# Question Video: Locus of Points Equidistant from Two Points Mathematics

A complex number π§ satisfies |π§ β 2 + 4π| = |π§ + 2 + 2π|. Describe the locus of π§ and give its Cartesian equation. What is the minimum value of |π§|?

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### Video Transcript

A complex number π§ satisfies the modulus of π§ minus two plus four π equals the modulus of π§ plus two plus two π. Describe the locus of π§ and give its Cartesian equation. What is the minimum value of the modulus of π§?

There are a number of ways of describing the locus of a complex number π§ by using the modulus. And in fact, we can quote one of these here. For two given constant complex numbers, π§ one and π§ two, the locus of a point π§ which satisfies the modulus of π§ minus π§ one equals the modulus of π§ minus π§ two is the perpendicular bisector of the line segment that joins π§ one to π§ two. Weβre going to rewrite our equation so it looks a little more like this. The modulus of π§ minus two plus four π can be written as the modulus of π§ minus two minus four π. And we can rewrite π§ plus two plus two π as π§ minus negative two minus two π.

So our equation is as shown. And so we compare our equation to the equation in our rule. And we see that π§ one is equal to two minus four π and π§ two, our second constant complex number, is negative two minus two π. We can therefore say that the locus of π§ is the perpendicular bisector of the line segment joining two minus four π and negative two minus two π. The second part of the first part of this question asks us to find the Cartesian equation of this. So letβs clear some space.

And what weβre going to do is actually plot π§ one and π§ two on an Argand diagram. π§ one has a real part of two and an imaginary part of negative four. So it has a coordinate of two, negative four. π§ two has a real part of negative two and an imaginary part of negative two. So its coordinates are negative two, negative two. Weβre looking to find the Cartesian equation of the perpendicular bisector of the line segment between these two points. So weβll use the equation of a straight line. Itβs π¦ minus π¦ one equals π times π₯ minus π₯ one. Here, π₯ one π¦ one is a point that the line passes through and π is its slope.

The equation for π is π¦ two minus π¦ one over π₯ two minus π₯ one, change in π¦ divided by change in π₯. So weβre going to begin by finding the slope of our line segment, the line segment that joins two, negative four to negative two, negative two. Itβs negative four minus negative two over two minus negative two. Thatβs negative two over four, which is negative one-half.

Now, if we call π sub one the slope of this line segment and π sub two the slope of its perpendicular bisector, we know that the product of these two, π sub one and π sub two, is negative one. This can alternatively be written as π sub two is equal to negative one over π sub one. In other words, the slope of our perpendicular bisector is the negative reciprocal of the slope of the original line. And so the slope of our perpendicular bisector is two.

But we need to find a point that it passes through. Since itβs the perpendicular bisector of that line segment, we know it must pass through the midpoint. And we find the midpoint by essentially finding the average of our two coordinates. So thatβs negative two plus two over two, negative two plus negative four over two. That gives us a midpoint of zero, negative three. And so we substitute what we have into our formula for the equation of a straight line. We get π¦ minus negative three equals two times π₯ minus zero, which is π¦ plus three equals two π₯, and then we rearrange by subtracting three from both sides. And we find that the Cartesian equation for the locus of π§ is π¦ equals two π₯ minus three.

And so weβve answered the first part of this question. The locus of π§ is the perpendicular bisector of the line segment between two minus four π and negative two minus two π. And its Cartesian equation is π¦ equals two π₯ minus three.

And so we go back to the second part of this question. This says, what is the minimum value of the modulus of π§? Well, remember, the modulus of π§ is the distance between a point π§ and the origin. So to find the minimum value of π§, we need to find a point on the locus of π§ which is closest to the origin. Now, we know then that the line segment that joins this point to the origin will be perpendicular to our line that represents the locus of π§. And so the slope of this line will be the same as the slope of the line segment joining our original two points, joining negative two, negative two and two, negative four. And, of course, that was negative one-half.

Letβs call this point π. And then the Cartesian equation for the line that passes through the origin and π is π¦ equals negative one-half π₯. π lies at the point of intersection of our line with the line that represents the locus of π§. So we need to solve the equations π¦ equals negative one-half π₯ and π¦ equals two π₯ minus three simultaneously. We can achieve this by equating π¦. And we get negative a half π₯ equals two π₯ minus three. We add one-half π₯ and three to both sides. And we find that three is equal to five over two π₯. Then we divide through by five over two, remembering that, to divide through by a fraction, we multiply by the reciprocal of that fraction. And we get π₯ equals six-fifths.

Weβre looking to find the coordinates of π. So we substituted this value into our earlier equation π¦ equals negative a half π₯. And we get π¦ equals negative a half times six-fifths which gives us π¦ equals negative three-fifths. We can therefore say that point π has Cartesian coordinates six-fifths, negative three-fifths. The minimum value of the modulus of π§ is found by finding the distance between this point and the origin. So we use the Pythagorean theorem. Itβs the square root of the sum of the squares of six-fifths and negative three-fifths. Thatβs the square root of 36 over 25 plus nine over 25 which is the square root of 45 over 25.

Now, we know that the square root of 25 is five, so we get a fifth times root 45. But the square root of 45 can be simplified as three root five since itβs the square root of nine times the square root of five, and the square root of nine is three. And we can therefore say that the minimum value of the modulus of π§ is three root five over five.