Question

ABC has vertices at A(3,4), B(-2,0),and C(5,0). Prove the area of the triangle formed by joining the midpoints of ABC is one-quarter the area of ABC:
A) find the distance from the point A(5,7) to the line joining B(-2,1) and C(4,-3)

Answer 1

$\displaystyle \textsf{Area of traingle joining the points}\, A, B \, \& C \textsf{is}\\ A_1 = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1\\ 3 & -2 & 5\\ 4 & 0 & 0 \end{vmatrix} = \frac{1}{2}\left(-1(-20_ + 1(8))\right) = \frac{28}{2} = 14\\ \textsf{Midpoint of}\, AB\,\textsf{is}\, \left(\frac{1}{2}, 2\right)\\ \textsf{Midpoint of}\, BC\,\textsf{is}\, \left(\frac{3}{2}, 0\right)\\ \textsf{Midpoint of}\, AC\,\textsf{is}\, \left(4, 2\right)\\ A_2 = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1\\ \frac{1}{2} & \frac{3}{2} & 4\\ 2 & 0 & 2 \end{vmatrix} = \frac{1}{2}\left(3 - 1(-7) + 1(-3)\right) = \frac{7}{2}\\ \textsf{It is clear that}\,\, A_2 = \frac{1}{4}A_1\\ \textsf{Equation of line joining}\, (-2, 1) \,\textsf{and}\, (4, -3)\\ \frac{y - 1}{x + 2} = -\frac{4}{6} = -\frac{2}{3}\\ 3y - 3 = -2x - 4\\ 2x + 3y + 1 = 0\\ \textsf{Distance}\, (D) = \frac{2(5) + 3(7) + 1}{\sqrt{2^2 + 3^2}} = \frac{32}{\sqrt{13}} = \frac{32\sqrt{13}}{13}$