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Question
Find the distance between the line 4x + 3y + 4 = 0, and a point (−2, 4)
Solution
The distance between the line ax + by + c = 0 and the point (x1, y1) is d = `("a"x_1 + "b"y_1 + "c")/(sqrt("a"^2 + "b"^2)`
Here (x1, y1) = (−2, 4) and the equation of the line is 4x + 3y + 4 = 0
∴ Required dostance = `(4(- 2) + 3(4) + 4)/sqrt(4^2 + 3^2)`
= `(- 8 + 12 + 4)/(sqrt(16 + 9)`
= `8/5`
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