Question

show that standard form, point - slope form, and slope - intercept form are equivalent for the arbitrary point ((x_1, y_1)) on a line with a slope (m) and (y) - intercept of ((0, b)).

Explanation:

Step1: Recall the three forms

• Slope - Intercept Form: The slope - intercept form of a line is given by $y=mx + b$, where $m$ is the slope and $b$ is the $y$-intercept.
• Point - Slope Form: The point - slope form of a line is given by $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope.
• Standard Form: The standard form of a line is given by $Ax+By = C$, where $A$, $B$, and $C$ are integers and $A\geq0$.

Step2: Derive from Slope - Intercept to Point - Slope

We know that the $y$-intercept is $(0,b)$, so this point lies on the line. Let $(x_1,y_1)=(0,b)$. Substitute into the point - slope form $y - y_1=m(x - x_1)$:
$y - b=m(x - 0)$
Simplify to get $y - b=mx$, and then $y=mx + b$, which is the slope - intercept form. So we can go from point - slope (with the $y$-intercept point) to slope - intercept.

Step3: Derive from Slope - Intercept to Standard Form

Start with the slope - intercept form $y=mx + b$. Subtract $mx$ from both sides:
$-mx+y=b$
Multiply both sides by - 1 (to make the coefficient of $x$ non - negative, following the standard form convention where $A\geq0$):
$mx - y=-b$
Here, $A = m$, $B=- 1$, and $C=-b$, which is in the standard form $Ax + By=C$.

Step4: Derive from Point - Slope to Standard Form

Start with the point - slope form $y - y_1=m(x - x_1)$. We know that $(x_1,y_1)=(0,b)$, so substitute:
$y - b=m(x - 0)$
$y - b=mx$
Subtract $mx$ and add $b$ to both sides:
$-mx+y=b$
Multiply by - 1:
$mx - y=-b$, which is in standard form.

Step5: Derive from Standard Form to Slope - Intercept Form

Start with the standard form $Ax+By = C$, where $B
eq0$ (if $B = 0$, the line is vertical and has no slope - intercept form in the traditional sense $y=mx + b$). Solve for $y$:
$By=-Ax + C$
$y=-\frac{A}{B}x+\frac{C}{B}$
Let $m =-\frac{A}{B}$ and $b=\frac{C}{B}$, then we get the slope - intercept form $y=mx + b$.

Step6: Derive from Standard Form to Point - Slope Form

From the standard form $Ax+By = C$, we can find a point on the line. When $x = 0$, $By=C$, so $y=\frac{C}{B}$ (assuming $B
eq0$), so the point $(0,\frac{C}{B})$ is on the line. Let $(x_1,y_1)=(0,\frac{C}{B})$. The slope $m=-\frac{A}{B}$ (from the slope - intercept form we derived earlier). Substitute into the point - slope form $y - y_1=m(x - x_1)$:
$y-\frac{C}{B}=-\frac{A}{B}(x - 0)$
Multiply both sides by $B$:
$By - C=-Ax$
$Ax+By = C$, which is the standard form. Also, if we simplify $y-\frac{C}{B}=-\frac{A}{B}x$, we get $y=-\frac{A}{B}x+\frac{C}{B}$, the slope - intercept form. And we can see the connection to point - slope.

Answer:

To show the equivalence:

1. Slope - Intercept ($y = mx + b$) to Point - Slope: Use the $y$-intercept point $(0,b)$ in $y - y_1=m(x - x_1)$ to get $y - b=m(x - 0)\implies y=mx + b$.
2. Slope - Intercept to Standard: Rearrange $y=mx + b$ to $mx - y=-b$ (standard form $Ax + By = C$ with $A = m,B=-1,C=-b$).
3. Point - Slope to Standard: Start with $y - b=m(x - 0)$ (using $(0,b)$), rearrange to $mx - y=-b$ (standard form).
4. Standard to Slope - Intercept: Solve $Ax+By = C$ for $y$: $y=-\frac{A}{B}x+\frac{C}{B}$ (slope - intercept with $m =-\frac{A}{B},b=\frac{C}{B}$).
5. Standard to Point - Slope: Use the $y$-intercept point $(0,\frac{C}{B})$ (from $x = 0$ in $Ax+By = C$) and slope $m=-\frac{A}{B}$ in $y - y_1=m(x - x_1)$ to get back to standard form.

Thus, the three forms are equivalent for a line with slope $m$ and $y$-intercept $(0,b)$.