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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

Guest Nov 9, 2017
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#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.