Geometry with coordinates
Tonight's idea
A point is a pair of numbers, so every geometry statement becomes an algebra statement. Four tools do almost all the work.
Distance. From \((x_1, y_1)\) to \((x_2, y_2)\), go across \(x_2 - x_1\) and up \(y_2 - y_1\). Those are the legs of a right triangle, so the distance is the hypotenuse: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
Midpoint. Average the coordinates: \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
Slope. If \(x_2\) differs from \(x_1\), \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\), rise over run; vertical lines have no slope. Non-vertical parallel lines have equal slopes. Perpendicular non-vertical lines have slopes whose product is \(-1\). Horizontal and vertical lines are perpendicular too. For nonzero \(m\), rotate the direction “across 1, up \(m\)” a quarter turn: it becomes “across \(-m\), up 1,” with slope \(-\dfrac1m\).
Circle. The points at distance \(r\) from the centre \((h, k)\). Write that with the distance formula and square both sides:
\((x - h)^2 + (y - k)^2 = r^2\)
Handed a circle in expanded form, complete the square in \(x\) and in \(y\) separately to read off the centre and radius. To prove a shape has a property: put coordinates on it, compute, conclude. Choose coordinates to make the arithmetic kind: a vertex at the origin, a side along an axis.
Watch one: is there a right angle at \(B\)?
\(A(1, 2)\), \(B(4, 6)\), \(C(8, 3)\). Slope of \(AB\): \(\frac{6 - 2}{4 - 1} = \frac43\). Slope of \(BC\): \(\frac{3 - 6}{8 - 4} = -\frac34\). Product: \(-1\). So \(AB \perp BC\): a right angle at \(B\). Check with lengths: \(AB = 5\), \(BC = 5\), and the hypotenuse should be \(5\sqrt2\). Indeed \(AC = \sqrt{7^2 + 1^2} = \sqrt{50} = 5\sqrt2\). Everything agrees.
Your turn
- \(P(-3, 4)\) and \(Q(5, -2)\). Find the length \(PQ\), the midpoint \(M\), the slope of \(PQ\), and the equation of the perpendicular bisector of \(PQ\) in the form \(ax + by + c = 0\).
- The circle \(x^2 + y^2 - 6x + 4y - 12 = 0\). Complete the square to find its centre and radius. Show that \((7, 1)\) lies on it, then find the tangent line at \((7, 1)\). (Hint: the tangent is perpendicular to the radius.)
- Find the point on the \(x\)-axis that is the same distance from \((2, 5)\) as from \((6, -1)\). Check your answer by computing both distances.
- Triangle with vertices \((0,0)\), \((6,0)\), \((0,8)\). Find the centre and radius of the circle through all three vertices, and write its equation. Which circle theorem told you where to look?
Stretch: a theorem for the price of a good coordinate choice
Prove that the diagonals of any parallelogram bisect each other. Place it as \(A(0,0)\), \(B(a, 0)\), \(D(b, c)\), work out where \(C\) must be, and compare the midpoints of \(AC\) and \(BD\).
Answers
- \(PQ = \sqrt{8^2 + 6^2} = 10\). \(M = (1, 1)\). Slope \(= \frac{-6}{8} = -\frac34\). Perpendicular slope \(\frac43\), through \((1,1)\): \(y - 1 = \frac43(x - 1)\), so \(4x - 3y - 1 = 0\).
- \((x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4\), so \((x - 3)^2 + (y + 2)^2 = 25\): centre \((3, -2)\), radius 5. At \((7, 1)\): \(4^2 + 3^2 = 25\). Yes. Radius slope \(\frac{1 - (-2)}{7 - 3} = \frac34\), so tangent slope \(-\frac43\): \(y - 1 = -\frac43(x - 7)\), or \(4x + 3y = 31\).
- Call it \((x, 0)\). \((x - 2)^2 + 25 = (x - 6)^2 + 1\). Expand: \(x^2 - 4x + 29 = x^2 - 12x + 37\), so \(8x = 8\), \(x = 1\). The point is \((1, 0)\). Check: distances squared are \(1 + 25 = 26\) and \(25 + 1 = 26\).
- The right angle is at the origin, so the hypotenuse from \((6,0)\) to \((0,8)\) is a diameter (converse of Thales). Centre is its midpoint \((3, 4)\), radius half its length \(= 5\). Equation \((x - 3)^2 + (y - 4)^2 = 25\). Check \((0,0)\): \(9 + 16 = 25\).
Stretch. \(C\) is reached from \(B\) by the same move that takes \(A\) to \(D\), so \(C = (a + b, c)\). Midpoint of \(AC\): \(\left(\frac{a + b}{2}, \frac c2\right)\). Midpoint of \(BD\): \(\left(\frac{a + b}{2}, \frac c2\right)\). Same point, so each diagonal passes through the other's midpoint: they bisect each other.