Question

The equation of the line with x-intercept −2 and y-intercept 5 can be written in the form y=mx+b

what are m and b?

Answer 1

m is the gradient/slop
b is the y-intercept

Answer 2

*slope

Answer 3

what are the figures for m and b in this case?

Answer 4

well, b is 5 (says in the equation)

now you need to find m

Answer 5

the gradient formula is:
m=(y2-y1)/(x2-x1)

Answer 6

but i dont have points. i just have x and y intercepts

Answer 7

they are points
x intercept: (-2,0)
y intercept: (0,5)

Answer 8

you need to realise that when it says x-intercept, it means that y is zero, and vice versa. This is because when you draw it on a plane. You'll find the x-intercept is where y=0 and x is on the axis

Answer 9

oh ok

Answer 10

one sec let me attempt

Answer 11

5/2

Answer 12

yes

Answer 13

thanks can you help me with this last question please

Answer 14

Find an equation y=mx+b of the perpendicular bisector of the line segment joining the points A(7,7) and B(13,1). i just need to find the constant b

Answer 15

I do not understand the whole thing, I mean I think I do but I can't be sure.
I'll tell you what I know though.
to find b, you need to use the gradient and one of the points (any one will do)

Answer 16

so using this equation: \[y-y_1=m(x-x_1)\] you can find out what b is by subing in the values of m, and one of the coordinate points

Answer 17

wait, I think that's wrong. Ignore that

Answer 18

i got it nvm

Answer 19

so, here's what i think you have to do. Find the midpoint of that line segment using:
\[M=(\frac{ x_1+x_2 }{ 2 },\frac{ y_1+y_2 }{ 2 })\]
then find the gradient of the line with:
\[m=\frac{ y_2-y_1 }{ x_2-x_1 }\]
then using that coordinate point you just found use:
\[y-y_1=m(x-x_1)\] to find the equation of the line