Example

Let \(P(a,b)\) and \(Q(c,d)\) be two points in the plane. Find the equation of the line \(l\) that is the perpendicular bisector of the line segment \(PQ\).

Solution

First assume that \(a \ne c\) and \(b \ne d\). The gradient of \(PQ\) is \(\dfrac{b-d}{a-c}\), and so the gradient of the perpendicular line \(l\) is

\[ m = \dfrac{c-a}{b-d}. \]

The line \(l\) goes through the midpoint \(\Bigl(\dfrac{a+c}{2},\dfrac{b+d}{2}\Bigr)\) of \(PQ\). Thus the equation of \(l\) is

\begin{align*} y-\dfrac{b+d}{2} &= m\Bigl(x-\dfrac{a+c}{2}\Bigr)\\ (b-d)(2y-b-d) &= (c-a)(2x-a-c)\\ 2(b-d)y+(d^2-b^2) &= 2(c-a)x+(a^2-c^2)\\ 2(b-d)y+2(a-c)x &= a^2+b^2-c^2-d^2. \end{align*}

The final equation above is a sensible symmetric form for the equation of the perpendicular bisector of \(PQ\). This equation is also correct in the case that \(a = c\) or \(b = d\).

Alternatively, the question can be answered by finding the locus of all points \(X(x,y)\) equidistant from \(P(a,b)\) and \(Q(c,d)\).

Detailed description

\begin{align*} & PX = QX\\ \iff \ {}& PX^2 = QX^2\\ \iff \ {}& (x-a)^2+(y-b)^2 = (x-c)^2+(y-d)^2\\ \iff \ {}& -2ax-2by+a^2+b^2 = -2cx-2dy+c^2+d^2\\ \iff \ {}& 2(b-d)y+2(a-c)x = a^2+b^2-c^2-d^2. \end{align*}

Note that the locus method involves less algebra.

Example

What is the locus of all points distance \(d\) from the fixed interval \(AB\)?

Solution

This is not meant to be a problem in coordinate geometry!

The locus is two intervals parallel to \(AB\) together with two semicircles of radius \(d\) with centres \(A\) and \(B\).

Detailed description

Example

What is the area of the triangle with vertices \(A(x_1,y_1)\), \(B(x_2,y_2)\) and \(C(x_3,y_3)\)? Assume that \(C\) is above the interval \(AB\) as shown.

Detailed description

Solution

Let \(P(x_1,0)\), \(Q(x_2,0)\) and \(R(x_3,0)\) be the projections of \(A\), \(B\) and \(C\) onto the \(x\)-axis. We express the area of the triangle \(ABC\) in terms of three trapezia:

\begin{align*} \text{Area } \triangle ABC &= \text{Area } APRC + \text{Area } CRQB - \text{Area } APQB\\ &=\dfrac{1}{2}\bigl((y_1+y_3)(x_3-x_1)+(y_3+y_2)(x_2-x_3)-(y_1+y_2)(x_2-x_1)\bigr)\\ &=\dfrac{1}{2}\bigl(y_1(x_3-x_1-x_2+x_1)+y_2(x_2-x_3-x_2+x_1)+y_3(x_3-x_1+x_2-x_3)\bigr)\\ &=\dfrac{1}{2}\bigl(y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)\bigr). \end{align*}

Example

Use coordinate geometry to show that the altitudes of a triangle \(ABC\) are concurrent (at the orthocentre of the triangle).

Solution

By choosing the \(x\)-axis to be the line \(AB\) and the \(y\)-axis to be the line \(OC\), we can assume \(A(a,0)\), \(B(b,0)\) and \(C(0,c)\) are the vertices of the triangle \(ABC\), as shown in the following diagram.

The gradient of \(BC\) is \(-\dfrac{c}{b}\), so the gradient of the altitude through \(A\) is \(\dfrac{b}{c}\). Thus the equation of the altitude through \(A\) is

\[ y = \dfrac{b}{c}(x-a). \]

To find the \(y\)-intercept of the altitude, take \(x=0\), which implies \(y=-\dfrac{ba}{c}\). By swapping \(a\) and \(b\), we get \(-\dfrac {ab}{c}=-\dfrac{ba}{c}\). So the altitude through \(B\) has the same \(y\)-intercept. The altitude through \(C\) is the \(y\)-axis. Thus \(H=\Bigl(0,-\dfrac{ab}{c}\Bigr)\) lies on all three altitudes.