The line passing through \( (-2,5) \) and \( (2, p) \) has a gradient of \( -\frac{1}{2} \). Find the value of \( p \). \( \theta p=\square \) Submit Answer

To find the value of \( p \), we can use the formula for the gradient (slope) of a line that passes through two points \((x_1, y_1)\) and \((x_2, y_2)\): \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \] Here, \((x_1, y_1) = (-2, 5)\) and \((x_2, y_2) = (2, p)\). The slope is given as \(-\frac{1}{2}\). Plugging in the values: \[ -\frac{1}{2} = \frac{p - 5}{2 - (-2)} \] This simplifies to: \[ -\frac{1}{2} = \frac{p - 5}{4} \] Now, we can cross-multiply to eliminate the fraction: \[ -2(p - 5) = 4 \] Expanding the left side gives: \[ -2p + 10 = 4 \] Now, move \(10\) to the right: \[ -2p = 4 - 10 \] \[ -2p = -6 \] Dividing both sides by \(-2\): \[ p = 3 \] Thus, the value of \( p \) is \( \boxed{3} \).