Coordinate Geometry uses algebra to answer questions about points, lines and shapes on a plane. Starting from the Class 9 idea of abscissa and ordinate, it builds two tools: the distance formula, which comes straight from Pythagoras theorem, and the section formula, which gives the point dividing a segment internally in a given ratio, with the mid-point formula as its special case. With them you can test collinearity, name triangles and quadrilaterals, find equidistant points, and locate missing vertices of a parallelogram, all with short, checkable working.
Key Concepts
1. Coordinates, Axes and Distances Along an Axis
To locate a point on a plane we use a pair of perpendicular lines called the coordinate axes. The horizontal line is the x-axis, the vertical line is the y-axis, and they meet at the origin O(0, 0). Every point is then named by an ordered pair (x, y).
- The x-coordinate, or abscissa, is the distance of the point from the y-axis.
- The y-coordinate, or ordinate, is the distance of the point from the x-axis.
- A point on the x-axis has the form (x, 0), because it is at zero distance from the x-axis.
- A point on the y-axis has the form (0, y), because it is at zero distance from the y-axis.
The signs of the coordinates tell you the quadrant. Measured anticlockwise from the positive x-axis:
| Quadrant | Sign of x | Sign of y | Example |
|---|---|---|---|
| I | + | + | (4, 6) |
| II | − | + | (−3, 5) |
| III | − | − | (−5, −3) |
| IV | + | − | (6, −2) |
Distance along an axis. When two points lie on the same axis, their distance is found by subtraction. The NCERT chapter uses A(4, 0) and B(6, 0) on the x-axis: OA = 4 and OB = 6, so AB = 6 − 4 = 2 units. For C(0, 3) and D(0, 8) on the y-axis, CD = 8 − 3 = 5 units.
One point on each axis. A(4, 0) and C(0, 3) form a right triangle with the origin, with the right angle at O. By Pythagoras theorem AC = √(42 + 32) = √25 = 5 units. In the same way B(6, 0) and D(0, 8) give BD = √(36 + 64) = 10 units.
2. The Distance Formula
For two points that are not on an axis, drop perpendiculars to the x-axis and build a right triangle.
Worked illustration (NCERT). Take P(4, 6) and Q(6, 8), both in the first quadrant. Draw PR and QS perpendicular to the x-axis, so R is (4, 0) and S is (6, 0). Draw PT perpendicular to QS. Then RS = 6 − 4 = 2, QS = 8 and TS = PR = 6. So QT = 8 − 6 = 2 and PT = RS = 2. Triangle PTQ has a right angle at T, so PQ2 = PT2 + QT2 = 4 + 4 = 8, and PQ = 2√2 units.
Points in different quadrants (NCERT). For P(6, 4) and Q(−5, −3), the horizontal gap is 6 − (−5) = 11 and the vertical gap is 4 − (−3) = 7. So PT = 11, QT = 7 and PQ = √(112 + 72) = √(121 + 49) = √170 units. Subtracting a negative coordinate makes the gap larger.
General case. Let P(x1, y1) and Q(x2, y2) be any two points. Draw PR and QS perpendicular to the x-axis and PT perpendicular to QS. Then OR = x1 and OS = x2, so PT = RS = x2 − x1. Also SQ = y2 and ST = PR = y1, so QT = y2 − y1. Pythagoras theorem in triangle PTQ gives
PQ2 = PT2 + QT2 = (x2 − x1)2 + (y2 − y1)2
PQ = √[(x2 − x1)2 + (y2 − y1)2]
This is the distance formula. Distance is never negative, so we take only the positive square root.
- Remark 1: the distance of P(x, y) from the origin O(0, 0) is OP = √(x2 + y2).
- Remark 2: PQ = √[(x1 − x2)2 + (y1 − y2)2] also works, because (a − b)2 = (b − a)2. The order of the points does not matter.
Quick check (the two towns in NCERT Section 7.2). Town B is 36 km east and 15 km north of town A. With A as the origin and 1 km as one unit, B is (36, 15). AB = √(362 + 152) = √(1296 + 225) = √1521 = 39 km.
3. Using Distances: Triangles and Collinear Points
Once you can find the three sides of a triangle, you can name the triangle and test whether it exists at all.
| What you find | Conclusion |
|---|---|
| Sum of the two smaller distances equals the largest | Points are collinear (no triangle) |
| Sum of any two distances is greater than the third | Points form a triangle |
| Two sides equal | Isosceles triangle |
| All three sides equal | Equilateral triangle |
| Square of the longest side = sum of squares of the other two | Right triangle (converse of Pythagoras theorem) |
Worked illustration (NCERT Example 1). Do P(3, 2), Q(−2, −3) and R(2, 3) form a triangle?
