In order to convert an equation in slope-intercept form (\(y=mx+b\)) into standard form, you must understand a few rules. I will list them for you:
1. \(Ax+Bx=C\) is the form of the linear function
2. \(A,B,C\in\mathbb{Z}\) (A, B, and C must be integers)
3. \(A\geq0\)
4. A,B, and C must be co-prime
Knowing these rules, let's now convert \(y=-\frac{2}{5}x-5\) into standard form:
\(y=-\frac{2}{5}x-5\) | Because standard form does not allow there to be fractions, let's eliminate the current fraction by multiplying all sides by 5. |
\(5y=-2x-25\) | Add 2x to both sides. |
\(2x+5y=-25\) | We have converted the equation from slope-intercept form while meeting all the conditions. |
This process is more or less the same when you have 2 arbitrary points (such as (4,6) and (-9,1)). This time, however, we must take into account that there are variables involved. Let's remind you of slope-intercept form of a line.
\(y=mx+b\)
m = slope of the line
b = y-intercept
In this particular case, we know the x- and y-intercepts because those points are given in the original problem. We know that the y-intercept is located at \((0,b)\). Since b is the y-intercept, fill that in! That's the easy bit, I think you'd agree.
\(y=mx+b\)
We know that the x-intercept is at the point when y=0, so plug that in:
\(0=mx+b\) | Now, solve for x by subtracting b on both sides. |
\(-b=mx\) | Divide by m on both sides. |
\(x=\frac{-b}{m}\) | |
We have determined, with the above algebraic work that when \(y=0,\hspace{1mm}x=\frac{-b}{m}\), which means that the x-intercept is located at \(\left(\frac{-b}{m},0\right)\). However, we also know that the x-intercept is located at \((a,0)\), which means that \(a=\frac{-b}{m}\):
\(a=\frac{-b}{m}\) | Now, we must solve for m because that is the slope of this linear equation after all. |
\(ma=-b\) | Divide by a on both sides. |
\(m=\frac{-b}{a}\) | |
We now know the value for b and for m, so fill it in to get the equation.
\(y=\frac{-b}{a}x+b\)
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