Chapter 4 Quadratic Functions and Factoring - 4.3 Solve x(squared) + bx + c = 0 - 4.2 Exercises - Mixed Review - Page 258: 90

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Given: $A=(x_A,y_A)\\B=(x_B,y_B)\\C=(x_C,y_C)$ are the verticles of the triangle. Since we have $(0,0);(14,3);(6,25)$, then $x_A=0,y_A=0\\x_B=14,y_B=3\\x_C=6,y_C=25$ Formula of the area of the triangle is given by: $\frac{1}{2}|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\\ =\frac{1}{2}|0(3-25)+14(25-0)+6(0-3)\\ =\frac{1}{2}|350-18|\\ =\frac{1}{2}|332|\\ =166$