- PQ = √[(3 + 2)2 + (2 + 3)2] = √(25 + 25) = √50 ≈ 7.07
- QR = √[(−2 − 2)2 + (−3 − 3)2] = √(16 + 36) = √52 ≈ 7.21
- PR = √[(3 − 2)2 + (2 − 3)2] = √(1 + 1) = √2 ≈ 1.41
The sum of any two distances is greater than the third, so P, Q, R form a triangle. Also PQ2 + PR2 = 50 + 2 = 52 = QR2, so by the converse of Pythagoras theorem ∠P = 90°. PQR is a right triangle.
Collinearity (NCERT Example 3). Ashima, Bharti and Camella sit at A(3, 1), B(6, 4) and C(8, 6). AB = √(9 + 9) = √18 = 3√2, BC = √(4 + 4) = √8 = 2√2 and AC = √(25 + 25) = √50 = 5√2. Since AB + BC = 3√2 + 2√2 = 5√2 = AC, the points are collinear, so the three girls are seated in a line.
4. Naming a Quadrilateral from Its Vertices
Find all four sides and both diagonals, then use the properties below. Always take the vertices in the order given.
| Shape | Sides | Diagonals |
|---|---|---|
| Square | All four equal | Equal |
| Rhombus (not a square) | All four equal | Unequal |
| Rectangle (not a square) | Opposite sides equal | Equal |
| Parallelogram (not a rectangle) | Opposite sides equal | Unequal |
Worked illustration (NCERT Example 2). Show that A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4) are the vertices of a square.
- AB = √[(1 − 4)2 + (7 − 2)2] = √(9 + 25) = √34
- BC = √[(4 + 1)2 + (2 + 1)2] = √(25 + 9) = √34
- CD = √[(−1 + 4)2 + (−1 − 4)2] = √(9 + 25) = √34
- DA = √[(1 + 4)2 + (7 − 4)2] = √(25 + 9) = √34
- AC = √[(1 + 1)2 + (7 + 1)2] = √(4 + 64) = √68
- BD = √[(4 + 4)2 + (2 − 4)2] = √(64 + 4) = √68
All four sides are equal and the diagonals are equal, so ABCD is a square.
Alternative method. Find the four sides and one diagonal. AD2 + DC2 = 34 + 34 = 68 = AC2, so ∠D = 90° by the converse of Pythagoras theorem. A quadrilateral with four equal sides and one right angle is a square.
Watch out. Four equal sides alone only prove a rhombus. You need equal diagonals, or one right angle, to reach a square. Also check that no three of the points are collinear: if they are, the four points do not form a quadrilateral at all.
5. Points Equidistant from Two Given Points
If P(x, y) is equidistant from A and B, then PA = PB. Square both sides to remove the roots: PA2 = PB2. The x2 and y2 terms cancel, leaving a linear equation.
Worked illustration (NCERT Example 4). Find a relation between x and y such that (x, y) is equidistant from A(7, 1) and B(3, 5).
(x − 7)2 + (y − 1)2 = (x − 3)2 + (y − 5)2
x2 − 14x + 49 + y2 − 2y + 1 = x2 − 6x + 9 + y2 − 10y + 25
−14x − 2y + 50 = −6x − 10y + 34
−8x + 8y = −16, so x − y = 2.
The graph of x − y = 2 is a line. A point equidistant from A and B lies on the perpendicular bisector of AB, so this line is the perpendicular bisector of AB. You can check: the mid-point of AB is (5, 3), and 5 − 3 = 2.
Point on an axis (NCERT Example 5). Find a point on the y-axis equidistant from A(6, 5) and B(−4, 3). A point on the y-axis is (0, y).
(6 − 0)2 + (5 − y)2 = (−4 − 0)2 + (3 − y)2
36 + 25 + y2 − 10y = 16 + 9 + y2 − 6y
61 − 10y = 25 − 6y, so 4y = 36 and y = 9.
The point is (0, 9). Check: AP = BP = √52. It is where the perpendicular bisector of AB meets the y-axis.
6. The Section Formula
A point P on segment AB divides it internally in the ratio m1 : m2 when PA : PB = m1 : m2.
The relay tower (NCERT). A tower P is to stand on the line joining towns A(0, 0) and B(36, 15), with its distance from B twice its distance from A, so P divides AB in the ratio 1 : 2. Similar triangles (AA criterion) give x/(36 − x) = 1/2 and y/(15 − y) = 1/2, so x = 12 and y = 5. The tower is at P(12, 5): AP = √(144 + 25) = 13 km and PB = √(576 + 100) = 26 km, a ratio of 1 : 2.
