Set up the equation

Since $Q$ is the mid - point of $PR$, $PQ=QR$. So, $3x + 14=7x-10$.

Step2: Solve for $x$

Subtract $3x$ from both sides: $14=4x-10$. Add 10 to both sides: $24 = 4x$. Divide both sides by 4: $x = 6$.

Answer:

$3\sqrt{10}$

12.