Coordinate Geometry Class 9: notes and important questions

Coordinate Geometry: revision notes

1. The Cartesian plane

Two perpendicular number lines, the \(x\)-axis (horizontal) and the \(y\)-axis (vertical), meet at the origin \(O(0, 0)\). A point is written \(P(x, y)\): \(x\) is the abscissa (signed distance from the \(y\)-axis) and \(y\) the ordinate (signed distance from the \(x\)-axis). The order matters: \((2, 5)\) and \((5, 2)\) are different points.

2. Quadrants and the axes

• Quadrant I \((+, +)\), II \((-, +)\), III \((-, -)\), IV \((+, -)\), counted anticlockwise from the positive \(x\)-axis.
• A point on the \(x\)-axis has \(y = 0\); a point on the \(y\)-axis has \(x = 0\). Points on an axis are in no quadrant.
• The distance of \(P(x, y)\) from the \(x\)-axis is \(|y|\), and from the \(y\)-axis is \(|x|\).

3. Distance formula

For \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the horizontal and vertical legs are \(|x_2 - x_1|\) and \(|y_2 - y_1|\), so by the Baudhayana-Pythagoras theorem

\[AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

• Distance from the origin: \(OP = \sqrt{x^2 + y^2}\).
• Points on a horizontal line: \(AB = |x_2 - x_1|\); on a vertical line: \(AB = |y_2 - y_1|\).

4. Mid-point of a segment

The mid-point of \(AB\) is \(M\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\): the average of the \(x\)-coordinates and the average of the \(y\)-coordinates.

5. Using the formulas in geometry

• Isosceles / equilateral triangle: compare the three side lengths.
• Right angle: check whether the squares of two sides add up to the square of the third.
• Parallelogram: the diagonals bisect each other, so the diagonals have the same mid-point. Rhombus: all sides equal; rectangle: diagonals equal too; square: all sides equal and diagonals equal.
• Equidistant points: write \(PA^2 = PB^2\) (squaring removes the roots).

Worked example 1

Find the distance between \(A(-1, 3)\) and \(B(5, -5)\).

\(AB = \sqrt{6^2 + (-8)^2} = \sqrt{100} = 10\).

Worked example 2

Find the point on the \(x\)-axis that is equidistant from \(A(1, 4)\) and \(B(5, 2)\).

Take \(P(x, 0)\). \(PA^2 = PB^2\): \((x - 1)^2 + 16 = (x - 5)^2 + 4 \Rightarrow 8x = 12 \Rightarrow x = \dfrac32\). So \(P\left(\dfrac32, 0\right)\).

Worked example 3

\(A(1, 2)\), \(B(5, 3)\), \(C(6, 7)\) are three vertices of parallelogram \(ABCD\). Find \(D\).

Mid-point of \(AC\) is \(\left(\dfrac72, \dfrac92\right)\), which is also the mid-point of \(BD\): \(D = (7 - 5, 9 - 3) = (2, 6)\).

Common errors

• Writing \((y, x)\) instead of \((x, y)\), or reading the ordinate as the distance from the \(y\)-axis.
• Squaring a negative difference wrongly: \((-8)^2 = 64\), not \(-64\).
• Stopping at \(\sqrt{52}\) instead of simplifying to \(2\sqrt{13}\), or "simplifying" \(\sqrt{a^2 + b^2}\) to \(a + b\).
• Forgetting the second value when an equation such as \((a - 5)^2 = 9\) gives \(a - 5 = \pm 3\).
• Calling a four-sided figure a square after checking only the four sides (a rhombus has equal sides too): check the diagonals.

Exam tips

• Write the formula first, then substitute with brackets around negative numbers.
• For "show that" questions, finish with a sentence that states the property and the reason.
• Work with squared distances when comparing lengths; take square roots only at the end.