General derivation. Let P(x, y) divide the join of A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2. Draw AR, PS and BT perpendicular to the x-axis, and AQ and PC parallel to the x-axis. By AA similarity, ΔPAQ ~ ΔBPC, so PA/BP = AQ/PC = PQ/BC. Here AQ = x − x1, PC = x2 − x, PQ = y − y1 and BC = y2 − y. So
m1/m2 = (x − x1)/(x2 − x) = (y − y1)/(y2 − y)
Solving each equation for x and y gives the result.
P(x, y) = ((m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2))
A memory aid: cross-multiply. m1 (the part next to A) multiplies the coordinate of B, and m2 (the part next to B) multiplies the coordinate of A.
The k : 1 form. If P divides AB in the ratio k : 1, then P = ((kx2 + x1)/(k + 1), (ky2 + y1)/(k + 1)). Use this whenever the ratio is unknown, because it leaves only one unknown, k.
Worked illustration (NCERT Example 6). The point dividing the join of (4, −3) and (8, 5) internally in the ratio 3 : 1 is x = (3 × 8 + 1 × 4)/(3 + 1) = 28/4 = 7 and y = (3 × 5 + 1 × (−3))/4 = 12/4 = 3, that is (7, 3).
Finding the ratio (NCERT Example 7). In what ratio does (−4, 6) divide the join of A(−6, 10) and B(3, −8)? Let the ratio be k : 1. Then −4 = (3k − 6)/(k + 1), so −4k − 4 = 3k − 6, 7k = 2 and k = 2/7. The ratio is 2 : 7. The y-coordinate gives the same ratio.
Division by an axis (NCERT Example 9). In what ratio does the y-axis divide the join of (5, −6) and (−1, −4)? A point on the y-axis has abscissa 0. With ratio k : 1, the point is ((−k + 5)/(k + 1), (−4k − 6)/(k + 1)). Setting (−k + 5)/(k + 1) = 0 gives k = 5, so the ratio is 5 : 1. The point is (0, (−20 − 6)/6) = (0, −13/3).
Rule for axes: for the x-axis set the y-coordinate to 0; for the y-axis set the x-coordinate to 0.
External division. The NCERT chapter notes that if P lies on line AB but outside the segment, P divides AB externally. Its formula comes in higher classes, so every ratio in this chapter is internal and a correct k is positive.
7. Mid-point, Trisection and Parallelogram Problems
The mid-point divides a segment in the ratio 1 : 1. Putting m1 = m2 = 1 in the section formula gives
Mid-point of A(x1, y1) and B(x2, y2) = ((x1 + x2)/2, (y1 + y2)/2)
Points of trisection (NCERT Example 8). Points of trisection divide a segment into three equal parts. For A(2, −2) and B(−7, 4), let P and Q be the points with AP = PQ = QB.
- P divides AB in 1 : 2: P = ((1 × (−7) + 2 × 2)/3, (1 × 4 + 2 × (−2))/3) = (−3/3, 0/3) = (−1, 0).
- Q divides AB in 2 : 1: Q = ((2 × (−7) + 1 × 2)/3, (2 × 4 + 1 × (−2))/3) = (−12/3, 6/3) = (−4, 2).
Q is also the mid-point of PB: ((−1 − 7)/2, (0 + 4)/2) = (−4, 2). Same answer.
Missing vertex of a parallelogram (NCERT Example 10). The diagonals of a parallelogram bisect each other, so the mid-point of AC equals the mid-point of BD. For A(6, 1), B(8, 2), C(9, 4) and D(p, 3) taken in order: mid-point of AC = (15/2, 5/2) and mid-point of BD = ((8 + p)/2, 5/2). So (8 + p)/2 = 15/2, which gives p = 7.
Other uses. The centre of a circle is the mid-point of any diameter, so the other end of a diameter is 2 × centre − known end, coordinate by coordinate. For four equal parts, find the mid-point first, then the mid-points of the two halves.
Formula and Theorem Sheet
| Result | Statement | Remember |
|---|---|---|
| Distance formula | PQ = √[(x2 − x1)2 + (y2 − y1)2] | Comes from Pythagoras theorem; take the positive root |
| Distance from origin | OP = √(x2 + y2) | Special case with (0, 0) |
| Collinear points | AB + BC = AC | Sum of the two smaller equals the largest |
| Right triangle test | Longest2 = sum of squares of the other two | Converse of Pythagoras theorem |
| Equidistant point | PA2 = PB2 | Gives the perpendicular bisector of AB |
| Section formula | ((m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)) | PA : PB = m1 : m2, internal division |
| k : 1 form | ((kx2 + x1)/(k + 1), (ky2 + y1)/(k + 1)) | Best for finding an unknown ratio |
| Mid-point formula | ((x1 + x2)/2, (y1 + y2)/2) | Ratio 1 : 1 |
| Trisection | Ratios 1 : 2 and 2 : 1 | Second point is also the mid-point of the first point and B |
| Parallelogram | Mid-point of AC = mid-point of BD | Diagonals bisect each other |
| Area of a rhombus | (1/2) × d1 × d2 | Find the diagonals with the distance formula |
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Important Definitions
- Coordinate axes: a pair of perpendicular number lines, the x-axis and the y-axis, used to locate points on a plane.
