equation of the line passing points `(3,5) and (-2,7)`

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equation of the line passing points `(3,5) and (-2,7)`

Postby barnamah » Thu 31 March, 2011 18:06

Most often we are given the points such as `P_1` (3,5) and `P_2` (-2,7) and we are asked to find the equation of the line in the form of y=mx+b

first we have to find the slope of the line.

`P_1` (3,5)
`x_1=3` and `y_1=5`

and
`P_2` (-2,7)
`x_2=-2` and `y_2=7`

now let's use the slope equation `m=(y_2-y_1)/(x_2-x_1)`. Sub in the values above and we have:
`m=(7-5)/(-2-3)=2/-5`

so our slope or m is `1/-5`

We have the standard for m equation `y=mx+b`.
We can use one of the points values of x and y `P_1` (3,5) x=3 and y=5 in this equation including the m and we have:
`y=mx+b`
`5=(2/-5)(3)+b`
`5=(6/-5)+b`
`5=-6/5+b`
Now add `6/5` to both side and we have:
`5+6/5=-6/56/5+b`
`5+6/5=b`

so we got `b=5+6/5` and simplify it `b=31/5`

Now put them all together in our y=mx+b as we have `m=2/-5` and `b=31/5`
`:.` our equation of the line is `y=2/-5x+31/5`
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