Math Questions?

Question

These are for slopes and formula y=mx+b

1. Parallel to x=4 and goes through point (-2, 3)
2. Parallel to line 2X+Y and goes through point (5,0)
3. Perpendicular to line -3X+6 y intercept is -2
4. Goes through point (2,5) and (1,6).

How did you do this? Thanks!

Answer 1

1) x= 4 is a vertical line through the point at which x is = 4. therefore your line is always a vertical line as it is parallel to x=4. It goes through (-2,3) A vertical line contains infinite many y values but only 1 x value, in this case is -2.

so you line is x=-2

4) the gradient of the line passing through the 2 points can be found by m=[y2-y1] / [x2-x1]

[2-1] / [5-6] = -1 so m= -1

pick one of the points to use as a base. i'll use (2,5)

m=[y-y1] / [x-x1]
-1= [y-5]/[x-2]
y-5 = -1(x-2)
y-5 = -x +2
y= -x+7


i can't solve 2 and 3 as you didn't write the correct equations to the line (there are no equal signs)

Answer 2

1. y= mx+c
I agree with Wendy for number one. It is a straight line where x= 4. Therefore, x=-2

2. Assuming y+2x=c, y= -2x+c. 3= -2(-2)+c where c=-1
y=-2x-1

3. y=mx-2

The line -3x+6y would have a gradient of -3/6=-1/2.
Therefore, y= -1/2x-2

4. m=[y2-y1] / [x2-x1]

[2-1] / [5-6] = -1 so m= -1

pick one of the points to use as a base. i'll use (2,5)

m=[y-y1] / [x-x1]
-1= [y-5]/[x-2]
y-5 = -1(x-2)
y-5 = -x +2
y= -x+7


I am not sure about 2 and 3 beacuse of the same reason as Wendy, but I hope this helps. :)