Question

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 `sqrt(6)/3` from the given point.

Answer 1

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

Slope of line ‘l’ is given by m = `(b - 2)/(a - 1)` = tan θ  .....(i)

Given that AB = `sqrt(6)/3`

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

`sqrt((a - 1)^2 + (b - 2)^2) = sqrt(6)/3`

On squaring both the side

a² + 1 – 2a + b² + 4 – 4b = `6/9`

a² + b² – 2a – 4b + 5 = `2/3`  .....(ii)

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

∴ a + b = 4   .....(iii)

On solving (ii) and (iii)

We get a = `(3sqrt(3) + 1)/(2sqrt(3))`

b = `(5sqrt(3) - 1)/(2sqrt(3))`

Putting values of a and b in eqn. (i), we have

tan θ = `((5sqrt(3) - 1)/(2sqrt(3)))/((3sqrt(3) + 1)/(2sqrt(3))`

= `(5sqrt(3) - 1 - 4sqrt(3))/(3sqrt(3) + 1 - 2sqrt(3))`

= `(sqrt(3) - 1)/(sqrt(3) + 1)`

∴ tan θ = tan 15°

⇒ θ = 15°