How do you find the equation of the straight line joining these: ...?

Question

a) (-1, 2) and (3, -2)

and

b) (4, 4) and (2, 0)

Answer 1

a)

gradient, m = (y1-y2)/(x1-x2)

m = (2-(-2))/(-1-3) = 4/-4 = -1

(y-y1)=m(x-x1)

=> y - 2 = -1(x - (-1))

y = - x + 1 (ans)


b)

m = (y1-y2)/(x1-x2) = (4 - 0) / (4 - 2) = 2

(y-y1)=m(x-x1)

y - 4 = 2(x - 4)
y = 2x - 4
y = 2(x-2) (ans)

Answer 2

get slope (y2-y1)/(x2-x1)

a) ((2- (-2))/(-1-3)=4/-4 = -1 y=-1x + b b=x+y at point (-1,2)
=1
y=-x+1

b) m=(4-0)/(4-2)=2
y=2x+b
y=0 when x=2 so b=-4
y=2x-4

Answer 3

Use two points equation which is as under
((x-x1)/(y-y1) = ((x2-x1)/y2-y1))
Select any point and say it is P1 its co-ordinates will be x1 and x2 and similarly, P2 with x2 and y2. Now put in above said equation, solve and you will get an equation of straight line.

Answer 4

eqn of a line is y=mx+b
m=(y2-y1)/(x2-x1)=(-2-2)/(3-(-1))=-4/4=-1
so now we know m, so y=-1x+b
now get x and y from any one of the twop ponts. use (-1,2)
y=-1x+b
2=-1*-1+b
2=1+b
1=b
now you know b
y=-1x+1

you can use this procedure for the next one

Answer 5

use slope formula:

slope = (y1-y2)/(x1-x2)

then stick the slope in the point slope formula

(y-y1)=m(x-x1)

*note* y1 & x1 can either be the first set of points or the 2nd set of points...

Answer 6

you can try using microsoft excel which help a lot in plotting graph/line.