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Question: How do you write the equation \(y-2=3\left( x-1 \right)\) in the slope-intercept form ?...

How do you write the equation y2=3(x1)y-2=3\left( x-1 \right) in the slope-intercept form ?

Explanation

Solution

We recall the three forms of writing a linear equation which are the general formAx+By+C=0Ax+By+C=0, the slope intercept form y=mx+cy=mx+c and the standard formAx+By=CAx+By=C. We add 2 both sides of the given equation and simplify the right hand side to convert the given equation into slope point form.

Complete step by step solution:
We know from the Cartesian coordinate system that every linear equation Ax+By+C=0Ax+By+C=0can be represented as a line. If the line is inclined with positive xx-axis at an angle θ\theta then its slope is given by m=tanθm=\tan \theta and if it cuts yy-axis at a point (0,c)\left( 0,c \right) from the origin the yy-intercept is given by cc. The slope-intercept form of equation is given by
y=mx+c....(1)y=mx+c....\left( 1 \right)
We know that the standard form of linear equation otherwise also known as intercept form is written with constant CCon the right side of equality sign as
Ax+By=C...(2)Ax+By=C...\left( 2 \right)
We are given in the following equation in the question
y2=3(x1)y-2=3\left( x-1 \right)
We add 2 both sides of the given equation to have
y=3(x1)+2\Rightarrow y=3\left( x-1 \right)+2
We simplify the right hand side of the above equation to have;

& \Rightarrow y=3x-3+2 \\\ & \Rightarrow y=3x-1 \\\ \end{aligned}$$ We see that the above equation is in slope-point form. We compare it with general slope point equation $y=mx+c$ to have the slope and intercept as $$\begin{aligned} & m=3 \\\ & c=-1 \\\ \end{aligned}$$ **Note:** We note that the given equation is in the Cartesian slope point equation. If $\left( {{x}_{1}},{{y}_{1}} \right)$ is a point on the line with slope $m$ then the equation is given by $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. Here the given equation $y-2=3\left( x-1 \right)$ has slope 3 and passes through $\left( 2,1 \right)$. The other Cartesian equation of line is $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are two points on the line.