Question

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

Answer 1

Answer: Option D.

Step-by-step 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 $m=\frac{y_2-y_1}{x_2-x_1}$. Then:

$m=\frac{1-2}{8-4}=-\frac{1}{4}$

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

$y-2=-\frac{1}{4}(x-4)$

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

$y-2=-\frac{1}{4}(x-4)\\\\4(y-2)=x-4\\4y-8=-x+4\\x+4y=4+8\\x+4y=12$

Answer 2

Answer:

D. $x+4y=12$

Step-by-step 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:

$m=\frac{y_2-y_1}{x_2-x_1}$

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

$m=\frac{1-2}{8-4} =-\frac{1}{4}$

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

$y-2=-\frac{1}{4}(x-4)$

We multiply through by 4 to obtain;

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

Expand using the distributive property;

$4y-8=-x+4$

$4y+x=4+8$

$x+4y=12$