- Abscissa (x-coordinate): the distance of a point from the y-axis.
- Ordinate (y-coordinate): the distance of a point from the x-axis.
- Collinear points: three or more points that lie on one straight line.
- Equidistant point: a point whose distances from two given points are equal. All such points lie on the perpendicular bisector of the segment joining the two points.
- Internal division: a point P on segment AB divides it internally in the ratio m1 : m2 when PA : PB = m1 : m2.
- Mid-point: the point dividing a segment in the ratio 1 : 1, found by averaging the x-coordinates and the y-coordinates.
- Points of trisection: the two points that divide a segment into three equal parts, in the ratios 1 : 2 and 2 : 1.
Solved Examples (NCERT-Based)
Example 1: Distance between pairs of points (NCERT Exercise 7.1)
Q. Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (−a, −b)
Solution.
- (i) d = √[(4 − 2)2 + (1 − 3)2] = √(4 + 4) = √8 = 2√2 units.
- (ii) d = √[(−1 + 5)2 + (3 − 7)2] = √(16 + 16) = √32 = 4√2 units.
- (iii) d = √[(−a − a)2 + (−b − b)2] = √(4a2 + 4b2) = 2√(a2 + b2) units.
Example 2: Collinear or not (NCERT Exercise 7.1)
Q. Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Solution. Let A(1, 5), B(2, 3), C(−2, −11).
- AB = √[(2 − 1)2 + (3 − 5)2] = √(1 + 4) = √5 ≈ 2.24
- BC = √[(−2 − 2)2 + (−11 − 3)2] = √(16 + 196) = √212 ≈ 14.56
- AC = √[(−2 − 1)2 + (−11 − 5)2] = √(9 + 256) = √265 ≈ 16.28
The largest is AC. AB + BC ≈ 16.80, which is not equal to AC ≈ 16.28. No sum of two distances equals the third, so the points are not collinear.
Example 3: Isosceles triangle (NCERT Exercise 7.1)
Q. Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
Solution. Let A(5, −2), B(6, 4), C(7, −2). AB = √(1 + 36) = √37, BC = √(1 + 36) = √37, AC = √(4 + 0) = 2. AB + AC > BC and the other sums also exceed the third side, so a triangle exists. Since AB = BC = √37, the triangle is isosceles.
Example 4: Name the quadrilateral (NCERT Exercise 7.1)
Q. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (−1, −2), (1, 0), (−1, 2), (−3, 0) (ii) (−3, 5), (3, 1), (0, 3), (−1, −4) (iii) (4, 5), (7, 6), (4, 3), (1, 2)
(i) A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0). AB = √(4 + 4) = 2√2, BC = √(4 + 4) = 2√2, CD = √(4 + 4) = 2√2, DA = √(4 + 4) = 2√2. AC = √(0 + 16) = 4 and BD = √(16 + 0) = 4. Four equal sides and equal diagonals: square.
(ii) A(−3, 5), B(3, 1), C(0, 3), D(−1, −4). AB = √(36 + 16) = √52 = 2√13, BC = √(9 + 4) = √13, AC = √(9 + 4) = √13. So AC + CB = √13 + √13 = 2√13 = AB, which means A, C and B are collinear. Three of the points lie on one line, so no quadrilateral is formed.
(iii) A(4, 5), B(7, 6), C(4, 3), D(1, 2). AB = √(9 + 1) = √10, BC = √(9 + 9) = √18, CD = √(9 + 1) = √10, DA = √(9 + 9) = √18. AC = √(0 + 4) = 2 and BD = √(36 + 16) = √52. Opposite sides are equal and the diagonals are unequal: parallelogram.
Example 5: Point on the x-axis (NCERT Exercise 7.1)
Q. Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).
Solution. Let the point be P(x, 0). PA2 = PB2:
(x − 2)2 + (0 + 5)2 = (x + 2)2 + (0 − 9)2
x2 − 4x + 4 + 25 = x2 + 4x + 4 + 81
−8x = 56, so x = −7.
The point is (−7, 0). Check: PA2 = 81 + 25 = 106 and PB2 = 25 + 81 = 106.
Example 6: Values of y (NCERT Exercise 7.1)
Q. Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.
