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Question
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is (`7/2`, y). Find the value of y
Solution
Let `A(x_1,y_1) = A(2,1), B(x_2, y_2) = B(3,-2)` and `C(x_3,y_3) = C(7/2,y)`
Now
Area(ΔABC) =`1/2|x_1(y_2 - y_3) + x_2(y_3-y_1) + x_3(y_1 - y_2)|`
`=>5 = 1/2|2(-2-y) + 3(y - 1) + 7/2(1+2)|`
`=> 10 = |-4-2y+3y - 3 + 21/2|`
`=>10 = |y + 7/2|`
`=> 10 = y + 7/2` or `-10 = y + 7/2`
`=> y = 13/2` or `y = (-27)/2`
Hence `y = 13/2 or (-27)/2`
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