Find the equation that has slope [latex]m = 2[/latex] and passes through the point [latex](2, 5)[/latex].

First, replace [latex]m[/latex] with 2 and [latex]x_1, y_1[/latex] with [latex](2, 5)[/latex].

[latex]m=\dfrac{y_2-y_1}{x_2-x_1} \text{ becomes } 2=\dfrac{y-5}{x-2}[/latex]

Next, reduce the resulting equation.

First, multiply both sides by the denominator to eliminate the fraction:

[latex](x-2)2=\dfrac{y-5}{x-2}(x-2)[/latex]

This leaves:

[latex](x-2)2=y-5[/latex]

Which simplifies to:

[latex]2x-4=y-5[/latex]

This can be written in the [latex]y[/latex]-intercept form by isolating the variable [latex]y[/latex]:

[latex]\begin{array}{rrrrrrr} 2x&-&4&=&y&-&5 \\ &+&5&&&+&5 \\ \hline 2x&+&1&=&y&& \\ \\ &&&\text{or}&&& \\ \\ &&y&=&2x&+&1 \end{array}[/latex]

It is also useful to write the equation in the general form of [latex]Ax + By + C = 0[/latex], where [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] are integers and [latex]A[/latex] is positive.

In general form, [latex]2x-4=y-5[/latex] becomes:

[latex]\begin{array}{rrrrrrrrr} 2x&-&4&&&=&y&-&5 \\ &-&y&+&5&&-y&+&5 \\ \hline 2x&-&y&+&1&=&0&& \end{array}[/latex]

The standard form of a linear equation is written as [latex]Ax + By = C[/latex].

In standard form, [latex]2x-4=y-5[/latex] becomes:

[latex]\begin{array}{rrrrrrr} 2x&-&4&=&y&-&5 \\ -y&+&4&&-y&+&4 \\ \hline 2x&-&y&=&-1&& \end{array}[/latex]

The three common forms that a linear equation can be written in are:

[latex]\begin{array}{rl} \text{Slope-intercept form:}& y = mx + b \\ \\ \text{General form:}&Ax + By + C = 0 \\ \\ \text{Standard form:}&Ax + By = C \end{array}[/latex]

Find the equation having slope [latex]m = -\dfrac{1}{2}[/latex] that passes though the point [latex](1, -4).[/latex]

Write the solutions in slope-intercept form and in both general and standard forms.

First, replace [latex]m[/latex] with [latex]-\dfrac{1}{2}[/latex] and [latex]x_1, y_1[/latex] with [latex](1, -4).[/latex]

[latex]m=\dfrac{y_2-y_1}{x_2-x_1}\text{ becomes }-\dfrac{1}{2}=\dfrac{y--4}{x-1}[/latex]

Multiplying both sides by [latex]2(x - 1)[/latex] to eliminate the denominators yields:

[latex]-1 (x - 1) = 2 (y + 4)[/latex]

Which simplifies to:

[latex]-x + 1 = 2y + 8[/latex]

Writing this solution in all three forms looks like:

[latex]\begin{array}{rl} \text{Slope-intercept form:}&y=-\dfrac{1}{2}x-7 \\ \\ \text{General form:}&x+2y+7=0 \\ \\ \text{Standard form:}&x+2y=-7 \end{array}[/latex]

Find the equation of the line that runs through [latex](-1, -2)[/latex] and [latex](3, 8)[/latex].

First, find the slope:

[latex]\begin{array}{rrl} m&=&\dfrac{y_2-y_1}{x_2-x_1} \\ \\ m&=&\dfrac{8--2}{3--1} \text{ or } \dfrac{8+2}{3+1} \\ \\ m&=&\dfrac{10}{4} \text{ or } \dfrac{5}{2} \end{array}[/latex]

Now, treat this as a problem of finding a line with a given slope [latex]m[/latex] running through a point [latex](x, y)[/latex].

The slope is [latex]\dfrac{5}{2}[/latex], and there are two points to choose from: [latex](-1, -2)[/latex] and [latex](3, 8)[/latex]. Choose the simplest point to work with. For this problem, either point works. For this example, choose [latex](3, 8)[/latex].

[latex]\begin{array}{rrl} m&=&\dfrac{y-y_1}{x-x_1} \\ \\ \dfrac{5}{2}&=&\dfrac{y-8}{x-3} \end{array}[/latex]

Eliminate the fraction by multiplying both sides by [latex]2(x - 3)[/latex].

This leaves [latex]5(x-3)=2(y-8),[/latex] which must be simplified:

[latex]\begin{array}{rrrrrrr} 5(x&-&3)&=&2(y&-&8) \\ 5x&-&15&=&2y&-&16 \\ -2y&+&15&&-2y&+&15 \\ \hline 5x&-&2y&=&-1&& \end{array}[/latex]

This answer is in standard form, but it can easily be converted to the [latex]y[/latex]-intercept form or general form if desired.