Solution. PQ2 = 100, so (10 − 2)2 + (y + 3)2 = 100. Then 64 + (y + 3)2 = 100, (y + 3)2 = 36 and y + 3 = ±6. So y = 3 or y = −9. Both values are correct: there are two points, (10, 3) and (10, −9), at distance 10 from P.
Example 7: Q equidistant from P and R (NCERT Exercise 7.1)
Q. If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution. QP2 = 25 + 16 = 41. QR2 = x2 + 25. Setting them equal: x2 = 16, so x = 4 or x = −4.
- QR = √41 in both cases.
- If x = 4, R is (4, 6) and PR = √[(4 − 5)2 + (6 + 3)2] = √(1 + 81) = √82.
- If x = −4, R is (−4, 6) and PR = √[(−4 − 5)2 + (6 + 3)2] = √(81 + 81) = √162 = 9√2.
Example 8: Relation for an equidistant point (NCERT Exercise 7.1)
Q. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).
Solution. (x − 3)2 + (y − 6)2 = (x + 3)2 + (y − 4)2
x2 − 6x + 9 + y2 − 12y + 36 = x2 + 6x + 9 + y2 − 8y + 16
−6x − 12y + 45 = 6x − 8y + 25
−12x − 4y + 20 = 0, so 3x + y − 5 = 0.
Check with the mid-point (0, 5): 3(0) + 5 − 5 = 0. The mid-point lies on the line, as it must.
Example 9: Section formula and trisection (NCERT Exercise 7.2)
Q1. Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
x = (2 × 4 + 3 × (−1))/(2 + 3) = (8 − 3)/5 = 1; y = (2 × (−3) + 3 × 7)/5 = (−6 + 21)/5 = 3. The point is (1, 3).
Q2. Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Let A(4, −1), B(−2, −3). P divides AB in 1 : 2: x = (1 × (−2) + 2 × 4)/3 = 6/3 = 2; y = (1 × (−3) + 2 × (−1))/3 = −5/3. So P(2, −5/3). Q divides AB in 2 : 1: x = (2 × (−2) + 1 × 4)/3 = 0; y = (2 × (−3) + 1 × (−1))/3 = −7/3. So Q(0, −7/3).
Example 10: Finding the ratio (NCERT Exercise 7.2)
Q1. Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6).
Let the ratio be k : 1. From x: (6k − 3)/(k + 1) = −1, so 6k − 3 = −k − 1, 7k = 2 and k = 2/7. The ratio is 2 : 7. Check with y: (−8 × 2/7 + 10)/(2/7 + 1) = (54/7)/(9/7) = 6. Correct.
Q2. Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Let the ratio be k : 1. On the x-axis the y-coordinate is 0: (5k − 5)/(k + 1) = 0, so k = 1. The ratio is 1 : 1, so the point is the mid-point: ((1 − 4)/2, 0) = (−3/2, 0).
Example 11: Parallelogram and diameter (NCERT Exercise 7.2)
Q1. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Let A(1, 2), B(4, y), C(x, 6), D(3, 5). Diagonals AC and BD bisect each other, so their mid-points are equal: ((1 + x)/2, (2 + 6)/2) = ((4 + 3)/2, (y + 5)/2). Then 1 + x = 7 gives x = 6, and 8 = y + 5 gives y = 3.
Q2. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
The centre is the mid-point of AB. Let A be (x, y): (x + 1)/2 = 2 gives x = 3, and (y + 4)/2 = −3 gives y = −10. So A(3, −10).
Example 12: P with AP = (3/7) AB (NCERT Exercise 7.2)
Q. If A and B are (−2, −2) and (2, −4), respectively, find the coordinates of P such that AP = (3/7) AB and P lies on the line segment AB.
Solution. AP = (3/7) AB means PB = (4/7) AB, so AP : PB = 3 : 4. Using the section formula with m1 = 3, m2 = 4: x = (3 × 2 + 4 × (−2))/7 = (6 − 8)/7 = −2/7; y = (3 × (−4) + 4 × (−2))/7 = (−12 − 8)/7 = −20/7. So P(−2/7, −20/7).
Example 13: Four equal parts and a rhombus (NCERT Exercise 7.2)
Q1. Find the coordinates of the points which divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.
Let P, Q, R divide AB into four equal parts. Q is the mid-point of AB: (0, 5). P is the mid-point of AQ: ((−2 + 0)/2, (2 + 5)/2) = (−1, 7/2). R is the mid-point of QB: ((0 + 2)/2, (5 + 8)/2) = (1, 13/2). The points are (−1, 7/2), (0, 5) and (1, 13/2).
