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Question
Find the distance of the point (4, 1) from the line 3x – 4y + 12 = 0.
Solution
The length of perpendicular from a point (x1, y1) to the line ax + by + c = 0 is d = `|(ax_1 + by_1 + c)/(sqrt(a^2 + b^2))|`
∴ The distance of the point (4, 1) to the line 3x – 4y + 12 = 0 is
d = `|(3(4) - 4(1) + 12)/(sqrt(3^2 + (- 4)^2))|`
`= |(12 - 4 + 12)/(sqrt(9 + 16))|`
`= |20/5|`
= 4 units
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