Hong Kong

Stage 4 - Stage 5

Lesson

A locus is a pathway (perhaps a line or a curve) formed by a collection of points, each of which has a location that satisfies some given condition.

For example, a straight line can be defined as the pathway that consists of all points satisfying an equation of the form $y=mx+b$`y`=`m``x`+`b`. Points that are not on this line will not satisfy this equation and so are not part of the locus.

In the diagram the red points satisfy the line $y=\frac{1}{2}x+3$`y`=12`x`+3 as exemplified by the point $\left(12,9\right)$(12,9), whereas the blue points, such as the point $\left(2,2\right)$(2,2), don't.

In the study of cartesian geometry, we are often tasked with constructing an equation of a line or a curve that is based on some imposed condition.

For example, suppose we wish to find the locus of all points $\left(x,y\right)$(`x`,`y`) that are at a distance of exactly $5$5 units from the origin. Just like a child on a play horse on the outer rim of a merry-go-round the locus becomes a circle with centre $\left(0,0\right)$(0,0) and radius $5$5.

From the distance formula, we can easily locate a point, say $\left(3,4\right)$(3,4), that satisfies the imposed condition:

$d$d |
$=$= | $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$√(x2−x1)2+(y2−y1)2 |

$=$= | $\sqrt{\left(3-0\right)^2+\left(4-0\right)^2}$√(3−0)2+(4−0)2 | |

$=$= | $\sqrt{9+16}$√9+16 | |

$=$= | $\sqrt{25}$√25 | |

$\therefore$∴ $d$d |
$=$= | $5$5 |

But what about any point $\left(x,y\right)$(`x`,`y`) on the locus?

Again, using the distance formula, and setting $d=5$`d`=5, we have:

$5$5 | $=$= | $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$√(x2−x1)2+(y2−y1)2 |

$5$5 | $=$= | $\sqrt{\left(x-0\right)^2+\left(y-0\right)^2}$√(x−0)2+(y−0)2 |

$5$5 | $=$= | $\sqrt{x^2+y^2}$√x2+y2 |

$\therefore$∴ $x^2+y^2$x2+y2 |
$=$= | $25$25 |

This is such an interesting result. For example, the left hand side of this equation is symmetrical - we could swap values for $x$`x` and $y$`y` around and it would not make any difference to the sum. Hence we know $\left(3,4\right)$(3,4) *and* $\left(4,3\right)$(4,3) will both be on the circle.

Because the left hand side is the sum of two *squares*, we also know that $\left(-3,4\right)$(−3,4), ** **$\left(-3,-4\right)$(−3,−4), $\left(-4,3\right)$(−4,3) and $\left(-4,-3\right)$(−4,−3) satisfy the equation. The shape of the locus reflects this symmetric relationship between $x$`x` and $y$`y`.

Of course there are an infinite set of points $\left(x,y\right)$(`x`,`y`) that satisfy $x^2+y^2=25$`x`2+`y`2=25. Points like $\left(0,5\right)$(0,5), $\left(0,-5\right)$(0,−5) and $\left(\sqrt{10},\sqrt{15}\right)$(√10,√15) for example.

In general terms, the locus of all points lying a distance $r$`r` from the origin is given by $x^2+y^2=r^2$`x`2+`y`2=`r`2. Even more generally, the locus of all points lying a distance $r$`r` units from the fixed point $\left(h,k\right)$(`h`,`k`) is given, through the same process, by $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(`x`−`h`)2+(`y`−`k`)2=`r`2.

Find the equation of the locus of all points $7$7 units from the $x$`x` axis.

Here the locus is simply the two lines that are parallel to the $x$`x` axis and a distance of $7$7 away from it. The locus includes all points on both of the lines given by $y=\pm7$`y`=±7.

Sometimes we talk about a variable point $\left(x,y\right)$(`x`,`y`) that *moves* according to the imposed condition, just like a train moves according to where the train tracks are pointing. Here is an example:

Find the locus of a variable point which moves so that its distance from the $y$`y` axis is always $3$3 times its distance from the $x$`x` axis.

The diagram makes it clear that the value of the $x$`x` component must remain $3$3 times the value of the $y$`y` component. That is to say, for the point $\left(x,y\right)$(`x`,`y`), we have $x=3y$`x`=3`y`, and so the locus becomes the line given by $y=\frac{1}{3}x$`y`=13`x`.

Find the locus of the point which moves such that it remains equidistant from the two fixed points $\left(5,3\right)$(5,3) and $\left(2,8\right)$(2,8).

Here, we have to keep the variable point $\left(x,y\right)$(`x`,`y`) the same distance away from the given fixed points.

Using the distance formula, this means setting:

$\sqrt{\left(x-5\right)^2+\left(y-3\right)^2}=\sqrt{\left(x-2\right)^2+\left(y-8\right)^2}$√(`x`−5)2+(`y`−3)2=√(`x`−2)2+(`y`−8)2 and simplifying:

$\sqrt{\left(x-5\right)^2+\left(y-3\right)^2}$√(x−5)2+(y−3)2 |
$=$= | $\sqrt{\left(x-2\right)^2+\left(y-8\right)^2}$√(x−2)2+(y−8)2 |

$\left(x-5\right)^2+\left(y-3\right)^2$(x−5)2+(y−3)2 |
$=$= | $\left(x-2\right)^2+\left(y-8\right)^2$(x−2)2+(y−8)2 |

$x^2-10x+25+y^2-6y+9$x2−10x+25+y2−6y+9 |
$=$= | $x^2-4x+4+y^2-16y+64$x2−4x+4+y2−16y+64 |

$-10x-6y+34$−10x−6y+34 |
$=$= | $-4x-16y+68$−4x−16y+68 |

$10y$10y |
$=$= | $6x+34$6x+34 |

$y$y |
$=$= | $\frac{3}{5}x+3\frac{2}{5}$35x+325 |

Thus the locus is a line with gradient $m=\frac{3}{5}$`m`=35 and $y$`y`-intercept $3\frac{2}{5}$325 with the general form given as $3x-5y+17=0$3`x`−5`y`+17=0.