The Art of Derivation: A Path to Mathematical Clarity

This method, commonly found in grade 11 math books, involves deriving a formula by equating two expressions for the area of a single triangle. To initiate this process, consider a line represented by \(ax+by+c=0\) intersecting the \(x\)-axis at point R with coordinates \((-c/a, 0)\) and the \(y\)-axis at point \(Q\) with coordinates \((0, -c/b)\). Now, suppose we have a point \(P(x_1, y_1)\) from which the distance, denoted as $d,” is to be determined. Let’s delve into the derivation process.

Here is the triangle whose vertices are P, Q and R. The most obvious approach for finding the area of this triangle is to use the well known area formula ( for triangle of course!) half of the product of of base and height. So if the area of this triangle is denoted by A, it will be found by the equation

\[A = \frac{1}{2} QR \times d\] where QR serves as the base and d as the altitude of the triangle. As QR is the length of the hypotenuse of the triangle ORQ, it is given by \[\sqrt{(c/a)^2 + (c/b)^2}\]  Thus, the area \(A\) will now be

                       \[A= \frac{1}{2} \sqrt{(c/a)^2 + (c/b)^2} \times d\]

And that can be simplified as  to

\[A= \frac{1}{2} |\frac{c}{ab}|\sqrt{a^2 + b^2} \times d \tag{1}\]

Notice that this equation involves \(d\) that we have to discover.

So our next step will be to find another expression for the area (may be off \(d\)) so that it can be set equal to the area we have just found.  Notice that vertices of the triangle are known to us. Can’t we use them? Yes we can! The area of a triangle can be found if the co-ordinates of the its vertices is known. We use this formula

\[A = \frac{1}{2}|x_1(y_2 -y_3)+x_2(y_3 -y_1)+x_3(y_1-y_2)|\]

Now! In the formula for the area, Let us replace \(x_1, y_1\) with the coordinates of the vertex P that is \((x_1, y_1)\) itself. Replace \((x_2, y_2)\) with \((0, -c/b)\) the coordinate of vertex Q and \((x_3, y_3)\) with \((-c/b, 0)\) the coordinates of vertex R. And on simplification we have,

\[A = \frac{1}{2} \bigg|x_{1} (0 + \frac{c}{b}) – \frac{c}{a}(- \frac{c}{b}- y_{1} ) + 0 (y_{1} – 0) \bigg|\]

\[A = \frac{1}{2} \bigg| \frac{cx_{1}}{b} +\frac{cy_{1}}{a} + \frac{c^{2}}{ab} \bigg|\]

\[A=\frac{1}{2}\bigg|\frac{c}{2ab}(a x_{1} + b y_{1} + c )\bigg| \tag{2}\]

Now on equating these two expressions for the same area from (1) and (2)

\[\frac{1}{2} \bigg|\frac{c}{ab}\bigg|\sqrt{a^2 + b^2} \times d = \frac{1}{2}\bigg|\frac{c}{2ab}(a x_{1} + b y_{1} + c )\bigg|\]

and solving further for \(d\) we get,

\[d = \frac{ax_1 + by_1 +c}{\sqrt{a^2 + b^2}}\]

In this method, we construct a triangle PMN which is similar to the the right triangle ROQ such that the side PM is parallel to the \(x-\)axis and the vertex M is the point where it meets the line \(ax+by+c=0\). As the point M shares the same \(y-\)co-ordinate with the point P, its \(x-\)coordinate can be found by plugging \(y_1\)  into the equation of the line.

And, thus the vertex M will have the co-ordinates \(\bigg(\frac{-by_1 -c}{a}, y_1 \bigg)\).  Now this will make easy for us to find the length of the side PM, which happens to be the hypotenuse of the triangle PMN, and is given by
\[PM = \bigg|x_1 – \frac{-by_1 -c}{a}\bigg| \] so,
\[PM = \bigg|\frac{a x_1+ by_1 +c}{a} \bigg| \]
Now, since this triangle PMN is similar to the big triangle ROQ they have their corresponding sides proportional and so
\[\frac{PN}{OQ} = \frac{PM}{QR}\]
\[ \frac{d}{\bigg|\frac{-c}{b}\bigg| } = \frac{\frac{a x_1+ by_1 +c}{a}}{\sqrt{(c/a)^2 + (c/b)^2}}\]
which, when simplified further and solved for d, we get the desired result

\[d = \dfrac{ax_1 + by_1 +c}{\sqrt{a^2 + b^2}}.\]

Method 3: Using Trigonometric Identity

Again, we have the line and point and we have to find the length of the perpendicular d. Let us draw two line segments through the point P — the first one, parallel to the \(x-\) axis that meets our line at M.  Notice that, we have the coordinates of the point M as well as the length of PM from the previous method, \[PM = \bigg| \frac{a x_1+ by_1 +c}{a} \bigg| \]

Now we draw a vertical line segment through the point P that meets the line at S.  Notice that we can find the length of segment PS by finding the coordinates of the point S.

Since S share the \(x\) coordinate with P its \(y-\)coordinate will be found by plugging \(x_1\) into the equation of the line. We get the position of the point S as \((x_1, \frac{-ax_1-c}{b}))\. And thus we have the length of PS is
\[PS = \bigg|y_1 – \frac{-ax_1 -c}{b}\bigg|\]
\[PS = \bigg| \frac{a x_1+ by_1 +c}{b} \bigg|\]

Now let us assume, the angle at R that is \(\angle ORQ=\theta\). So angle at M will be \(theta\) (because they are alternate interior angles). That makes angle \(\angle MPN = 90 – \theta\) which in turn gives \(\angle NPS = \theta\).

Now we use some trig. In \(\triangle PMN\)
\[ \sin \theta = \frac{d}{PM} = \frac{d}{\frac{a x_1+ by_1 +c}{a}}\]
so \[\sin \theta = \frac{ad}{ax_1 + by_1 + c}\]
similarly in \(\triangle PSN\)
\[ \cos \theta = \frac{d}{PS} = \frac{d}{\frac{a x_1+ by_1 +c}{b}}\]
so \[\cos \theta = \frac{bd}{ax_1 + by_1 + c}\]

Now we use the identity, \(\sin^2 \theta + \cos^2 \theta =1\) which relates the right hand sides of the just found equations as follow
\[\bigg(\frac{ad}{ax_1 + by_1 + c}\bigg)^2 +\bigg(\frac{bd}{ax_1 + by_1 + c}\bigg)^2 = 1\]
which can be simplified to

\[d^2 (a^2 + b^2) = \bigg(ax_1 + by_1 + c\bigg)^2\]

and further to

\[d = \frac{ax_1 + by_1 +c}{\sqrt{a^2 + b^2}}. \]  That is what we were looking for.

Derive to Thrive

In conclusion, the diverse approaches to derivation not only enrich our understanding of mathematical concepts but also underscore the versatility of this crucial tool in problem-solving. Delving into the derivation of formulas from various angles not only reinforces our comprehension but also fosters a deeper appreciation for the interconnectedness of mathematical principles. By exploring different pathways to derive the same result, we illuminate the elegance and adaptability inherent in mathematical thinking. Keep Deriving!

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