Question

3 write the equation of the line that is parallel to y = -\frac{1}{2}x - 3 and passes through point (5, 2).
4 write the equation of the line that is perpendicular to y = 5x + 1 that passes through the point (3, -1).
5 consider quadrilateral jklm.
a determine the length of each side of quadrilateral jklm.
b determine the slope of each line segment of quadrilateral jklm.
c determine whether you can best describe quadrilateral jklm as a trapezoid, a rhombus, a rectangle, or a square. explain your reasoning.

Explanation:

3.

Step1: Find the slope of the parallel line

Parallel lines have equal slopes. The slope of the line $y =-\frac{1}{2}x - 3$ is $m =-\frac{1}{2}$.

Step2: Use the point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(5,2)$ and $m =-\frac{1}{2}$. Substitute these values: $y - 2=-\frac{1}{2}(x - 5)$.

Step3: Convert to slope - intercept form

Expand the right - hand side: $y-2=-\frac{1}{2}x+\frac{5}{2}$. Then add 2 to both sides: $y=-\frac{1}{2}x+\frac{5}{2}+2=-\frac{1}{2}x+\frac{5 + 4}{2}=-\frac{1}{2}x+\frac{9}{2}$.

Step1: Find the slope of the perpendicular line

The slope of the line $y = 5x+1$ is $m_1 = 5$. For a line perpendicular to it, the slope $m_2$ satisfies $m_1m_2=-1$. So $5m_2=-1$, then $m_2=-\frac{1}{5}$.

Step2: Use the point - slope form

The point - slope form with $(x_1,y_1)=(3,-1)$ and $m =-\frac{1}{5}$ is $y-(-1)=-\frac{1}{5}(x - 3)$.

Step3: Convert to slope - intercept form

Simplify the left - hand side: $y + 1=-\frac{1}{5}(x - 3)$. Expand the right - hand side: $y+1=-\frac{1}{5}x+\frac{3}{5}$. Subtract 1 from both sides: $y=-\frac{1}{5}x+\frac{3}{5}-1=-\frac{1}{5}x+\frac{3 - 5}{5}=-\frac{1}{5}x-\frac{2}{5}$.

Step1: Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ for side $JK$

$x_1=-8,y_1 = 8,x_2=-4,y_2 = 4$. Then $d_{JK}=\sqrt{(-4+8)^2+(4 - 8)^2}=\sqrt{4^2+(-4)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}$.

Step2: For side $KL$

$x_1=-4,y_1 = 4,x_2=-8,y_2 = 0$. Then $d_{KL}=\sqrt{(-8 + 4)^2+(0 - 4)^2}=\sqrt{(-4)^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$.

Step3: For side $LM$

$x_1=-8,y_1 = 0,x_2=-12,y_2 = 4$. Then $d_{LM}=\sqrt{(-12 + 8)^2+(4 - 0)^2}=\sqrt{(-4)^2+4^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}$.

Step4: For side $MJ$

$x_1=-12,y_1 = 4,x_2=-8,y_2 = 8$. Then $d_{MJ}=\sqrt{(-8 + 12)^2+(8 - 4)^2}=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$.

Answer:

$y =-\frac{1}{2}x+\frac{9}{2}$

4.