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)
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Expert's answer
2020-11-18T18:29:55-0500
Area of traingle joining the pointsA,B&CisA1=21∣∣1341−20150∣∣=21(−1(−20+1(8)))=228=14Midpoint ofABis(21,2)Midpoint ofBCis(23,0)Midpoint ofACis(4,2)A2=21∣∣12121230142∣∣=21(3−1(−7)+1(−3))=27It is clear thatA2=41A1Equation of line joining(−2,1)and(4,−3)x+2y−1=−64=−323y−3=−2x−42x+3y+1=0Distance(D)=22+322(5)+3(7)+1=1332=133213
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