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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 "$y = mx + b$."

 
Guest Nov 10, 2017
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3+0 Answers

 #1
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IDK how to solve this so yeah, have fun.

 
Guest Nov 10, 2017
 #2
avatar+5227 
+2

1)   First we need to find the slope of  l1 . Let's get its equation into slope-intercept form.

 

5x + 8y  =  -9          Subtract  5x  from both sides of the equation.

 

8y  =  -5x - 9           Divide through by  8 .

 

y  =  -\(\frac58\)x - \(\frac98\)

 

Now we can see that the slope of  l1  =  -\(\frac58\)  .

Line  l2  is perpendicular to  l1  , so the slope of  l2  =  +\(\frac85\)  .

 

Line  l2  has a slope of  \(\frac85\)  and passes through the point  (10, 10) .

So.... in point - slope form, the equation of  l2  is

 

y - 10  =  \(\frac85\)(x - 10)             We want this in the form  y = mx + b  . Distribute the  \(\frac85\) .

 

y - 10  =  \(\frac85\)x - \(\frac85\)(10)

 

y - 10  =  \(\frac85\)x - 16               Add  10  to both sides.

 

y  =  \(\frac85\)x - 6

 

Now the equation for  l2  is in the form  y = mx + b  , where  m = \(\frac85\)  and  b = -6 .

 

We can look at a graph to verify that  l1  and  l2  are perpendicular, and  l2  passes through (10, 10) .

 

m + b   =  \(\frac85\) + -6   =   -4.4

 
hectictar  Nov 10, 2017
edited by hectictar  Nov 10, 2017
 #3
avatar+5227 
+2

2)

The midpoint of  (1, 2)  and  (7, 4)  \(=\,(\frac{1+7}{2},\frac{2+4}{2}) \,=\,(\frac82,\frac62)\)  =  (4, 3)

 

The slope between  (1, 2)  and  (7, 4)  \(=\,\frac{4-2}{7-1}\,=\,\frac26\,=\,\frac13\)

 

So the slope of the perpendicular bisector   =   -\(\frac31\)   =   -3

 

We want an equation of a line that passes through  (4, 3)  with a slope of  -3 .

In point-slope form, that is

 

y - 3  =  -3(x - 4)         This is the equation of the line. We need it in  y = mx + b  form.

                                   Distribute the  -3 .

y - 3  =  -3x + 12

                                   Add  3  to both sides.

y  =  -3x + 15

 
hectictar  Nov 10, 2017

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