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In the post made 2 yrs ago 'Explain why perpendicular lines have negative slopes', Explain why perpendicular lines have negative reciprocal slopes

How does the distance formula get applied to obtain

$$ \frac{(c-b)}{(m-n)}\sqrt{(1+m^2)}$$ for the distance between $$(\frac{(c-b)}{(m-n)},\frac{m(c-b)}{m-n)} +b) $$ to $$(0,b)$$

The distance formula (general form) has the $\sqrt{(1+m^2)}$ term in the denominator.

i.e. $$ \frac{ax + by +c}{\sqrt{(a^2 + b^2)}}$$ yet the answer has the $\sqrt{(1+m^2)}$ term in the numerator.

Community
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J132
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  • are these two points? – Dr. Sonnhard Graubner Nov 07 '15 at 01:21
  • Yes, the term $(0,b)$ is on the y-axis where the line $f(x) = mx + b$ crosses it and $(\frac{(c-b)}{(m-n)},\frac{m(c-b)}{m-n)} +b) $ is the intersection between the lines $f(x) = mx + b$ and $g(x) = nx + c$. For clarification there is a link to the previous post made 2 yrs ago above. – J132 Nov 07 '15 at 01:27

2 Answers2

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The distance formula you state gives the distance from a point $(x, y)$ to the line with equation $ax + by + c = 0$.

The distance formula you ask about is the diatance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$, which is usually expressed $$ \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}. \tag{1} $$ For the points $$ (x_{1}, y_{1}) = (0, b),\qquad (x_{2}, y_{2}) = \left(\frac{c - b}{m - n}, \frac{m(c - b)}{m - n} + b\right) $$ in question, $$ (x_{2} - x_{1}, y_{2} - y_{1}) = \left(\frac{c - b}{m - n}, \frac{m(c - b)}{m - n}\right) = \frac{c - b}{m - n} (1, m). $$ The distance formula (1) gives $$ \left|\frac{c - b}{m - n}\right| \sqrt{1 + m^{2}}. $$

Andrew D. Hwang
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  • We answered the question about the same time, it made sense when the distance formula for 2 points was used. We just set the calculations out differently. Thanks for your response. – J132 Nov 07 '15 at 11:10
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An alternative version of the distance formula can be used:

$$ d= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

So we have, for $x_1=0$, $y_1=b$, $x_2=\frac{(c-b)}{(m-n)}$ ($=x$, from the previous post) and $y_2=m\frac{(c-b)}{(m-n)} + b = mx+b,$ $$d = \sqrt{(x-0)^2 + (mx+b - b)^2}$$ which gives

$$d = \sqrt{x^2 + (mx)^2} = x\sqrt{(1+m^2)}$$

Substituting for $x$ gives:

$$d = \frac{(c-b)}{(m-n)}\sqrt{(1+m^2)}$$

J132
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