Question

A line is in point-slope form and and passes through points (-4,-3) and (12,1) and is y-1= 1/4(x-12) what is the standard form of this equation

Answer 1

Answer:

x - 4y = 8

Step-by-step explanation:

The point-slope form of an equation of a line:

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

m - slope

The formula of a slope:

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

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We have the points (-4, -3) and (12, 1).

Substitute:

$m=d\frac{1-(-3)}{12-(-4)}=\dfrac{4}{16}=\dfrac{1}{4}$

Put the volume of a slope and the coordinates of the point (12, 1) to the equation of a line:

$y-1=\dfrac{1}{4}(x-12)$

The standard formula of an equation of a line:

$Ax+By=C$

Convert:

$y-1=\dfrac{1}{4}(x-12)$           multiply both sides by 4

$4y-4=x-12$             add 4 to both sides

$4y=x-8$             subtract x from both sides

$-x+4y=-8$         change the signs

$x-4y=8$