Prove that 2-5 √ 3 is an irrational number given that √ 3 is irrational.

Let \(\displaystyle 2-5 \sqrt{3}\) be a rational number. \(\displaystyle \therefore 2-5 \sqrt{3}=\frac{\mathrm{a}}{\mathrm{b}}\) where a and b are integers and \(\displaystyle \mathrm{b} \neq 0\). \(\displaystyle \sqrt{3}=\frac{2 \mathrm{~b}-\mathrm{a}}{5 \mathrm{~b}}\) RHS is rational but LHS is an irrational which is a contradiction to our supposition. Hence \(\displaystyle 2-5 \sqrt{3}\) is an irrational number. Sol. \[\begin{aligned} & \mathrm{BC}=\sqrt{(-8-1)^{2}+(7-1)^{2}}=\sqrt{117}=3 \sqrt{13} \\ & \mathrm{AB}=\sqrt{(3-1)^{2}+(4-1)^{2}}=\sqrt{13} \\ & \tan \mathrm{~A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 \sqrt{13}}{\sqrt{13}}=3 \end{aligned} \] Sol. \[\begin{aligned} & \mathrm{AB}=\sqrt{(-7-2)^{2}+(0-3)^{2}}=\sqrt{90}=3 \sqrt{10} \\ & \mathrm{BC}=\sqrt{(-1+7)^{2}+(2-0)^{2}}=\sqrt{40}=2 \sqrt{10} \\ & \mathrm{AC}=\sqrt{(-1-2)^{2}+(2-3)^{2}}=\sqrt{10}=\sqrt{10} \\ & \mathrm{AC}+\mathrm{BC}=\mathrm{AB} \\ & \therefore \mathrm{~A}, \mathrm{~B}, \mathrm{C} \text { are collinear. } \end{aligned} \]