Question

The straight line $3x+4y-12=0$ meets the coordinate axes at $A$ and $B.$ An equilateral triangle $ABC$ is constructed. The possible coordinates of vertex $C$ are

• A. $(2(1-\frac{3\sqrt{3}}{4}),\frac{3}{2}(1-\frac{4}{\sqrt{3}}))$
• B. $(-2(1+\sqrt{3}),\frac{3}{2}(1-\sqrt{3}\left)\right)$
• C. $\left(2\right(1+\sqrt{3}),\frac{3}{2}(1+\sqrt{3}\left)\right)$
• D. $(2(1+\frac{3\sqrt{3}}{4}),\frac{3}{2}(1+\frac{4}{\sqrt{3}}))$

Answer 1

Answer: (A), (D)

$(2(1-\frac{3\sqrt{3}}{4}),\frac{3}{2}(1-\frac{4}{\sqrt{3}}))$
$(2(1+\frac{3\sqrt{3}}{4}),\frac{3}{2}(1+\frac{4}{\sqrt{3}}))$

Given side

$3x+4y=12$

$\frac{x}{4}+\frac{y}{3}=1$

On comparing above eq with intercept form of line $\frac{x}{a}+\frac{y}{b}=1$

$a=4,b=3$

Points on coordinate axis $A(4,0),B(0,3)$

Let Point $C(x,y)$

The given triangle is equilateral

Hence $AB=BC=AC$

$AB=\sqrt{4^2+3^2}=5$

$AB=BC$

$5=\sqrt{(x^2+(y-3{)}^2)}$

On squaring both sides

$25=x^2+y^2+9-6y$

$x^2+y^2-6y=16$----(1)

$AB=AC$

$5=\sqrt{(x-4{)}^2+y^2)}$

On squaring both sides

$25=x^2+y^2+16-8x$

$x^2+y^2-8x=9$----(2)

On substraction (1) and (2)

$8x-6y=7$

$y=\frac{8x-7}{6}$

Putting y in eq (1)

$x^2+{\left(\frac{8x-7}{6}\right)}^2-6\left(\frac{8x-7}{6}\right)=16$

$x^2+\left(\frac{64x^2+49-112x}{36}\right)-\left(\frac{48x-42}{6}\right)=16$

$36x^2+64x^2+49-112x-288x+252=576$

$100x^2-400x=275$

$4x^2-16x-11=0$

$x=\frac{16\pm \sqrt{256+176}}{8}$

$x=\frac{16\pm \sqrt{432}}{8}$

$x=\frac{16\pm 12\sqrt{3}}{8}$

$x=2\pm \frac{3\sqrt{3}}{2}$

$x=2(1\pm \frac{3\sqrt{3}}{4})$

$x=2(1+\frac{3\sqrt{3}}{4})$ and $x=2(1-\frac{3\sqrt{3}}{4})$

$y=\frac{16(1+\frac{3\sqrt{3}}{4})-7}{6}$ and $y=\frac{16(1-\frac{3\sqrt{3}}{4})-7}{6}$

$y=\frac{3}{2}(1+\frac{4}{\sqrt{3}})$ and $y=\frac{3}{2}(1-\frac{4}{\sqrt{3}})$

$C(2(1+\frac{3\sqrt{3}}{4}),\frac{3}{2}(1+\frac{4}{\sqrt{3}}))$

and

$C(2(1-\frac{3\sqrt{3}}{4}),\frac{3}{2}(1-\frac{4}{\sqrt{3}}))$