Question 1190941
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Part a)


I'm assuming you meant to say AC instead of AV.


A = (1,1)
B = (-1,4)
Gradient = slope
Slope = (change in y)/(change in x)
Slope = (y2-y1)/(x2-x1)
Slope of AB = (4-1)/(-1-1)
Slope of AB = (3)/(-2)
Slope of AB = -3/2


Set this equal to the stated slope of AB (which was -3m) and solve for m
-3/2 = -3m
m = (-3/2)*(-1/3)
<font color=red>m = 1/2</font>


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Part b)


Let C be located at (x,y). 
The goal is to find the numbers that replace x and y.


A = (1,1)
C = (x,y)
Slope of AC = (y2-y1)/(x2-x1)
Slope of AC = (y-1)/(x-1)
Set this equal to 3m. Then plug in m = 1/2


Slope of AC = 3m
(y-1)/(x-1) = 3m
(y-1)/(x-1) = 3(1/2)
(y-1)/(x-1) = 3/2


Let's now cross multiply and isolate y
(y-1)/(x-1) = 3/2
2(y-1) = 3(x-1)
2y-2 = 3x-3
2y = 3x-3+2 
2y = 3x-1 
y = (3x-1)/2 
y = (3/2)x - (1/2)
y = 1.5x - 0.5
This is the equation of line AC.


Next, we'll find the slope of line BC
B = (-1,4)
C = (x,y)
Slope of BC = (y2-y1)/(x2-x1)
Slope of BC = (y-4)/(x-(-1))
Slope of BC = (y-4)/(x+1)


Set this equal to the stated slope of m
m = (y-4)/(x+1)


Then we'll plug in these items
m = 1/2 = 0.5
y = 1.5x - 0.5


We end up with this
m = (y-4)/(x+1)
0.5 = (1.5x-0.5-4)/(x+1)
0.5 = (1.5x-4.5)/(x+1)


It's a lot of steps to take in (there might be a faster more efficient way, so I'll let another tutor step in for that). 
However, it shouldn't be too bad once you get enough practice with algebra.


From here, we isolate x to get...
0.5 = (1.5x-4.5)/(x+1)
0.5(x+1) = 1.5x-4.5
0.5x+0.5 = 1.5x-4.5
0.5+4.5 = 1.5x-0.5x
5 = x
x = 5
Then use this x value to find its paired y value
y = 1.5x-0.5
y = 1.5*5-0.5
y = 7.5-0.5
y = 7
Point C is located at (x,y) = <font color=red>(5, 7)</font>
Don't forget to use parenthesis when writing the coordinates.


We have these three point locations
A = (1,1)
B = (-1,4)
C = (5,7)
I'll let you use the slope formula to confirm that -3m, 3m and m represent the slopes of AB, AC, and BC respectively (where m = 1/2 = 0.5 calculated earlier).
A tool like GeoGebra is very handy to quickly confirm we have the correct values so far.


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Part c)


Focus on points A and C. Use the distance formula to find the length of the segment AC.
A = (x1,y1) = (1,1)
C = (x2,y2) = (5,7)


*[tex \large d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}]


*[tex \large d = \sqrt{(1-5)^2 + (1-7)^2}]


*[tex \large d = \sqrt{(-4)^2 + (-6)^2}]


*[tex \large d = \sqrt{16 + 36}]


*[tex \large d = \sqrt{52}]


*[tex \large d = \sqrt{4*13}]


*[tex \large d = \sqrt{4}*\sqrt{13}]


*[tex \large d = 2\sqrt{13}]


*[tex \large d \approx 7.21]
Segment AC is exactly *[tex 2\sqrt{13}] units long.
If we can show that segment AB is exactly *[tex \large \sqrt{13}] units long, then we'd be able to wrap things up.


Use the distance formula for points A and B
A = (x1,y1) = (1,1)
B = (x2,y2) = (-1,4)


*[tex \large d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}]


*[tex \large d = \sqrt{(1-(-1))^2 + (1-4)^2}]


*[tex \large d = \sqrt{(1+1)^2 + (1-4)^2}]


*[tex \large d = \sqrt{(2)^2 + (-3)^2}]


*[tex \large d = \sqrt{4 + 9}]


*[tex \large d = \sqrt{13}]


*[tex \large d \approx 3.61]
We see that AB is exactly *[tex \large \sqrt{13}] units long, so the proof is concluded.
We have shown that AC is twice as long as AB.


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Answers:
a) <font color=red>m = 1/2</font>
b) Point C is at <font color=red>(5, 7)</font>
c) Proof is shown above
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