Qhe line l passing through points \( A(-1,1), 8(5,5) \), and point \( C(4,2) \) is not on line. Write the equation of the line \( A B \) in normal form and also find they area of triangle \( A B C \)

To find the equation of the line \(AB\) in normal form, we first determine the slope using the coordinates of points \(A(-1, 1)\) and \(B(5, 5)\). The slope \(m\) is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 1}{5 - (-1)} = \frac{4}{6} = \frac{2}{3} \] Next, we use the point-slope form to write the equation of line \(AB\): \[ y - y_1 = m(x - x_1) \implies y - 1 = \frac{2}{3}(x + 1) \] Simplifying this equation gives: \[ y = \frac{2}{3}x + \frac{5}{3} \] To convert it to normal form \(Ax + By + C = 0\), we multiply through by 3 to eliminate the fraction: \[ 2x - 3y + 5 = 0 \] Now for the area of triangle \(ABC\), we can use the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \] Substituting the coordinates \(A(-1, 1)\), \(B(5, 5)\), and \(C(4, 2)\): \[ \text{Area} = \frac{1}{2} \left| -1(5-2) + 5(2-1) + 4(1-5) \right| = \frac{1}{2} \left| -3 + 5 - 16 \right| = \frac{1}{2} \left| -14 \right| = 7 \] Thus, the area of triangle \(ABC\) is \(7\) square units.