Question

Find the slope and y intercept of ${\displaystyle \sqrt{3x}+(1-\sqrt{3})y}$ = 3

Answer 1

The equation of a line is ${\displaystyle \sqrt{3x}+(1-\sqrt{3})y}$ = 3

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

y = ${\displaystyle \frac{-\sqrt{3}x}{1-\sqrt{3}}+\frac{3}{1-\sqrt{3}}}$

Slope (m) = ${\displaystyle \frac{-\sqrt{3}}{1-\sqrt{3}}}$

= ${\displaystyle \frac{-\sqrt{3}(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}}$

= ${\displaystyle \frac{-\sqrt{3}-3}{1-3}}$

= ${\displaystyle \frac{-(\sqrt{3}+3)}{-2}}$

= ${\displaystyle \frac{\sqrt{3}+3}{2}}$

y intercept = ${\displaystyle \frac{3}{1-\sqrt{3}}}$

= ${\displaystyle \frac{3(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}}$

= ${\displaystyle \frac{3+3\sqrt{3}}{1-3}}$

= ${\displaystyle \frac{3+3\sqrt{3}}{-2}}$

= ${\displaystyle -\frac{3(1+\sqrt{3})}{2}}$