Question

*#4.) write the equation of the perpendicular bisector of the line ab given: a(1,3) & b(-3,5)
y = 2x + 4
y = 2x + 6
y = -2x + 6
y = 2x - 6
**#5.)
find the value of \x\.
5
90
25
10
**#6.) in the figure below, \\(\overline{kn}\\) bisects \\(\angle{jkl}\\). if \\(m\angle{jkl}=(9y + 15)^{circ}\\), and \\(m\angle{jkn}=(5y + 2)^{circ}\\), find \\(m\angle{mkl}).
#7.) determine if the lines 5y - 6x = 30 and 12x - 10y = 10 are parallel, perpendicular or neither. explain how you know.
i) parallel
ii) perpendicular
iii) neither
justification:

Explanation:

Step1: Find the mid - point of line segment AB

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $A(1,3)$ and $B(-3,5)$, the mid - point $M$ is $(\frac{1+( - 3)}{2},\frac{3 + 5}{2})=(-1,4)$.

Step2: Find the slope of line AB

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$. For $A(1,3)$ and $B(-3,5)$, the slope of $AB$ is $m_{AB}=\frac{5 - 3}{-3 - 1}=\frac{2}{-4}=-\frac{1}{2}$.

Step3: Find the slope of the perpendicular bisector

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the perpendicular bisector be $m_p$. Then $m_p\times(-\frac{1}{2})=-1$, so $m_p = 2$.

Step4: Use the point - slope form to find the equation of the perpendicular bisector

The point - slope form of a line is $y - y_0=m(x - x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope. Using the point $M(-1,4)$ and $m = 2$, we have $y-4=2(x + 1)$.

Step5: Simplify the equation

Expand the right - hand side: $y-4=2x+2$. Then $y=2x + 6$.

for #5:

Step1: Set up the equation based on the angle relationship

Since the two angles $5x + 25$ and $x + 5$ are complementary (they form a right - angle, so their sum is $90^{\circ}$), we have the equation $(5x + 25)+(x + 5)=90$.

Step2: Combine like terms

$5x+x+25 + 5=90$, which simplifies to $6x+30 = 90$.

Step3: Solve for $x$

Subtract $30$ from both sides: $6x=90 - 30=60$. Then divide both sides by $6$: $x = 10$.

for #6:

Step1: Use the angle - bisector property

Since $\overrightarrow{KN}$ bisects $\angle{JKL}$, then $m\angle{JKN}=m\angle{NKL}$ and $m\angle{JKL}=2m\angle{JKN}$. So $9y + 15=2(5y + 2)$.

Step2: Expand the right - hand side

$9y + 15=10y+4$.

Step3: Solve for $y$

Subtract $9y$ from both sides: $15=y + 4$. Then subtract $4$ from both sides to get $y = 11$.

Step4: Find $m\angle{M KL}$

Since $m\angle{JKL}=9y + 15$ and $y = 11$, $m\angle{JKL}=9\times11+15=99 + 15=114^{\circ}$. And $m\angle{MKL}=\frac{1}{2}m\angle{JKL}=57^{\circ}$.

for #7:

Step1: Rewrite the equations in slope - intercept form ($y=mx + b$)

For the line $5y-6x=30$, solve for $y$: $5y=6x + 30$, so $y=\frac{6}{5}x+6$.
For the line $12x-10y=10$, solve for $y$: $-10y=-12x + 10$, then $y=\frac{6}{5}x - 1$.

Step2: Compare the slopes

The slopes of both lines are $\frac{6}{5}$. Since the slopes are equal, the lines are parallel.

Answer:

$y = 2x+6$