The equation of one of the straight lines which passes through the point (1, 3) and makes an angles tan⁻¹(√2) with the straight line, y + 1 = 3√2 x is :

Solution

Maths sir · 4 steps

Put both slopes into the tan formula and open the modulus both ways: you get two possible slopes.

1. Slope of given line. Write y + 1 = 3√2x as y = 3√2x - 1, so its slope is m₁ = 3√2. Let the required line have slope m, with tan θ = √2.

   √2 = |(m - 3√2)/(1 + 3√2 m)|

2. Open the modulus. With the plus sign, m - 3√2 = √2 + 6m, so m = -4√2/5. With the minus sign, m - 3√2 = -√2 - 6m, so m = 2√2/7.
3. Write the line. Take m = -4√2/5 through (1, 3): y - 3 = -4√2/5(x - 1). Multiplying by 5 and rearranging gives 4√2x + 5y - (15 + 4√2) = 0.

   Option	Slope	Left side at (1, 3)	Verdict
   (a)	-4√2/5	4√2 + 15 - 15 - 4√2 = 0	right slope, passes through
   (b)	-5√2/4	√2 - 3 ≠ 0	wrong slope
   (c)	-4√2/5	15 ≠ 0	misses the point
   (d)	4√2/5	-20 ≠ 0	wrong slope

4. So. Only (a) has an allowed slope and also passes through (1, 3). The other line, with slope 2√2/7, is not among the options.

The idea to keep. For a line at a given angle to another, tan θ = |(m - m₁)/(1 + m m₁)| gives two slopes, one for each sign. Then check that the option actually passes through the given point.

Answer from the NTA final answer key. Solution by PrepWiser.