If vertices of a triangle have coordinates \((-1, 1)\), \((4, 1)\), and \((x, y)\), what is the area of the triangle?


(1) \(y^2 - 2y - 3 = 0\)

(2) \(x^2 = y^2\)
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I think this is a high-quality question and I agree with explanation.

What a beautiful question!
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Lets use the determinant method to find out the area from the question stem.

Area of triangle = 1/2 |-1(1-y)-1(4-x) +1 (4y-x) = 1/2(y-1 = x-4 =4y-x) = (5y-5)/2 = 5 (y-1)/2
A=5(y-1)/2

Lets look at statement I and II

statement I)
y2-2y-3=0--> y2-3y+y-3 = 0 --> (y-3)(y+1) =0--> y=-1 or y=3
For y=(-1), A= 5/2(-1-1)= -5

For y=3, A=5/2(3-1)= +5

area is always positive , therefore for both cases area is same that is A=5 units

A is sufficient, so cancel out answer B,C,E; only possibilities are A and D

statement II)
x2=y2
x2-y2=0
(x+y)(x-y)=0
either x=-y or x=y
Area is dependent on value of y, since value of y is not defined on basis of a variable x, therefore statement II is not sufficient

Answer= A
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Yes, I agree, this question is beautiful! Thanks Bunuel!
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