Coordinate Geometry

Abscissa

The x-coordinate of a point, equal to its distance from the y-axis. On the x-axis the ordinate is zero, so every point there has the form (x, 0).

Ordinate

The y-coordinate of a point, equal to its distance from the x-axis. On the y-axis the abscissa is zero, so every point there has the form (0, y).

Distance formula

The distance between P(x₁, y₁) and Q(x₂, y₂) is PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]. It comes from applying Pythagoras theorem to the right triangle whose legs are the horizontal and vertical gaps between the points.

Distance from the origin

The distance of P(x, y) from O(0, 0) is OP = √(x² + y²), the distance formula with one point at the origin.

Why does it not matter whether you use (x₂ − x₁) or (x₁ − x₂) in the distance formula, and why is only the positive root taken?

Each difference is squared, and a number and its negative have the same square, so both orders give the same value. The positive square root is taken because a distance can never be negative.

How do you check whether (3, 2), (−2, −3) and (2, 3) form a triangle and name its type?

Find all three distances: PQ = √50, QR = √52 and PR = √2. Any two add to more than the third, so the points form a triangle. Since PQ² + PR² = QR², the converse of Pythagoras theorem gives a right angle at P.

How do you use the distance formula to show that four points are the vertices of a square?

Find all four sides and both diagonals. In the chapter's example each side is √34 and each diagonal √68; equal sides with equal diagonals make a square. Alternatively, four equal sides with AD² + DC² = AC² give a right angle, which also proves it.

Collinear points

Points that lie on one straight line. With the distance formula, three points are collinear when two of the distances add up to the third, as with AB = 3√2, BC = 2√2 and AC = 5√2 in the chapter's classroom example.

How do you find a relation between x and y so that P(x, y) is equidistant from (7, 1) and (3, 5)?

Set AP² = BP², giving (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)², and expand. The x² and y² terms cancel, leaving x − y = 2. This line is the perpendicular bisector of AB, where every equidistant point must lie.

How do you find the point on the y-axis that is equidistant from A(6, 5) and B(−4, 3)?

Any point on the y-axis is (0, y), so write AP² = BP²: 6² + (5 − y)² = (−4)² + (3 − y)². The y² terms cancel, giving 4y = 36 and y = 9, so the point is (0, 9). Check that both distances come to √52.

Section formula

The point dividing the segment from A(x₁, y₁) to B(x₂, y₂) internally in the ratio m₁ : m₂ is ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)). Here PA : PB = m₁ : m₂.

Section formula for a ratio k : 1

The point dividing the join of A(x₁, y₁) and B(x₂, y₂) in the ratio k : 1 is ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1)). Using one unknown k makes it easy to find an unknown ratio.

Midpoint formula

The mid-point of the join of A(x₁, y₁) and B(x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). It is the section formula with the ratio 1 : 1.

How do you find the ratio in which (−4, 6) divides the segment joining A(−6, 10) and B(3, −8)?

Let the ratio be k : 1 and equate the x-coordinate from the section formula: −4 = (3k − 6)/(k + 1). This gives 7k = 2, so the ratio is 2 : 7. Then check that the y-coordinate also comes out as 6.

How do you find the ratio in which the y-axis divides the segment joining (5, −6) and (−1, −4)?

Take the ratio as k : 1; the dividing point's x-coordinate is (−k + 5)/(k + 1). Any point on the y-axis has abscissa 0, so −k + 5 = 0, k = 5 and the ratio is 5 : 1. Putting k = 5 into the y-coordinate gives the point (0, −13/3).

How do you find the points of trisection of the segment joining A(2, −2) and B(−7, 4)?

The trisection points P and Q split AB into three equal parts, so P divides it in the ratio 1 : 2 and Q in the ratio 2 : 1. The section formula then gives P(−1, 0) and Q(−4, 2); Q is also the mid-point of PB.

If A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are vertices of a parallelogram in order, how do you find p?

The diagonals of a parallelogram bisect each other, so AC and BD share a mid-point. Equate the x-coordinates of the two mid-points: (6 + 9)/2 = (8 + p)/2, giving p = 7.

What is the difference between a point dividing a segment internally and externally?

Internal division means the point lies between A and B on the segment, which is the case the section formula here covers. If the point lies on line AB but outside the segment, it divides AB externally; that case is left for higher classes.

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