25. \( (3,-2) ; y=x+4 \)

It appears you're working on a problem involving the point \( (3, -2) \) and the line defined by the equation \( y = x + 4 \). Let's explore the relationship between the point and the line. ### **1. Determining if the Point Lies on the Line** To check whether the point \( (3, -2) \) lies on the line \( y = x + 4 \), substitute the \( x \)-coordinate of the point into the equation and see if the corresponding \( y \)-value matches. \[ \begin{align*} \text{Given point: } & (3, -2) \\ \text{Line equation: } & y = x + 4 \\ \text{Substitute } x = 3 \text{ into the line equation:} \\ & y = 3 + 4 \\ & y = 7 \end{align*} \] The calculated \( y \)-value is 7, but the \( y \)-coordinate of the given point is -2. Since \( -2 \neq 7 \), **the point \( (3, -2) \) does not lie on the line \( y = x + 4 \).** ### **2. Finding the Distance from the Point to the Line** If you're interested in finding the shortest distance from the point \( (3, -2) \) to the line \( y = x + 4 \), you can use the formula for the distance \( D \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \): \[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] First, rewrite the line equation in the standard form: \[ y = x + 4 \quad \Rightarrow \quad x - y + 4 = 0 \] Here, \( A = 1 \), \( B = -1 \), and \( C = 4 \). Plugging in the point \( (3, -2) \): \[ \begin{align*} D &= \frac{|1 \cdot 3 + (-1) \cdot (-2) + 4|}{\sqrt{1^2 + (-1)^2}} \\ &= \frac{|3 + 2 + 4|}{\sqrt{1 + 1}} \\ &= \frac{|9|}{\sqrt{2}} \\ &= \frac{9}{\sqrt{2}} \\ &= \frac{9\sqrt{2}}{2} \quad \text{(rationalizing the denominator)} \end{align*} \] **Therefore, the shortest distance from the point \( (3, -2) \) to the line \( y = x + 4 \) is \( \frac{9\sqrt{2}}{2} \) units.** ### **3. Additional Consideration: Equation of the Perpendicular Line** If the problem requires finding the equation of the line perpendicular to \( y = x + 4 \) that passes through \( (3, -2) \), follow these steps: 1. **Find the slope of the given line:** The slope \( m \) of \( y = x + 4 \) is 1. 2. **Determine the slope of the perpendicular line:** The slope \( m_{\perp} \) is the negative reciprocal of 1, which is \( -1 \). 3. **Use the point-slope form to find the equation:** \[ y - y_1 = m_{\perp}(x - x_1) \] \[ y - (-2) = -1(x - 3) \] \[ y + 2 = -x + 3 \] \[ y = -x + 1 \] **So, the equation of the line perpendicular to \( y = x + 4 \) that passes through \( (3, -2) \) is \( y = -x + 1 \).** --- **Summary:** - The point \( (3, -2) \) does **not** lie on the line \( y = x + 4 \). - The shortest distance from the point to the line is \( \frac{9\sqrt{2}}{2} \) units. - The equation of the perpendicular line through \( (3, -2) \) is \( y = -x + 1 \). If you have a specific aspect of the problem you'd like to explore further, feel free to ask!