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Question
Find the area of ΔABC with A(1, -4) and midpoints of sides through A being (2, -1) and (0, -1).
Solution
Let ( x2, y2) and (x3, y3) be the coordinates of B and C respectively. Since, the coordinates of A are (1,-4) , therefore
`(1+x_2)/2= 2⇒x_2 = 3`
`(-4+y_2)/2=-1⇒y_2 = 2`
`(1+x_2)/2 =0⇒x_3=-1`
`(-4+y_3)/2 = -1 ⇒ y_3 = 2`
` " let " A (x_1,y_1) = A (1,-4) , B (x_2,y_2) = B(3,2) and C(x_3,y_3) = C (-1,2) Now`
`"Area " (Δ ABC) = 1/2 [x_1(y_2-y_3) +x_2 (y_3-y_1)+x_3(y_1-y
_2)]`
`=1/2 [1(2-2)+(2+4)-1(-4-2)]`
`=1/2[0+18+6]`
=12 sq. units
Hence, the area of the triangle ΔABCis 12 sq. units
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