Figure 5.47 9. If whe line \( L \) passing through the point \( \mathrm{P}(-2,1) \) and parallel to the vector \( 2 \mathbf{i}+3 \mathbf{j} \), when find a. vector equation of the line. b. parametric equation of the line. 10. Find the equation of the line tangent to the circle \( x^{2}+y^{2}+x-y=7 \) at point \( \begin{array}{ll}\text { a. A }(2-5) & \text { b. } B(-1,4)\end{array} \)

To define the line \( L \) that passes through the point \( P(-2, 1) \) and is parallel to the vector \( 2\mathbf{i} + 3\mathbf{j} \), we can first find the vector equation. The vector equation can be represented as: \[ \mathbf{r}(t) = \begin{pmatrix} -2 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \end{pmatrix}, \quad t \in \mathbb{R} \] For the parametric equations, we can break this down into its components: \[ x = -2 + 2t, \quad y = 1 + 3t, \quad t \in \mathbb{R} \] For the tangent line to the circle defined by \( x^2 + y^2 + x - y = 7 \) at the points \( A(2, -5) \) and \( B(-1, 4) \), we first need to convert the equation into standard form by completing the square, which gives us the center and radius. The derivative (and subsequently the slope of the tangent line at a particular point) can be found using implicit differentiation. For the point \( A(2, -5) \) you would find: 1. Substitute \( A \) into the circle's equation, then differentiate to find \( \frac{dy}{dx} \). 2. Use the slope to find the line's equation through point \( A \). For point \( B(-1, 4) \), you’d repeat the steps similarly, ensuring you calculate the slope correctly at this point to derive the tangent equation. You'll end up with two distinct tangent line equations based on the slopes derived from those points! Happy calculating!