Question

A line passes through the point (4,-2) and has a slope of 1/2. what is an equation for this line in standard form?

Answer 1

To find the equation of a line given a point and a slope, we can use the point-slope form of the equation of a line, which is:

\[
y - y_1 = m(x - x_1)
\]

where \((x_1, y_1)\) is the given point and \(m\) is the slope. Here, \((x_1, y_1) = (4, -2)\) and \(m = \frac{1}{2}\).

Substituting in these values:

\[
y - (-2) = \frac{1}{2}(x - 4)
\]

This simplifies to:

\[
y + 2 = \frac{1}{2}(x - 4)
\]

Next, distribute \(\frac{1}{2}\):

\[
y + 2 = \frac{1}{2}x - 2
\]

Now, subtract \(2\) from both sides to isolate \(y\):

\[
y = \frac{1}{2}x - 4
\]

Now we want to convert this equation to standard form, which is generally written as \(Ax + By = C\) where \(A\), \(B\), and \(C\) are integers. To convert \(y = \frac{1}{2}x - 4\) into standard form, we can first eliminate the fraction by multiplying every term by \(2\):

\[
2y = x - 8
\]

Now, we can rearrange this into standard form:

\[
-x + 2y = -8
\]

To make \(A\) positive, we can multiply through by \(-1\):

\[
x - 2y = 8
\]

Thus, the equation of the line in standard form is:

\[
x - 2y = 8
\]