Question

Find the equation of the line passes through the point (1,-2) and the point of intersection of the lines 3x-4y=10 and 2y+3x+8=0?

Answer 1

`y=3/5x-13/5`

Explanation:

At first we compute the solution of the System
`3x-4y=10`
`3x+2y=-8`

Multiplying the second equation by `2` and adding to the first we get.

`x=-2/3`
and then we get `y` by

`3*(-2/3)-4y=10`

so `y=-3`
the slope `m` of our searched line is given by

`m=(y_2-y_1)(x_2-x_1)=(-3+2)/(-2/3-1)=3/5`
so we have
`y=3/5x+n`

From

`-2=3/5+n`
we obtain

`y=3/5x-13/5`

Answer 2

`y=3/5 x - 13/5`

Explanation:

First find the intersection of the two lines

From the first line equation, `3x=4y+10`; substitute for `3x` into the second line equation:
`2y+(4y+10)+8=0`
`6y+18=0`
`y=-3`
Thus `3x=4(-3)+10=-12+10=-2` and so `x=-2/3` and the intersection is at `(x,y)=(-2/3,-3)`

Now find the equation of the line passing through the two points

Knowing that a straight line has an equation of the form `y=mx+c` we create two simultaneous equations from the two points `(1,-2)` and `(-2/3,-3)`.

Eq 1: `-2 = m+c`
Eq 2: `-3=-2/3m+c`

Subtract eq 2 from eq 1:
`-2-(-3)=m(1-(-2/3))+ (1-1)c`
`1=5/3 m`
`m=3/5`

Go back to eq 1:
`-2=3/5+c`
`c=-13/5`

Thus the equation of the line is
`y=3/5 x - 13/5`