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)
m = 1/2
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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) = (5, 7)
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)
Segment AC is exactly
If we can show that segment AB is exactly
Use the distance formula for points A and B
A = (x1,y1) = (1,1)
B = (x2,y2) = (-1,4)
We see that AB is exactly
We have shown that AC is twice as long as AB.
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Answers:
a) m = 1/2
b) Point C is at (5, 7)
c) Proof is shown above