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1)

Let line $l_1$ be the graph of $5x + 8y = -9$. Line $l_2$ is perpendicular to line $l_1$ and passes through the point $(10,10)$. If line $l_2$ is the graph of the equation $y=mx +b$, then find $m+b$.

 

2)

The perpendicular bisector of the line segment $\overline{AB}$ is the line that passes through the midpoint of $\overline{AB}$ and is perpendicular to $\overline{AB}$.
Find the equation of the perpendicular bisector of the line segment joining the points $(1,2)$ and $(7,4).$ Enter your answer in the form "$y = mx + b$."

 

3)

The lines $3y + 2x = 7$ and $y = mx - 11$ are perpendicular. Find $m.$

Guest Nov 9, 2017
 #1
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1) To find a perpendicular line, the slope is the negative reciprocal.

Get the equation into y=mx+b form.

5x+8y=-9

Subtract 5x from both sides.

\(8y=-5x-9\)

Divide both sides by 8.

\(y=-\frac{5}{8}x-\frac{9}{8}\)

Take the negative reciprocal of m and use it with (10,10) in point-slope form, y-y1=m(x-x1).

\(y-10=\frac{8}{5}(x-10)\)

Multiply the 8/5 through and add 10 to both sides.

\(y=\frac{8}{5}x-16+10\)

\(y=\frac{8}{5}x-6\)

Then, m+b would be 8/5-6. This is 1.6-6, or -5.6.

 

2) The slope of a line is \(\frac{y_2-y_1}{x_2-x_1}\).

Plug in the points (1,2) and (7,4)

\(\frac{4-2}{7-1}=\frac{2}{6}=\frac{1}{3}\)

So the slope is 1/3. Now use point-slope form.

\(y-2=\frac{1}{3}(x-1)\)

Distribute 1/3 and add 2 to both sides.

\(y={1\over3}x-\frac{1}{3}+2\)

Combine.

\(y=\frac{1}{3}x-\frac{1}{3}+\frac{6}{3}\)

\(y=\frac{1}{3}x+\frac{5}{3}\)

 

3) Put 3y+2x=7 into y=mx+b form.

\(3y=-2x+7\)

Divide by 3.

\(y=-\frac{2}{3}+\frac{7}{3}\)

The slope of a line perpendicular to another is the negative reciprocal of the first slope.

\(m=\frac{3}{2}\)

So 3/2 is the answer to #3.

AdamTaurus  Nov 9, 2017

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