Notes · Unit 7
Coordinate Geometry
Twelve short slides on the coordinate plane, the distance and mid-point formulas, collinear points, and identifying triangles and quadrilaterals.
Press Next (or the right arrow key) to move on. On most slides, each press reveals one more idea.
Coordinate planeDistance formulaMid-pointCollinear pointsTriangles and quadrilaterals
1 · The coordinate plane
Every point has an address
A point is written (x, y). x is the abscissa (first coordinate) and y is the ordinate (second coordinate). The axes meet at the origin (0, 0).
On the axes. Every point on the x-axis has y = 0, so it looks like (x, 0). Every point on the y-axis has x = 0. So (0, −5) lies on the y-axis.
Order matters. (2, 3) and (3, 2) are different points. They are the same only when x = y.
Quadrant I (+, +)
Quadrant II (−, +) e.g. (−3, 4)
Quadrant III (−, −)
Quadrant IV (+, −)
2 · Distance on a number line
Subtract, then take the absolute value
d(A, B) = |b − a|
Example. Points −3 and 5: |5 − (−3)| = |5 + 3| = 8.
Reversing the points only changes the sign inside the bars, so d(A, B) = d(B, A). A distance is never negative.
Two points with the same y-coordinate lie on a horizontal line, so only the x-coordinates matter.
3 · The distance formula
Pythagoras on the plane
The horizontal and vertical differences are the two legs of a right-angled triangle. The distance is its hypotenuse.
d = √((x2 − x1)2 + (y2 − y1)2)
(0, 0) to (3, 4): √(9 + 16) = 5
(2, −1) to (14, 4): √(122 + 52) = √169 = 13
(5, 3) to (−1, −5): √(36 + 64) = 10. And (−8, 15) from the origin: √(64 + 225) = 17.
Either point can come first. Squaring removes the sign, so reversing the points gives the same result.
Unknown coordinate. The distance between (1, 2) and (4, k) is 5 and k > 2. Then 9 + (k − 2)2 = 25, so k − 2 = ±4. Since k > 2, k = 6.
4 · Collinear points
When three points lie on one line
If B lies on the segment AC, the two parts add up to the whole: AB + BC = AC.
Example. AB = √5, BC = √5 and AC = 2√5. Since √5 + √5 = 2√5, the three points are on one line.
Not collinear. If AB + BC > AC, the points form a triangle. For AB = 3, BC = 4 and AC = 5, the sum 7 is greater than 5, so they are not collinear.
5 · The mid-point
Average the coordinates
M = ((x1 + x2)/2, (y1 + y2)/2)
(4, 2) and (10, 8): ((4 + 10)/2, (2 + 8)/2) = (7, 5)
(−3, 4) and (5, −2): ((−3 + 5)/2, (4 − 2)/2) = (1, 1)
Finding an end point. M(2, 3) is the mid-point of A(−1, 5) and B(x, y).
(−1 + x)/2 = 2 gives x = 5. (5 + y)/2 = 3 gives y = 1. So B = (5, 1).
(−1 + x)/2 = 2 gives x = 5. (5 + y)/2 = 3 gives y = 1. So B = (5, 1).
The mid-point splits a segment into two equal parts, so LM : MN = 1 : 1.
6 · Mid-point in action
Centres and bisectors
Centre of a circle. The centre is the mid-point of any diameter. For a diameter with ends (4, 5) and (6, 9), the centre is (5, 7).
Bisectors. Every bisector of a segment cuts it into two equal parts, so all bisectors pass through its mid-point.
Bisecting each other. Two segments that bisect each other are cut at the same point, their common mid-point. That is why the diagonals of a parallelogram share a mid-point.
Mid-point theorem. The segment joining the mid-points of two sides of a triangle is parallel to the third side and half its length.
7 · Triangles from coordinates
Find the three sides, then decide
Isosceles: two equal sides. Equilateral: three equal sides.
Right-angled: the squares of two sides add up to the square of the third (converse of Pythagoras theorem).
Example. Vertices (0, 0), (4, 0), (2, 3). The sides are 4, √13 and √13. Two sides are equal, so it is isosceles. It is not right-angled, because 42 = 16 but 13 + 13 = 26.
Perimeter. Vertices (0, 0), (3, 0), (0, 4). The sides are 3, 4 and √(9 + 16) = 5, so the perimeter is 12.
8 · Quadrilaterals from coordinates
What to show for each shape
Parallelogram. Opposite sides parallel and equal. The diagonals bisect each other.
Rectangle. A parallelogram with equal diagonals. The diagonals of a rectangle are equal and bisect each other.
Rhombus. It is enough to show that all four sides are equal.
Trapezium. Exactly one pair of parallel sides.
9 · Common mistakes
Four traps to avoid
Adding the differences. The distance between (1, 2) and (4, 6) is not 3 + 4 = 7. It is the hypotenuse, √(32 + 42) = 5.
Forgetting to take the square root in the distance formula.
Mixing up the coordinates. The abscissa is x and the ordinate is y.
Keeping both roots when a condition rules one out, such as k > 2 above.
10 · Quick check
Try these, then tap to check
11 · Your turn
Practise Unit 7
Three timed MCQ tests with an explanation after every answer.
Practice questions are independently written and are not official FBISE papers.