Question

1. if $rs = 23 - 2x$, $st = 9x - 5$, and $rt = 39$, find $rs$.
2. given the points below, find $xy$. round to the nearest hundredth.

$x(-9,2)$ and $y(5,-4)$

1. find the mid - point of $overline{ab}$ if $a(-3,8)$ and $b(-7,-6)$.
2. if line $n$ bisects $overline{ce}$, find $cd$.

Explanation:

Step1: Solve for \(x\) in problem 9

Since \(RS + ST=RT\), we substitute the given expressions: \((23 - 2x)+(9x - 5)=39\). Combine like - terms: \(23-5+(9x - 2x)=39\), which simplifies to \(18 + 7x=39\). Then subtract 18 from both sides: \(7x=39 - 18=21\), so \(x = 3\).

Step2: Find \(RS\) in problem 9

Substitute \(x = 3\) into the expression for \(RS\): \(RS=23-2x=23-2\times3=23 - 6=17\).

Step3: Find the distance \(XY\) in problem 10

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for points \(X(-9,2)\) and \(Y(5,-4)\). Here \(x_1=-9,y_1 = 2,x_2=5,y_2=-4\). Then \(d=\sqrt{(5-(-9))^2+((-4)-2)^2}=\sqrt{(5 + 9)^2+(-6)^2}=\sqrt{14^2+(-6)^2}=\sqrt{196 + 36}=\sqrt{232}\approx15.23\) (rounded to the nearest hundredth).

Step4: Find the mid - point in problem 11

Use the mid - point formula \((\frac{x_1+x_2}{2},\frac{y_1 + y_2}{2})\) for points \(A(-3,8)\) and \(B(-7,-6)\). So \(\frac{-3+( - 7)}{2}=\frac{-3-7}{2}=\frac{-10}{2}=-5\) and \(\frac{8+( - 6)}{2}=\frac{8 - 6}{2}=1\). The mid - point is \((-5,1)\).

Step5: Solve for \(x\) in problem 12

Since line \(n\) bisects \(\overline{CE}\), then \(CD=DE\), so \(x + 6=4x-21\). Subtract \(x\) from both sides: \(6=4x-x-21\), which gives \(6 = 3x-21\). Add 21 to both sides: \(3x=6 + 21=27\), so \(x = 9\). Then \(CD=x + 6=9+6=15\).

Answer:

1. \(RS = 17\)
2. \(XY\approx15.23\)
3. \((-5,1)\)
4. \(CD = 15\)