The equation of one of the straight lines which passes through the point (1,3) and makes an angle tan−1(2) with the straight line y+1=32x is:
Solutions · 1
Official solution
Given line: 32x−y−1=0, slope =32. Let m be the required slope; tanθ=2:
\[\left|\frac{m-3\sqrt{2}}{1+3\sqrt{2}m}\right|=\sqrt{2}\]
Squaring: m2−62m+18=2(1+18m2+62m)⟹35m2+182m−16=0
\[m = \frac{-18\sqrt{2}\pm\sqrt{648+2240}}{70} = \frac{-18\sqrt{2}\pm38\sqrt{2}}{70}\]
\[m=\frac{2\sqrt{2}}7 \text{ or } -\frac{4\sqrt{2}}5\]
For m=−542 (matching the given options), through (1,3):
\[y-3=-\frac{4\sqrt{2}}5(x-1) \implies 5y-15=-4\sqrt{2}x+4\sqrt{2} \implies 4\sqrt{2}x+5y-(15+4\sqrt{2})=0\]