Question

Two lines from the family of lines $(1+2\lambda )x-(1+\lambda )y+1=0$ and the line $x+y=5$ form an equilateral triangle, the equation of two lines can be

• A. $y-2=(2+\sqrt{3})(x-1)$
• B. $y-2=(2-\sqrt{3})(x-1)$
• C. $y-2=(3-\sqrt{2})(x-1)$
• D. $y-1=(3-\sqrt{2})(x-1)$

Answer 1

Answer: (A), (B)

$y-2=(2-\sqrt{3})(x-1)$
$y-2=(2+\sqrt{3})(x-1)$

Family of lines passes through: $\left(1+2\lambda \right)x-\left(1+\lambda \right)y+1=0$

$\lambda \left(2x-y\right)+x-y+1=0y=2xx-y+1=0x-2x+1=01=xy=2\left(1,2\right)$

$Slopeoflines\left(m\right)=\tan \left(\alpha \perp {60}^0\right)or\tan \alpha =-1=\frac{-1+\sqrt{3}}{1+\sqrt{3}}or=\frac{-1-\sqrt{3}}{1-\sqrt{3}}=\frac{{\left(\sqrt{3}-1\right)}^2}{2}or=\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=2-\sqrt{3}or2+\sqrt{3}$

$equationoflinesare:\left(y-2\right)=\left(2-\sqrt{3}\right)\left(x-1\right)y-2=\left(2+\sqrt{3}\right)\left(x-1\right)$

Option $A,B$ is correct answer.