1, 2, 3, 4, 5

Geometry Level 1

The point ( 1 , 2 ) (1 , 2) and the line 3 x − 4 y − 5 = 0 3x - 4y - 5 = 0 are on the x y xy -plane.

What is the shortest distance between the line and the point?


The answer is 2.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

Given the point ( m , n ) (m , n) and the line A x + B y + C = 0 Ax + By + C = 0 , the distance between the point and the line can be formulated as: ∣ A m + B n + C ∣ A 2 + B 2 \dfrac{|Am + Bn + C|}{\sqrt{A^2 + B^2}}

As a result, the distance between the point ( 1 , 2 ) (1 , 2) and the line 3 x − 4 y − 5 = 0 3x - 4y - 5 = 0 equals to ∣ 3 × 1 − 4 × 2 − 5 ∣ 3 2 + 4 2 \dfrac{|3\times 1 - 4\times 2 - 5|}{\sqrt{3^2 + 4^2}} = ∣ − 10 ∣ 5 = 2 \dfrac{|-10|}{5} = \boxed{2}

2 Helpful 1 Interesting 1 Brilliant 0 Confused

Can you explain to me how that formula came out to the world?

. . - 1 month, 2 weeks ago

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