Question

Find the equation of a line perpendicular toy, plus, 2, equals, one half, xy+2=
2
1
​
xthat passes through the point left parenthesis, minus, 4, comma, minus, 2, right parenthesis(−4,−2).

Answer 1

To find the equation of a line perpendicular to the given equation and passing through the point \((-4, -2)\), we first need to rewrite the given equation \( y + 2 = \frac{1}{2}xy + 2 \) in a more manageable form.

1. Rearrange the given equation:

\[
y + 2 = \frac{1}{2}xy + 2
\]

Subtract 2 from both sides:

\[
y = \frac{1}{2}xy
\]

Rearranging this gives:

\[
y - \frac{1}{2}xy = 0
\]

Factoring out \(y\):

\[
y(1 - \frac{1}{2}x) = 0
\]

This results in two equations: \(y = 0\) and \(1 - \frac{1}{2}x = 0\), but since we are looking for a slope, focusing on \(1 - \frac{1}{2}x = 0\) gives:

2. Solve for \(y\):

Rearranging gives:

\[
y = \frac{2}{x}
\]

The slope of this curve when expressed in general form is not constant, but to find the perpendicular line, we need the slope of the linear approximation or evaluation at a point.

3. Find the slope of a tangent line:

To find the slope at \(\left(-4, -2\right)\), we can take the derivative of \(y\):

\[
\frac{dy}{dx} = -\frac{2}{x^2}
\]

Substitute \(x = -4\):

\[
\frac{dy}{dx} \bigg|_{x=-4} = -\frac{2}{(-4)^2} = -\frac{2}{16} = -\frac{1}{8}
\]

If the slope of the tangent line is \( -\frac{1}{8} \), then the slope of a line perpendicular to this line is the negative reciprocal:

\[
m_{\text{perpendicular}} = -\frac{1}{\left(-\frac{1}{8}\right)} = 8
\]

4. Now that we have the slope of the perpendicular line, use point-slope form to find the equation of the line that passes through the point \((-4, -2)\):

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

Here, \(m = 8\), \(x_1 = -4\), and \(y_1 = -2\):

\[
y - (-2) = 8(x - (-4))
\]

Simplifying this gives:

\[
y + 2 = 8(x + 4)
\]

\[
y + 2 = 8x + 32
\]

\[
y = 8x + 30
\]

Thus, the equation of the line perpendicular to the given equation and passing through the point \((-4, -2)\) is:

\[
\boxed{y = 8x + 30}
\]