SOLUTION: The perpendicular bisector of the interval CD has equation 4x-3y+16=0. C has coordinates (-9,10). find the coordinates of D

Algebra ->  Linear-equations -> SOLUTION: The perpendicular bisector of the interval CD has equation 4x-3y+16=0. C has coordinates (-9,10). find the coordinates of D      Log On


   



Question 1116921: The perpendicular bisector of the interval CD has equation 4x-3y+16=0. C has coordinates (-9,10). find the coordinates of D
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
A PICTURE IS USUALLY HELPFUL:
The line with equation 4x-3y%2B16=0 <--> y=%284%2F3%29x%2B16%2F3 has slope 4%2F3 .
It is easy to see that it passes through point(-4,0).
I can plot points of that line by adding 3 to the x-coordinate
(moving 3 spaces to the right),
and adding 4 to the y-coordinate
(moving 4 spaces up) from any point on the line, including point (-4,0).
That allows me to graph that line, along with point C%28-9%2C10%29 .


As the line with equation 4x-3y%2B16=0 <--> y=%284%2F3%29x%2B16%2F3 and slope 4%2F3 .
as it is the perpendicular bisector of segment CD,
it is perpendicular to the line CD containing segment CD, and points C and D.
Perpendicular lines have slopes whose product is -1 ,
so the slope m of the line containing points C and D is such that
m%284%2F3%29=-1 --> m=%28-1%29%283%2F4%29=-3%2F4 .
That slope would allow me to plot points of line CD (and graph line CD),
by adding 4 to the x-coordinate
(moving 4 spaces to the right),
and adding -3 to the y-coordinate
(moving 3 spaces down) from any point on the line, including point C(-9,10).
The points added and line CD are shown in green below.
.
From point to point on each line, I moved
3 units in one direction and 4 perpendicular to that direction,
for a distance (along the line) of sqrt%283%5E2%2B4%5E2%29=5 ,
so I see that point C is at a distance 10 from the blue perpendicular bisector line,
so the point at distance 10 on the other side of that line is point highlight%28D%287%2C-2%29%29 .

PROBABLY "FORMULAS" AND EQUATIONS ARE EXPECTED TO BE SHOWN:
The line with equation 4x-3y%2B16=0 <--> y=%284%2F3%29x%2B16%2F3 and slope 4%2F3 .
As it is the perpendicular bisector of segment CD,
it is perpendicular to the line CD containing segment CD, and points C and D.
Perpendicular lines have slopes whose product is -1 ,
so the slope m of the line containing points C and D is such that
m%284%2F3%29=-1 --> m=%28-1%29%283%2F4%29=-3%2F4 .

Knowing the coordinates of C%28-9%2C10%29 and the slope of the line,
we can write the equation of line CD in point-slope form as
y-10=%28-3%2F4%29%28x-%28-9%29%29 or y-10=%28-3%2F4%29%28x%2B9%29 .
Multiplying both sides of the equal sign times 4 , and rearranging, we get
4y-40=-3x-27 and 3x%2B4y-13=0 .

The intersection of linr CD and the perpendicular bisector of segment CD
is the midpoint of CD, and is given by
system%283x%2B4y-13=0%2C4x-3y%2B16=0%29 .
Adding up the first equation times 3 plus the second times 4, we get
25x%2B25=0 --> x=-1 ,
and substituting that into 4x-3y%2B16=0 , we get
-4-3y%2B16=0 --> -3y%2B12=0 --> y=4 .
So, the midpoint of CD, M%28x%5BM%5D%2Cy%5BM%5D%29 has x%5BM%5D=-1 and y%5BM%5D=4 .
The coordinates of the midpoint of a segment CD, with M%28x%5BC%5D%2Cy%5BC%5D%29 and M%28x%5BD%5D%2Cy%5BD%5D%29 are given by
x%5BM%5D=%28x%5BC%5D%2Bx%5BD%5D%29%2F2 and y%5BM%5D=%28y%5BC%5D%2By%5BD%5D%29%2F2 .
Substituting the known coordinates of C and M,
-1=%28-9%2Bx%5BD%5D%29%2F2 --> -2=-9%2Bx%5BD%5D --> highlight%28x%5BD%5D=7%29
and
4=%2810%2By%5BD%5D%29%2F2 --> 8=10%2By%5BD%5D --> highlight%28y%5BD%5D=-2%29 .