Question #145053
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)
1
Expert's answer
2020-11-18T18:29:55-0500

Area of traingle joining the pointsA,B&CisA1=12111325400=12(1(20+1(8)))=282=14Midpoint ofABis(12,2)Midpoint ofBCis(32,0)Midpoint ofACis(4,2)A2=1211112324202=12(31(7)+1(3))=72It is clear thatA2=14A1Equation of line joining(2,1)and(4,3)y1x+2=46=233y3=2x42x+3y+1=0Distance(D)=2(5)+3(7)+122+32=3213=321313\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}


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