The point-slope form of the equation of the line that passes through (

1.	The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = y minus 1 equals StartFraction one-fourth EndFraction left-parenthesis x minus 12 right-parenthesis.(x – 12). What is the standard form of the equation for this line?
Answer» GIVEN : The point-slope FORM of the EQUATION of the line that PASSES through (-4, -3) and (12, 1) is y minus 1 equals Start FRACTION one-fourth End Fraction left-parenthesis x minus 12 right-parenthesis. TO FIND : The standard form of the equation for this given line SOLUTION : Given points (-4, -3) and (12, 1) are passes through the given line is $y-1=\frac{1}{4}(x-12)$ The Standard form of the line equation is ax+by=c Solve the line equation $y-1=\frac{1}{4}(x-12)$ $4(y-1)=x-12$ 4y-4=x-12 x-12-4y+4=0 x-4y-8=0 x-4y=8 ∴ the standard form of the equation for this given line is x-4y=8

The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = y minus 1 equals StartFraction one-fourth EndFraction left-parenthesis x minus 12 right-parenthesis.(x – 12). What is the standard form of the equation for this line?

Answer»

GIVEN :

The point-slope FORM of the EQUATION of the line that PASSES through (-4, -3) and (12, 1) is y minus 1 equals Start FRACTION one-fourth End Fraction left-parenthesis x minus 12 right-parenthesis.

TO FIND :

The standard form of the equation for this given line

SOLUTION :

Given points (-4, -3) and (12, 1) are passes through the given line is $y-1=\frac{1}{4}(x-12)$

The Standard form of the line equation is

ax+by=c

Solve the line equation $y-1=\frac{1}{4}(x-12)$

$4(y-1)=x-12$

4y-4=x-12

x-12-4y+4=0

x-4y-8=0

x-4y=8

∴ the standard form of the equation for this given line is x-4y=8

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