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 
+895 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.