Q2. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order. [Hint: Area of a rhombus = (1/2) (product of its diagonals)]
Let A(3, 0), B(4, 5), C(−1, 4), D(−2, −1). AC = √[(−1 − 3)2 + (4 − 0)2] = √32 = 4√2. BD = √[(−2 − 4)2 + (−1 − 5)2] = √72 = 6√2. Area = (1/2) × 4√2 × 6√2 = (1/2) × 48 = 24 square units.
Competency-Based Questions (with answers)
1. Case-based: Three places on a town map
On a town map with 1 unit = 100 m, the school is at S(1, 2), the library at L(7, 10) and the park at P(13, 2).
(a) Find SL, LP and SP. SL = √(36 + 64) = 10; LP = √(36 + 64) = 10; SP = √(144 + 0) = 12 units.
(b) What type of triangle do the three places form? SL = LP, so the triangle is isosceles. 102 + 102 = 200 ≠ 144, so it is not right-angled.
(c) A water cooler is placed exactly halfway between the school and the park. Where is it? Mid-point of SP = ((1 + 13)/2, (2 + 2)/2) = (7, 2).
(d) How far is the library from the water cooler, in metres? Distance from (7, 10) to (7, 2) = 8 units = 800 m.
2. Case-based: Checking a garden plot
A gardener marks the corners of a plot at A(1, 1), B(5, 4), C(8, 0) and D(4, −3), in that order, with 1 unit = 1 m. She claims the plot is a square.
(a) Find the four sides. AB = √(16 + 9) = 5, BC = √(9 + 16) = 5, CD = √(16 + 9) = 5, DA = √(9 + 16) = 5 m.
(b) Find the diagonals. AC = √(49 + 1) = √50 and BD = √(1 + 49) = √50.
(c) Is she right? Yes. All four sides are equal and the diagonals are equal, so ABCD is a square.
(d) Find the area of the plot. Side2 = 52 = 25 m2.
3. Source-based: The perpendicular bisector
The NCERT chapter states: “a point which is equidistant from A and B lies on the perpendicular bisector of AB.” Take A(−1, 3) and B(5, 7).
(a) Find the relation between x and y for a point (x, y) equidistant from A and B. (x + 1)2 + (y − 3)2 = (x − 5)2 + (y − 7)2 gives 2x + 1 − 6y + 9 = −10x + 25 − 14y + 49, so 12x + 8y = 64, that is 3x + 2y = 16.
(b) Show that the mid-point of AB lies on this line. Mid-point = (2, 5). 3(2) + 2(5) = 16. It lies on the line.
(c) Find the point on the x-axis equidistant from A and B. Put y = 0: 3x = 16, so the point is (16/3, 0).
4. Assertion-Reason
Assertion (A): The point (0, −7) lies on the y-axis.
Reason (R): The x-coordinate of every point on the y-axis is zero.
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). Points on the y-axis have the form (0, y). The abscissa of (0, −7) is 0, which is exactly why it lies on the y-axis.
5. Assertion-Reason
Assertion (A): The mid-point of the segment joining (4, −1) and (−2, 5) is (1, 2).
Reason (R): The mid-point of the join of (x1, y1) and (x2, y2) is ((x1 − x2)/2, (y1 − y2)/2).
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (c). ((4 − 2)/2, (−1 + 5)/2) = (1, 2), so A is true. The mid-point formula adds the coordinates, ((x1 + x2)/2, (y1 + y2)/2), so R is false.
6. Assertion-Reason
Assertion (A): The distance between (a, b) and (−a, −b) is 2(a + b).
Reason (R): The distance of a point P(x, y) from the origin is √(x2 + y2).
Choose: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (d). The distance is √(4a2 + 4b2) = 2√(a2 + b2), which is not 2(a + b) in general. For a = 3, b = 4 it is 10, while 2(a + b) = 14. R is the true remark from the chapter.
7. Error analysis: The lost minus sign
A student finds the distance between (−3, 2) and (5, −4) as follows: “d = √[(5 − 3)2 + (−4 − 2)2] = √(4 + 36) = √40.” Find the error and give the correct answer.
Answer: The student wrote 5 − 3 instead of 5 − (−3). Subtracting a negative coordinate adds its size. Correct: d = √[(5 + 3)2 + (−4 − 2)2] = √(64 + 36) = √100 = 10 units.
8. Competency MCQ: Which quadrant
The point dividing the join of A(−4, 6) and B(8, −6) internally in the ratio 1 : 3 lies in which quadrant? (a) I (b) II (c) III (d) IV
Answer: (b). x = (1 × 8 + 3 × (−4))/4 = (8 − 12)/4 = −1; y = (1 × (−6) + 3 × 6)/4 = 12/4 = 3. The point (−1, 3) has x negative and y positive, so it is in the second quadrant.
