Question

Find an equation of the line satisfying the given conditions (standard form). 1.) Passing through the midpoint of the line joining the points (3/2 , -9/8) and (-15/2 / -9/2) and perpendicular to the line 3x-8y=10 ???

Answer 1

`48y+128x+303=0`

Explanation:

First up, we need to find the midpoint of `(3/2,-9/8)` and `(-15/2,-9/2)`

Let M be `(x,y)`

`x=(3/2-15/2)/2`

`x=-3`

`y=(-9/8+9/2)/2`

`y=27/16`

Therefore, M is `(-3,27/16)`

Now if the line is perpendicular to `3x-8y=10`, then we need to find the gradient of this line

`3x-8y=10`
`8y=3x-10`
`y=3/8x-5/4`
The gradient of this line is `m_1=3/8`

Remember the equation `m_1m_2=-1` which basically states that the gradient of two lines when multiplied together should equal to `-1` if they are perpendicular or at right angles to each other

Hence, `m_2=-8/3`

Using the point gradient formula,
`(y-27/16)=-8/3(x+3)`

`y-27/16=-8/3x-8`

`y=-8/3x-101/16`

`48y=-128x-303`

`48y+128x+303=0`