The distance from origin to $6 x - 8 y + 5 = 0$

\frac{1}{2}

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\frac{5}{2}

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\frac{7}{2}

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\frac{2}{5}

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Solution

The correct option is A $\frac{1}{2}$
Equation is $3 x + 4 y + 5 = 0$
Perpendicular distance is $\frac{| 5 |}{\sqrt{6^{2} + 8^{2}}} \frac{5}{\sqrt{100}} = \frac{1}{2}$

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Similar questions

x^{2} + y^{2} - 6 x + 8 y = 0

The distance between the lines 6x – 8y + 20 = 0 and 6x – 8y + k = 0 is 1/5, so what are the possible values of K?

3 x + 4 y = 25

x^{2} + y^{2} = 6 x - 8 y

What is the maximum and minimum distance of the point (9, 12) from the circle $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$

d_{1}

3 x + 4 y + 5 = 0

6 x + 8 y + 20 = 0

d_{2}

5 x + 12 y + 13 = 0

10 x + 24 y + 52 = 0

\frac{d_{1}}{d_{2}}

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