In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance from the given point.

Let the given line x + y = 4 and the required line ‘l’ intersect at B(a, b)

Slope of line ‘l’ is

and we also know that, m = tanθ

…(i)

Given that:

So, by distance formula for point A(1, 2) and B(a, b), we get

On squaring both the sides, we get

⇒ 2 = 3a² + 3 – 6a + 3b² + 12 – 12b

⇒ 2 = 3a² + 3b² – 6a – 12b + 15

⇒ 3a² + 3b² – 6a – 12b + 13 = 0 …(ii)

Point B(a, b) also satisfies the eq. x + y = 4

∴ a + b = 4

⇒ b = 4 – a …(iii)

Putting the value of b in eq. (ii), we get

3a² + 3(4 – a)² – 6a – 12(4 – a) + 13 = 0

⇒ 3a² + 3(16 + a² – 8a) – 6a – 48 + 12a + 13 = 0

⇒ 3a² + 48 + 3a² – 24a – 6a – 48 + 12a + 13 = 0

⇒ 6a² – 18a + 13 = 0

Now, we solve the above equation by using this formula

Putting the value of a in eq. (iii), we get

Now, putting the value of a and b in eq. (i), we get

We solve the eq. (A) to get the value of θ, we get

We know that,

We have,

θ = tan^-1(√3) - tan^-1(1)

θ = tan^-1(tan 60°) – tan^-1(tan 45°)

θ = 60° - 45°

θ = 15°

Now, we solve the eq. B.

We know that,

We have,

θ = tan^-1(√3) + tan^-1(1)

θ = tan^-1(tan 60°) + tan^-1(tan 45°)

θ = 60° + 45°

θ = 105°