The vertides of a triangle are \( A(2,4) \) B \( (-3,1) \) addc \( (3,-7) \). find the coordenates of hhe foot of the perpendiculen drawn from A to the Line BE

To find the coordinates of the foot of the perpendicular drawn from point A to line BE, we first need to determine the equation of line BE using the coordinates B(-3, 1) and E(3, -7). The slope \( m \) of line BE is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 1}{3 - (-3)} = \frac{-8}{6} = -\frac{4}{3}. \] Now, using point-slope form, the equation of line BE is: \[ y - 1 = -\frac{4}{3}(x + 3). \] This simplifies to: \[ y = -\frac{4}{3}x - 4 + 1. \] Thus, \[ y = -\frac{4}{3}x - 3. \] Next, to find the foot of the perpendicular from point A(2, 4), we find the slope of the perpendicular line, which is the negative reciprocal of \(-\frac{4}{3}\), that gives us \(\frac{3}{4}\). Using point-slope form again from point A, we have: \[ y - 4 = \frac{3}{4}(x - 2). \] Now, let's express this equation in slope-intercept form: \[ y - 4 = \frac{3}{4}x - \frac{3}{2} \\ y = \frac{3}{4}x + 4 - \frac{3}{2} \\ y = \frac{3}{4}x + \frac{8}{2} - \frac{3}{2} \\ y = \frac{3}{4}x + \frac{5}{2}. \] At this point, we can find the intersection by setting the two equations for y equal: \[ -\frac{4}{3}x - 3 = \frac{3}{4}x + \frac{5}{2}. \] Solving for x, we can multiply everything by 12 to eliminate the fractions: \[ -16x - 36 = 9x + 30 \\ -25x = 66 \\ x = -\frac{66}{25}. \] Now substituting back to find y: \[ y = -\frac{4}{3}(-\frac{66}{25}) - 3 \\ = \frac{88}{75} - 3 = \frac{88}{75} - \frac{225}{75} = -\frac{137}{75}. \] Thus, the coordinates of the foot of the perpendicular from A to line BE are \[ \left(-\frac{66}{25}, -\frac{137}{75}\right). \]