Question

directions: find the missing endpoint if s is the midpoint \\( \overline{rt} \\).

1. \\( r(-9, 4) \\) and \\( s(2, -1) \\); find \\( t \\).

1. \\( s(-4, -6) \\) and \\( t(-7, -3) \\); find \\( r \\).

\\( mp\left(\frac{x_2 + x_1}{2}, \frac{y_2 + y_1}{2}\
ight) \\) (handwritten)

1. \\( b \\) is the midpoint of \\( \overline{ac} \\) and \\( e \\) is the midpoint of \\( \overline{bd} \\). if \\( a(-9, -4) \\), \\( c(-1, 6) \\), and \\( e(-4, -3) \\), find the coordinates of \\( d \\).

directions: suppose \\( q \\) is the midpoint of \\( \overline{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 \\).

\\( \copyright \\) gina wilson (all things algebra®, llc), 2014-2019

Explanation:

Problem 10:

Step1: Recall midpoint formula

The midpoint formula for two points \( R(x_1, y_1) \) and \( T(x_2, y_2) \) with midpoint \( S(x_m, y_m) \) is \( x_m=\frac{x_1 + x_2}{2} \), \( y_m=\frac{y_1 + y_2}{2} \). Given \( R(-9,4) \) and \( S(2,-1) \), let \( T=(x,y) \).

Step2: Solve for \( x \)

Using \( x_m=\frac{x_1 + x_2}{2} \), substitute \( x_m = 2 \), \( x_1=-9 \):
\( 2=\frac{-9 + x}{2} \)
Multiply both sides by 2: \( 4=-9 + x \)
Add 9 to both sides: \( x = 13 \)

Step3: Solve for \( y \)

Using \( y_m=\frac{y_1 + y_2}{2} \), substitute \( y_m=-1 \), \( y_1 = 4 \):
\( -1=\frac{4 + y}{2} \)
Multiply both sides by 2: \( -2=4 + y \)
Subtract 4: \( y=-6 \)

Step1: Apply midpoint formula

Let \( R=(x,y) \), \( S(-4,-6) \), \( T(-7,-3) \). Midpoint formula: \( -4=\frac{x + (-7)}{2} \), \( -6=\frac{y + (-3)}{2} \)

Step2: Solve for \( x \)

\( -4=\frac{x - 7}{2} \)
Multiply by 2: \( -8=x - 7 \)
Add 7: \( x=-1 \)

Step3: Solve for \( y \)

\( -6=\frac{y - 3}{2} \)
Multiply by 2: \( -12=y - 3 \)
Add 3: \( y=-9 \)

Step1: Find midpoint \( B \) of \( AC \)

\( A(-9,-4) \), \( C(-1,6) \). Midpoint \( B \): \( x_B=\frac{-9 + (-1)}{2}=-5 \), \( y_B=\frac{-4 + 6}{2}=1 \), so \( B(-5,1) \)

Step2: Let \( D=(x,y) \), use midpoint \( E(-4,-3) \) of \( BD \)

Midpoint formula: \( -4=\frac{-5 + x}{2} \), \( -3=\frac{1 + y}{2} \)

Step3: Solve for \( x \)

\( -4=\frac{-5 + x}{2} \)
Multiply by 2: \( -8=-5 + x \)
Add 5: \( x=-3 \)

Step4: Solve for \( y \)

\( -3=\frac{1 + y}{2} \)
Multiply by 2: \( -6=1 + y \)
Subtract 1: \( y=-7 \)

Answer:

\( T(13, -6) \)

Problem 11: