Question

1. p(-4, 12), q(1, -3), r(-9, 3), s(-5, 4)
2. a(-3, 14), o(2, -1), r(4, 8), s(-2, -10)
3. a(2, -1), o(-3, -1), r(-11, 9), s(-7, 9)
4. given c(x, 16), d(2, -4), e(-6, 14), and f(-2, 4), find the value of x so that $overline{cd}paralleloverline{ef}$.
5. given j(x, -8) and k(-1, -5) and the graph of line l below, find the value of x so that $overline{jk}parallel l$.
6. given p(12, -2), q(5, -10), r(-4, 10), and s(4, y), find the value of y so that $overline{pq}perpoverline{rs}$.
7. given a(4, 2) and b(-1, y) and the graph of line t below, find the value of y so that $overline{ab}perp t$.

Explanation:

Step1: Recall slope - formula

The slope formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Also, if two lines are parallel, their slopes are equal (\(m_1=m_2\)), and if two lines are perpendicular, the product of their slopes is \(- 1\) (\(m_1\times m_2=-1\)).

Step2: Solve problem 17

For points \(C(x,16)\), \(D(2,-4)\), the slope of \(\overline{CD}\) is \(m_{CD}=\frac{-4 - 16}{2 - x}=\frac{-20}{2 - x}\). For points \(E(-6,14)\) and \(F(-2,4)\), the slope of \(\overline{EF}\) is \(m_{EF}=\frac{4 - 14}{-2+6}=\frac{-10}{4}=-\frac{5}{2}\). Since \(\overline{CD}\parallel\overline{EF}\), we set \(m_{CD}=m_{EF}\), so \(\frac{-20}{2 - x}=-\frac{5}{2}\). Cross - multiply: \(-20\times2=-5\times(2 - x)\), which simplifies to \(-40=-10 + 5x\). Add 10 to both sides: \(5x=-30\), and \(x=-6\).

Step3: Solve problem 18

First, find the slope of line \(l\) from the graph. Using two points on line \(l\) (e.g., \((0, - 2)\) and \((2,0)\)), \(m_l=\frac{0 + 2}{2-0}=1\). For points \(J(x,-8)\) and \(K(-1,-5)\), the slope of \(\overline{JK}\) is \(m_{JK}=\frac{-5 + 8}{-1 - x}=\frac{3}{-1 - x}\). Since \(\overline{JK}\parallel l\), we set \(m_{JK}=m_l\), so \(\frac{3}{-1 - x}=1\). Cross - multiply: \(3=-1 - x\), and \(x=-4\).

Step4: Solve problem 19

For points \(P(12,-2)\) and \(Q(5,-10)\), the slope of \(\overline{PQ}\) is \(m_{PQ}=\frac{-10 + 2}{5 - 12}=\frac{-8}{-7}=\frac{8}{7}\). For points \(R(-4,10)\) and \(S(4,y)\), the slope of \(\overline{RS}\) is \(m_{RS}=\frac{y - 10}{4 + 4}=\frac{y - 10}{8}\). Since \(\overline{PQ}\perp\overline{RS}\), \(m_{PQ}\times m_{RS}=-1\). So \(\frac{8}{7}\times\frac{y - 10}{8}=-1\). Simplify the left - hand side to \(\frac{y - 10}{7}=-1\). Multiply both sides by 7: \(y-10=-7\), and \(y = 3\).

Step5: Solve problem 20

First, find the slope of line \(t\) from the graph. Using two points on line \(t\) (e.g., \((-2,-2)\) and \((0,1)\)), \(m_t=\frac{1 + 2}{0+2}=\frac{3}{2}\). For points \(A(4,2)\) and \(B(-1,y)\), the slope of \(\overline{AB}\) is \(m_{AB}=\frac{y - 2}{-1 - 4}=\frac{y - 2}{-5}\). Since \(\overline{AB}\perp t\), \(m_{AB}\times m_t=-1\). So \(\frac{y - 2}{-5}\times\frac{3}{2}=-1\). Cross - multiply: \(3(y - 2)=10\), expand to \(3y-6 = 10\), add 6 to both sides: \(3y=16\), and \(y=\frac{16}{3}\).

Answer:

1. \(x=-6\)
2. \(x=-4\)
3. \(y = 3\)
4. \(y=\frac{16}{3}\)