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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.
Solution
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°
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