Question

How do you write an equation in standard form given a line that passes through (9,8) and (4,7)?

Answer 1

A lot of explanation given.

Standardised form type 1: `color(white)("d") x-5y=-31`

Standardised form type 2: `color(white)("d")y=1/5x+31/5`

Explanation:

The gradient (slope) is the amount of up or down for a given amount of along reading left to right on the x-axis.

Given the two points `(x,y)->(9,8) and (4,7)`

The `x` value of 4 comes before 9

Set point 1 as `P_1->(x_1,y_1)=(4,7)`
Set point 2 as `P_2->(x_2,y_2)=(9,8)`

So we are moving from `P_1" to "P_2`

Thus any change is `P_2-P_1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the gradient " -> m)`

Change in up or down is change in `y->y_2-y_1=8-7=+1`

As this is positive then the slope is up.

Change in along is change in `x->x_2-x_1=9-4=5`

Set gradient as `m=("change in up or down")/("change in along") =+1/5`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the full equation")`

Gradient `=m=(y_2-y_1)/(x_2-x_1) =("any "y-y_1)/("any "x-x_1) = (y-7)/(x-4)`

`m=1/5=(y-7)/(x-4)`

Multiply both sides by `(x-4)` giving:

`1/5 xx(x-4)=(y-7)`

Multiply both sides by 5 giving

`x-4=5y-35`

By further manipulation we have:

Standardised form type 1: `color(white)("d") x-5y=-31`

Standardised form type 2: `color(white)("d")y=1/5x+31/5`