Important Questions for Board Exams
1-Mark Questions
Q1. Find the distance of the point (−6, 8) from the origin.
OP = √(36 + 64) = √100 = 10 units.
Q2. Find the mid-point of the segment joining (2, −3) and (−4, 7).
((2 − 4)/2, (−3 + 7)/2) = (−1, 2).
Q3. What is the distance of the point (3, −4) from the x-axis?
The distance from the x-axis is the size of the ordinate: 4 units.
Q4. If the distance between (4, p) and (1, 0) is 5, find p.
9 + p2 = 25, so p2 = 16 and p = ±4.
Q5. In what ratio does the x-axis divide the join of (2, −3) and (5, 6)?
(6k − 3)/(k + 1) = 0 gives k = 1/2. The ratio is 1 : 2.
2-Mark Questions
Q6. Find the point on the y-axis which is equidistant from (5, −2) and (−3, 2).
Let P(0, y). 25 + (y + 2)2 = 9 + (y − 2)2 gives 25 + 4y + 4 = 9 − 4y + 4, so 8y = −16 and y = −2. The point is (0, −2). Check: both squared distances are 25.
Q7. Show that (−2, 3), (8, 3) and (6, 7) are the vertices of a right triangle.
AB = √(100 + 0) = 10, BC = √(4 + 16) = √20, AC = √(64 + 16) = √80. BC2 + AC2 = 20 + 80 = 100 = AB2. By the converse of Pythagoras theorem, the triangle is right-angled at C.
Q8. P(4, m) divides the join of A(2, 3) and B(6, −3). Find the ratio and m.
Let the ratio be k : 1. From x: (6k + 2)/(k + 1) = 4, so 6k + 2 = 4k + 4 and k = 1. Ratio 1 : 1, so P is the mid-point and m = (3 − 3)/2 = 0.
Q9. The mid-point of the segment joining A(x/2, (y + 1)/2) and B(x + 1, y − 3) is C(5, −2). Find x and y.
From x: (x/2 + x + 1)/2 = 5, so 3x/2 + 1 = 10 and x = 6. From y: ((y + 1)/2 + y − 3)/2 = −2, so (y + 1)/2 + y − 3 = −4. Multiply by 2: y + 1 + 2y − 6 = −8, so 3y = −3 and y = −1.
3-Mark Questions
Q10. Show that A(1, 1), B(−1, −1) and C(−√3, √3) are the vertices of an equilateral triangle.
AB2 = 4 + 4 = 8. BC2 = (−1 + √3)2 + (−1 − √3)2 = (4 − 2√3) + (4 + 2√3) = 8. AC2 = (1 + √3)2 + (1 − √3)2 = (4 + 2√3) + (4 − 2√3) = 8. All sides equal 2√2, so the triangle is equilateral.
Q11. Find the ratio in which the y-axis divides the join of A(−4, 2) and B(8, 5). Also find the point of division.
Let the ratio be k : 1. On the y-axis x = 0: (8k − 4)/(k + 1) = 0, so k = 1/2 and the ratio is 1 : 2. y = (1 × 5 + 2 × 2)/3 = 9/3 = 3. The point is (0, 3).
Q12. Find the coordinates of the points which divide the join of A(−4, 0) and B(0, 6) into four equal parts.
Mid-point Q of AB = (−2, 3). Mid-point of AQ = (−3, 3/2). Mid-point of QB = (−1, 9/2). The points are (−3, 3/2), (−2, 3) and (−1, 9/2).
5-Mark Questions
Q13. A(−2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of parallelogram ABCD. Find a and b. Find the lengths of the sides and diagonals, name the figure, and find its area.
Mid-point of AC = ((−2 + 4)/2, (1 + b)/2) = (1, (1 + b)/2). Mid-point of BD = ((a + 1)/2, 1). So a + 1 = 2 gives a = 1, and 1 + b = 2 gives b = 1. B is (1, 0) and C is (4, 1).
AB = √(9 + 1) = √10, BC = √(9 + 1) = √10, CD = √(9 + 1) = √10, DA = √(9 + 1) = √10. AC = √(36 + 0) = 6 and BD = √(0 + 4) = 2.
All sides are equal and the diagonals are unequal, so ABCD is a rhombus (not a square). Area = (1/2) × 6 × 2 = 6 square units.
Q14. P and Q trisect the segment joining A(−2, 0) and B(0, 8), with P nearer A. Find P and Q, and show that PQ = (1/3) AB.
P divides AB in 1 : 2: P = ((0 + 2 × (−2))/3, (8 + 0)/3) = (−4/3, 8/3). Q divides AB in 2 : 1: Q = ((0 + (−2))/3, (16 + 0)/3) = (−2/3, 16/3).
