Question

Write a linear equation in standard form for the line that goes through (4,2) and (8,1).

Answer 1

Answer: Option D.

Explanation:

The Standard form of the equation of then line is:

`Ax+By=C`

Where A, B and C are integers.

The Point-slope form of the equationof the line is:

`y-y_1=m(x-x_1)`

Where m is the slope of the line and (
`x_1,y_1`) is a point of the line.

Given the points (4,2) and (8,1), you can find the slope with the formula
`m=(y_2-y_1)/(x_2-x_1)`. Then:

`m=(1-2)/(8-4)=-(1)/(4)`

Substitute the slope and the point (4,2 ) into
`y-y_1=m(x-x_1)`:

`y-2=-(1)/(4)(x-4)`

To write in Standard form, move the variables to one sides of the equation. Then:

`y-2=-(1)/(4)(x-4)\\\\4(y-2)=x-4\\4y-8=-x+4\\x+4y=4+8\\x+4y=12`

Answer 2

D.
`x+4y=12`

Explanation:

The given equation passes through:

(4,2) and (8,1).

The equation can be obtained using the formula;

`y-y_1=m(x-x_1)`

The slope is given by:

`m=(y_2-y_1)/(x_2-x_1)`

Let
`(x_1,y_1)=(4,2)` and
`(x_2,y_2)=(8,1)`, then we have;

`m=(1-2)/(8-4) =-(1)/(4)`

we now plug in the slope and the point to obtain;

`y-2=-(1)/(4)(x-4)`

We multiply through by 4 to obtain;

`4(y-2)=-(x-4)`

Expand using the distributive property;

`4y-8=-x+4`

`4y+x=4+8`

`x+4y=12`