The straight line `3x+4y-12=0` meets the coordinate axes at `Aa n dB` . An equilateral triangle `A B C` is constructed. The possible coordinates of vertex `C` `(2(1-(3sqrt(3))/4),3/2(1-4/(sqrt(3))))` `(-2(1+sqrt(3)),3/2(1-sqrt(3)))` `(2(1+sqrt(3)),3/2(1+sqrt(3)))` `(2(1+(3sqrt(3))/4),3/2(1+4/(sqrt(3))))`

Options

A. `(2(1-(3sqrt(3))/(4)), (3)/(2) (1-(4)/(sqrt(3))))`
B. `(-2(1+sqrt(3)), (3)/(2) (1-sqrt(3)))`
C. `(2(1+sqrt(3)), (3)/(2) (1+sqrt(3)))`
D. `(2(1+(3sqrt(3))/(4)), (3)/(2) (1+(4)/sqrt(3)))`

Answer

A, D

Explanation

AB =5, `D-=(2,(3)/(2))`
`CD = 5 xx (sqrt(3))/(2) = (5sqrt(3))/(2)`
`"Slope of AB" =-(3)/(4)`
Slope of CD `=(3)/(4)`
If `C-=(h,k), "then"`
`(h-2)/(3//5) = (k-3//2)/(4//5) = +-(5sqrt(3))/(2)`
`"or " h=2(1-(3sqrt(3))/(4)), k=(3)/(2)(1-(4)/(sqrt(3)))`
`"or " h=2(1+(3sqrt(3))/(4)), k=(3)/(2)(1+(4)/(sqrt(3)))`