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y2 = 4x

The area of the square bounded by x=0, x=4, y=4 and y=0 will be 4×4=16 square units.

Let f: y2=4x and g: x2=4y

Let I1 be the area bounded by the curve f and y-axis and I2 be the area bounded by the curve g and x-axis and I3 be the area bound in between f and g-curve.

Then, I1+I2+I3=16 –(1)

Hence, -(2)

Hence, –(3)

Substituting values of (2) and (3) in (1), we get

or

Hence, and therefore it divides the square in 3 equal parts.

Or

The area of the triangle will be equal to

∆ABC= (area under line segment AB) + (area under line segment BC)- (area under line segment AC)

Let line AB, BC and AC be l, m and n respectively.

Equation of l, by two-point form, will be or y=2x-1.

Equation of m, by two-point form, will be or y=8-x.

Equation of n, by two-point form, will be or .

Hence,

Hence, the area of triangle whose vertices are (2,3), (3,5) and (4,4) is square units.

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