+0  
 
0
809
3
avatar

1. Line $l$ has equation $y = 4x - 7$, and line $m$ with equation $y = ax + b$ is perpendicular to line $l$ at $(2,1)$. What is the $y$-coordinate of the point on $m$ that has $x$-coordinate 6?

 

2. Find $B - A$ if the graph of $Ax + By = 7$ passes through $(2,1)$ and is parallel to the graph of $2x - 7y = 3$.

 

3.The "perpendicular bisector" of the line segment $\overline{AB}$ is the line that passes through the midpoint of $\overline{AB}$ and is perpendicular to $\overline{AB}$.  The equation of the perpendicular bisector of the line segment joining the points $(1,2)$ and $(-5,12)$ is $y = mx + b$. Find $m+b$.

Guest Dec 4, 2017
 #1
avatar+89791 
+1

1. Line $l$ has equation $y = 4x - 7$, and line $m$ with equation $y = ax + b$ is perpendicular to line $l$ at $(2,1)$. What is the $y$-coordinate of the point on $m$ that has $x$-coordinate 6?

 

The slope of the perpedicular line  is    - (1/4)

 

So...the equation of  line   m  is

 

y  = (-1/4)(x -2)  + 1

y  = (-1/4)x + 1/2 + 1

y  = (-1/4)x + 3/2 

 

So  on this line....when x  = 6

 

y  = (-1/4)(6) + 3/2

y  = -6/4 + 3/0

y = -3/2  + 3/2

y = 0

 

 

cool cool cool

CPhill  Dec 4, 2017
 #2
avatar+89791 
+1

2. Find $B - A$ if the graph of $Ax + By = 7$ passes through $(2,1)$ and is parallel to the graph of $2x - 7y = 3$.

 

 

The slope of the first line  =  -A/B  =  the slope of the second line  = 2/7

Which implies that   -7A  = 2B  ⇒  B  = (-7/2)A

 

So

 

A(2)  + B(1)  = 7

 

A(2)   +  (-7/2)A  = 7

 

-(3/2)A  =  7

 

A  =  -14/3

 

B =  (-7/2)(-14/3)  =  98/6  =  49/3

 

So.......the equation of the first line is

 

(-14/3)x  + (49/3)y  = 7

 

And B - A  =     49/3 + 14/3      =   63 / 3  =  21

 

Here's the graph showing this : https://www.desmos.com/calculator/qrhzf2f3op

 

 

cool cool cool

CPhill  Dec 4, 2017
edited by CPhill  Dec 4, 2017
 #3
avatar+89791 
+1

3.The "perpendicular bisector" of the line segment $\overline{AB}$ is the line that passes through the midpoint of $\overline{AB}$ and is perpendicular to $\overline{AB}$.  The equation of the perpendicular bisector of the line segment joining the points $(1,2)$ and $(-5,12)$ is $y = mx + b$. Find m + b

 

 

The midpoint of AB  is  [ (1 - 5) / 2, (12 + 2) / 2    =  ( - 4/2 14/2)  =  (-2, 7)

 

And the slope of AB  =  [ 12 - 2] / [ -5 - 1 ]   =  10 / -6   =   - 5/3

 

So....the perpedicular bisector will have the slope  3/5

 

And the equation of this bisector  is

 

y  = (3/5)(x - - 2) + 7

 

y  = (3/5)x + 6/5 + 7

 

 y = (3/5)x + 41/5

 

So.....m + b  =    3/5  + 41/5  =   44/5

 

 

cool cool cool

CPhill  Dec 4, 2017

30 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.