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Question: How do I change an \(Ax+By=C\) equation to a \(y=mx+b\) equation where \(m\) is the slope and b is t...

How do I change an Ax+By=CAx+By=C equation to a y=mx+by=mx+b equation where mm is the slope and b is the y−y- intercept like 4x+2y=84x+2y=8? $$$$

Explanation

Solution

We recall the three forms of writing a linear equation: the general form Ax+By+C=0Ax+By+C=0, the slope intercept form y=mx+by=mx+b and the standard form Ax+By=CAx+By=C. We take the term with which xx is multiplied to the right hand side and then divide both sides of the given equation Ax+By=CAx+By=C by a coefficient of yy to convert it into slope-intercept form. We use obtained m,bm,b in terms of A,B,CA,B,C to get the slope point from of 4x+2y=84x+2y=8.$$$$

Complete step by step answer:
We know from the Cartesian coordinate system that every linear equation Ax+By+C=0Ax+By+C=0 can be represented as a line. If the line is inclined with positive x−x-axis at an angle θ\theta then its slope is given by m=tan⁡θm=\tan \theta and if it cuts y−y-axis at a point (0,b)\left( 0,b \right) from the origin the y−y-intercept is given by bb. The slope-intercept form of equation is given by
y=mx+b....(1)y=mx+b....\left( 1 \right)
We know that the standard form of linear equation otherwise also known as intercept form is written with constant CC on the right side of equality sign as
Ax+By=C...(2)Ax+By=C...\left( 2 \right)
Let us subtract AxAx from both sides of the above equation to have;
By=−Ax+CBy=-Ax+C
We divided both side of above equation by BB to have
y=−ABx+CB.....(3)y=\dfrac{-A}{B}x+\dfrac{C}{B}.....\left( 3 \right)
We see that the above equation is in the slope-intercept form. We compare equation (1) and (3) to have
m=−AB,b=CBm=\dfrac{-A}{B},b=\dfrac{C}{B}
We are given the equation4x+2y=84x+2y=8. Here we have A=4,B=2,C=8A=4,B=2,C=8. The slope-point form of the equation 4x+2y=84x+2y=8 with slope m=−AB=−42=−2m=\dfrac{-A}{B}=\dfrac{-4}{2}=-2 and intercept b=CB=82=4b=\dfrac{C}{B}=\dfrac{8}{2}=4 is

& y=\left( -2 \right)x+4 \\\ & \Rightarrow y=-2x+4 \\\ \end{aligned}$$ **Note:** We note that if $A=0$ we have $m=\dfrac{-0}{B}=0$ and the line is parallel to the $x-$axis. If $B=0$ then we have $m=\dfrac{-A}{0}=\infty $ and the line is perpendicular to the $x-$axis. If we have intercept $C=0$ then the line passes through the origin. If two lines ${{A}_{1}}x+{{B}_{1}}y={{C}_{1}},{{A}_{2}}x+{{B}_{2}}y={{C}_{2}}$ are parallel the their slopes are equal which means $\dfrac{-{{A}_{1}}}{{{B}_{1}}}=\dfrac{-{{A}_{2}}}{{{B}_{2}}}$ where we can use alternedo to have $\dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{{{B}_{1}}}{{{B}_{2}}}$.