In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.


xy + 2x - 5y = 2, (3, 2)

Verified step by step guidance

1. First, we need to find the derivative of the given equation implicitly. Start with the equation: $xy + 2x - 5y = 2$. Differentiate both sides with respect to $x$. Remember to use the product rule for $xy$ and the chain rule for $-5y$.
2. Applying the product rule to $xy$, we get: $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$. For $2x$, the derivative is simply $2$. For $-5y$, using the chain rule, the derivative is $-5 \frac{dy}{dx}$. Set the derivative of the right side, which is a constant, to zero.
3. Combine all the derivatives: $x \frac{dy}{dx} + y + 2 - 5 \frac{dy}{dx} = 0$. Rearrange the terms to solve for $\frac{dy}{dx}$, which represents the slope of the tangent line.
4. Substitute the given point $(3, 2)$ into the equation to find the specific slope of the tangent line at that point. This will give you the value of $\frac{dy}{dx}$ at $(3, 2)$.
5. Once you have the slope of the tangent line, use the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point $(3, 2)$, to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope and apply the point-slope form again.