Mathematics

Locus of a point. The locus of a point in a plane is the path traced by the point (the set of all points, and only those points) that satisfy a given geometrical condition. For example, the locus of a point at a constant distance $r$ from a fixed point is a circle.

An equation in $x, y$ that is satisfied by every point of the locus, and by no other point, is called the equation of the locus.

(i) $y = mx + c$

This is of first degree in $x$ and $y$, so it represents a straight line with slope $m$ and $y$-intercept $c$ (it cuts the $y$-axis at $(0, c)$). This is the slope-intercept form.

(ii) $\dfrac{x}{a} + \dfrac{y}{b} = 1$

This is a straight line that cuts the $x$-axis at $(a, 0)$ and the $y$-axis at $(0, b)$, so $a$ and $b$ are its intercepts. This is the intercept form. Its slope is $-\dfrac{b}{a}$.

(iii) $x\cos\alpha + y\sin\alpha = p$

This is a straight line in normal (perpendicular) form. $p>0$ is the length of the perpendicular from the origin to the line, and $\alpha$ is the angle that this perpendicular makes with the positive $x$-axis. Its slope is $-\cot\alpha$.

All three loci are straight lines, since each equation is of the first degree.

Let $x^a = y^b = z^c = k$, where $k>0$, $k\neq 1$. Then

$x = k^{1/a},\qquad y = k^{1/b},\qquad z = k^{1/c}.$

Substitute these into $y^2 = xz$:

$\left(k^{1/b}\right)^2 = k^{1/a}\cdot k^{1/c}$

$k^{2/b} = k^{\frac{1}{a}+\frac{1}{c}}$

Since the bases are equal and $k \neq 1$, the powers must be equal:

$\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$

Hence proved.

Alternative (using logarithms). Taking logs of $x^a=y^b=z^c=k$ gives $a\log x = b\log y = c\log z = \log k$, so $\log x = \dfrac{\log k}{a}$, $\log y = \dfrac{\log k}{b}$ and $\log z = \dfrac{\log k}{c}$. From $y^2=xz$, $2\log y=\log x+\log z$, so $\dfrac{2\log k}{b}=\log k\left(\dfrac1a+\dfrac1c\right)$. Dividing by $\log k\ne0$ gives the same result.