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 \( \begin{array}{ll}\text { a. A }(2-5) & \text { b. } B(-1,4)\end{array} \)

Let's solve the problems step by step. ### Problem 9 **Given:** - Point \( P(-2, 1) \) - Direction vector \( \mathbf{d} = 2 \mathbf{i} + 3 \mathbf{j} \) #### a. Vector equation of the line The vector equation of a line can be expressed as: \[ \mathbf{r} = \mathbf{r_0} + t \mathbf{d} \] where \( \mathbf{r_0} \) is a position vector of a point on the line, \( t \) is a scalar parameter, and \( \mathbf{d} \) is the direction vector. 1. Convert point \( P(-2, 1) \) to a position vector: \[ \mathbf{r_0} = -2 \mathbf{i} + 1 \mathbf{j} \] 2. The direction vector \( \mathbf{d} \) is already given as \( 2 \mathbf{i} + 3 \mathbf{j} \). Thus, the vector equation of the line is: \[ \mathbf{r} = (-2 \mathbf{i} + 1 \mathbf{j}) + t(2 \mathbf{i} + 3 \mathbf{j}) \] #### b. Parametric equation of the line The parametric equations can be derived from the vector equation: \[ \begin{align*} x &= -2 + 2t \\ y &= 1 + 3t \end{align*} \] ### Problem 10 **Given:** - Circle equation: \( x^2 + y^2 + x - y = 7 \) - Points \( A(2, -5) \) and \( B(-1, 4) \) First, we need to rewrite the circle equation in standard form. 1. Rearranging the circle equation: \[ x^2 + y^2 + x - y - 7 = 0 \] 2. Completing the square for \( x \) and \( y \): - For \( x^2 + x \): \[ x^2 + x = (x + \frac{1}{2})^2 - \frac{1}{4} \] - For \( y^2 - y \): \[ y^2 - y = (y - \frac{1}{2})^2 - \frac{1}{4} \] 3. Substitute back into the equation: \[ (x + \frac{1}{2})^2 - \frac{1}{4} + (y - \frac{1}{2})^2 - \frac{1}{4} - 7 = 0 \] \[ (x + \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 7 + \frac{1}{2} = \frac{29}{4} \] The center of the circle is \( C(-\frac{1}{2}, \frac{1}{2}) \) and the radius is \( r = \sqrt{\frac{29}{4}} = \frac{\sqrt{29}}{2} \). #### a. Tangent line at point \( A(2, -5) \) 1. Find the slope of the radius \( CA \): \[ \text{slope} = \frac{-5 - \frac{1}{2}}{2 + \frac{1}{2}} = \frac{-\frac{11}{2}}{\frac{5}{2}} = -\frac{11}{5} \] 2. The slope of the tangent line is the negative reciprocal: \[ m = \frac{5}{11} \] 3. Using point-slope form: \[ y - (-5) = \frac{5}{11}(x - 2) \] \[ y + 5 = \frac{5}{11}x - \frac{10}{11} \] \[ y = \frac{5}{11}x - \frac{10}{11} - 5 \] \[ y = \frac{5}{11}x - \frac{65}{11} \] #### b. Tangent line at point \( B(-1, 4) \) 1. Find the slope of the radius \( CB \): \[ \text{slope} = \frac{4 - \frac{1}{2}}{-1 + \frac{1}{2}} = \frac{\frac{7}{2}}{-\frac{1}{2}} = -7 \] 2. The slope of the tangent line is the negative reciprocal: \[ m = \frac{1}{7} \] 3. Using point-slope form: \[ y - 4 = \frac{1}{7}(x + 1) \] \[ y - 4 = \frac{1}{7}x + \frac{1}{7} \] \[ y = \frac{1}{7}x + \frac{29}{7} \] ### Summary of Results - **Problem 9:** - a. Vector equation: \( \mathbf{r} = (-2 \mathbf{i} + 1 \mathbf{j}) + t(2 \mathbf{i} + 3 \mathbf{j}) \) - b. Parametric equations: \( x = -2 + 2t, y = 1 + 3t \) - **Problem 10:** - a. Tangent line at \( A(2, -5) \): \( y = \frac{5}{11}x - \frac{65}{11} \) - b. Tangent line at \( B(-1, 4) \): \( y = \frac{1}{7}x + \frac{29}{7} \)