Assignment Calculate the distance betwren the poind A(2,3) and the inters \( y=x+s \) and \( 2 x+y=0 \)

To calculate the distance between the point \( A(2, 3) \) and the intersection of the two lines: 1. \( y = x + s \) 2. \( 2x + y = 0 \) where \( s \) is a constant, follow these steps: ### 1. Find the Intersection Point of the Two Lines To find the intersection point \( (x, y) \) of the lines \( y = x + s \) and \( 2x + y = 0 \), substitute \( y = x + s \) into the second equation: \[ 2x + (x + s) = 0 \] \[ 3x + s = 0 \] \[ x = -\frac{s}{3} \] Now, substitute \( x = -\frac{s}{3} \) back into \( y = x + s \) to find \( y \): \[ y = -\frac{s}{3} + s = \frac{2s}{3} \] So, the intersection point is: \[ \left( -\frac{s}{3}, \frac{2s}{3} \right) \] ### 2. Calculate the Distance Between \( A(2, 3) \) and the Intersection Point The distance \( D \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula: \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Plugging in the coordinates of \( A(2, 3) \) and the intersection point \( \left( -\frac{s}{3}, \frac{2s}{3} \right) \): \[ D = \sqrt{\left( -\frac{s}{3} - 2 \right)^2 + \left( \frac{2s}{3} - 3 \right)^2} \] Simplify the expressions inside the square root: \[ D = \sqrt{\left( -\frac{s + 6}{3} \right)^2 + \left( \frac{2s - 9}{3} \right)^2} \] \[ D = \sqrt{\frac{(s + 6)^2 + (2s - 9)^2}{9}} \] \[ D = \frac{1}{3} \sqrt{(s + 6)^2 + (2s - 9)^2} \] Expand the squared terms: \[ (s + 6)^2 = s^2 + 12s + 36 \] \[ (2s - 9)^2 = 4s^2 - 36s + 81 \] Add them together: \[ s^2 + 12s + 36 + 4s^2 - 36s + 81 = 5s^2 - 24s + 117 \] Now, the distance becomes: \[ D = \frac{1}{3} \sqrt{5s^2 - 24s + 117} \] ### **Final Formula for the Distance** \[ D = \frac{\sqrt{5s^2 - 24s + 117}}{3} \] ### **Example Calculation** If a specific value for \( s \) is given, you can substitute it into the formula to find the numerical distance. For instance, let's assume \( s = 3 \): \[ D = \frac{\sqrt{5(3)^2 - 24(3) + 117}}{3} = \frac{\sqrt{45 - 72 + 117}}{3} = \frac{\sqrt{90}}{3} = \frac{3\sqrt{10}}{3} = \sqrt{10} \approx 3.162 \] So, when \( s = 3 \), the distance \( D \) is approximately \( 3.162 \) units. --- **Note:** If \( s \) is a known constant in your assignment, replace it with its value to compute the specific distance. If \( s \) is meant to represent something else (like a variable parameter), ensure you understand its intended use within the problem context.