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Find the distance from $3x-4y-10=0$ to the point $(2,0)$

my answer here is $ \dfrac{-4}{2}$ or $-2$ by substituting the given by the use of the formula but Im just wondering if there's a negative distance because it can be possibly rewritten to $ \dfrac{-4}{-2}$ or $2$. Because the square root of $4$ can be positive or negative $2$. What is the correct answer and explanation here?

Emilio Novati
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user219189
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2 Answers2

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I think you have the right formula, check Point-Line Distance (2D). The distance from $(x_0,y_0)$ to the line with equation $ax+by+c=0$ is given by $$d = \frac{\left| ax_0+by_0+c \right|}{\sqrt{a^2+b^2}}$$ Check your calculations and watch out for the absolute value in the numerator. The square root in the denominator is, by definition, (the) positive (square root).

In your case $(x_0,y_0) = (2,0)$ and $a=3$, $b=-4$, $c=-10$.

StackTD
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  • Alright; you're welcome! – StackTD Feb 25 '15 at 13:55
  • nice, thorough answer(+1) :) – Shobhit Feb 25 '15 at 13:58
  • ive got some questions If the distance is a fraction like 14/4. do I need to reduce it to its simplest form? If the distance has a radical number in the denominator like 25/square root of 10.Do I need to rationalize it? – user219189 Feb 25 '15 at 14:05
  • 'Have to' depends on who is asking and what they're asking... If your teacher wants you to simplify/rationalize, then yes: you should. It's a good habit to simplify when possible, but |-4|/sqrt(25), 4/5 or 0.8 all represent the same number (and in your case: the right answer :-)). – StackTD Feb 25 '15 at 14:07
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$$ d= \frac{|Ax + By + C|}{\sqrt{ A^2 + B^2}}=\dfrac{\left| 3\left( 2\right) -4\left( 0\right) -10\right| }{\sqrt{\left( 3^{2}\right) +\left( -4^{2}\right) }}=\dfrac{4}{5} $$

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