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1)

Let line be the graph of . Line is perpendicular to line and passes through the point . If line is the graph of the equation , then find .

2)

The perpendicular bisector of the line segment is the line that passes through the midpoint of and is perpendicular to .
Find the equation of the perpendicular bisector of the line segment joining the points and Enter your answer in the form " ."

3)

The lines and are perpendicular. Find 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.

Nov 9, 2017