Question

How do you find the equation in slope-intercept form of the line that is perpendicular to AB and passes through the midpoint of AB. Let A = (-6,2) and B = (4,-10)?

Answer 1

`y=5/6x-19/6`

Explanation:

The mid point is the mean value as you read left to right on the x-axis.

Point A `->P_a->(x_a,y_a)=(-6,2)`
Point B `->P_b->(x_b,y_b)=(4,-10)`

Let the mean point be `P_m->(x_m,y_m)`
Let the gradient of the line between `P_a and P_b" be "m`

Note that the gradient of the perpendicular line will be `-1/m`

We read from `x_a" to "x_b`

`color(blue)("Determine the mid point")`

`x_m=(x_b+x_a)/2=(4+(-6))/2=-1`

`y_m->(y_b+y_a)/2=( 2+(-10))/2=-4`

`P_m->(x_m,y_m)=(-1,-4)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the gradient (slope) of the perpendicular line")`

`m=("change in y")/("change in x")-> (y_b-y_a)/(x_b-x_a) = (-10-2)/(4-(-6))= (-12)/10 = -6/5`

Thus `-1/m=+5/6`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the equation of the perpendicular line")`
Found that `y=-1/m x+c` is:

`y=5/6x+c" ".................Equation(1)`

We know that this passes through the point
`P_m->(x_m,y_m)=(-1,-4)`

So by substitution into `Equation(1)` we have:

`-4=5/6(-1)+c`

`c=-4+5/6 = -3 1/6=-19/6` giving:

`y=5/6x-19/6`