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If the area of the triangle with vertices (1, 2), (2, 3) and (x, 4) is 40 sq. units then the value of $$\mid x - 3 \mid$$ is:
Consider a traingle ABC,Let A(1,2)=$$\ (x_1,y_1)$$
B(2,3)=$$\ (x_2,y_2)$$
C(x,4)=$$\ (x_3,y_3)$$
Area of traingle=
$$\ \frac{1 }{ 2}$$ [$$\ x_1*(y_2$$-$$y_3)$$+$$\ x_2*(y_3$$-$$y_1) $$+$$\ x_3*(y_1$$-$$y_2) $$]
40=$$\ \frac{1 }{2 }$$[1*(3$$\ -\ $$4)+2*(4$$\ -\ $$2)+x*(2$$\ -\ $$3)]
80=$$\ -\ $$1$$\ +\ $$4$$\ +\ $$x($$\ -\ $$1)
x=$$\ -\ $$77
$$\mid x-3 \mid$$ = $$\mid -77-3\mid$$=80
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