PQ = √[(2/3)2 + (8/3)2] = √(68/9) = (2√17)/3. AB = √(4 + 64) = √68 = 2√17. So PQ = (1/3) AB.
Common Mistakes and Examiner Tips
- Dropping the sign of a negative coordinate. 5 − (−3) is 8, not 2. Fix: write every negative coordinate in brackets before you subtract.
- Forgetting the square root. Many students stop at PQ2 = 50. Fix: finish with PQ = √50 = 5√2, unless the question only needs PQ2, as when comparing sides.
- Swapping m1 and m2 in the section formula. Fix: cross-multiply. m1 goes with B’s coordinates, m2 with A’s. A check: a ratio like 1 : 3 must give a point nearer A.
- Claiming a square from four equal sides. Four equal sides only prove a rhombus. Fix: also show equal diagonals, or one right angle by the converse of Pythagoras theorem.
- Ignoring collinear vertices. In NCERT Exercise 7.1 Q6 (ii), three points are collinear and no quadrilateral is formed. Fix: if one side equals the sum of two other lengths, check for collinearity before naming a shape.
- Using the wrong axis condition. On the x-axis the y-coordinate is zero, and on the y-axis the x-coordinate is zero. Fix: say the rule in words before you set up the equation.
- Keeping only one answer for a squared equation. (y + 3)2 = 36 gives y = 3 and y = −9. Fix: write ± and keep both values unless the question rules one out.
- Taking parallelogram vertices out of order. For ABCD, the diagonals are AC and BD. Fix: equate the mid-points of AC and BD, never AB and CD.
- Finding the ratio from x and never checking with y. Fix: substitute the ratio in the y-coordinate. If it does not match, either the arithmetic is wrong or the point is not on the segment.
- Leaving surds unsimplified in a collinearity test. 3√2 + 2√2 = 5√2 is easy to see; √18 + √8 = √50 is not. Fix: simplify each surd before adding.
Quick Revision Points
- The abscissa (x) is the distance from the y-axis; the ordinate (y) is the distance from the x-axis.
- A point on the x-axis is (x, 0); a point on the y-axis is (0, y).
- Quadrant signs: I (+, +), II (−, +), III (−, −), IV (+, −).
- Distance formula: PQ = √[(x2 − x1)2 + (y2 − y1)2]. It is Pythagoras theorem in coordinates.
- Distance from the origin: √(x2 + y2).
- Collinear: the sum of the two smaller distances equals the largest.
- Right triangle: longest side squared equals the sum of the squares of the other two.
- Square: four equal sides and equal diagonals. Rhombus: four equal sides, unequal diagonals.
- Equidistant point: set PA2 = PB2; the squares cancel and a linear relation remains.
- All points equidistant from A and B lie on the perpendicular bisector of AB.
- Section formula: ((m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)), where PA : PB = m1 : m2.
- Unknown ratio: take k : 1, solve from one coordinate, check with the other.
- Divided by the x-axis: set y = 0. Divided by the y-axis: set x = 0.
- Mid-point: ((x1 + x2)/2, (y1 + y2)/2), the section formula with ratio 1 : 1.
- Points of trisection: ratios 1 : 2 and 2 : 1.
- Parallelogram: mid-point of one diagonal = mid-point of the other.
- External division is left for higher classes; every ratio in this chapter is internal and positive.
Weightage in Board Exams
The 2026-27 NCERT chapter covers two formulas: the distance formula and the section formula, with the mid-point formula as a special case. Study from the current book, and check the current CBSE course structure for the marks given to this chapter.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Distance from the origin or an axis; mid-point; ratio in which an axis divides a segment; points on the axes |
| Short answers | Point on an axis equidistant from two points; unknown coordinate from a given distance; section formula with a given ratio |
| Longer answers | Type of triangle or quadrilateral from vertices; missing vertices of a parallelogram; trisection or four equal parts; rhombus area |
| Case-based | A map, classroom or field with positions as coordinates: find distances, a halfway point, or a point in a given ratio |
Every answer in this chapter can be checked by substituting it back. Solve every NCERT example and exercise question first, since board questions follow the same patterns.
Class 10 Maths Β· Chapter 7 β swipe through all 10 cards to understand the whole chapter.
Abscissa and Ordinate
Every point on the plane is an ordered pair (x, y).
Abscissa is x, the distance from the y-axis; ordinate is y, the distance from the x-axis.
- Quadrant signs: I (+, +), II (β, +), III (β, β), IV (+, β).
- A(4, 0) and B(6, 0) on the x-axis: AB = 6 β 4 = 2 units.
- A(4, 0) and C(0, 3): AC = β(16 + 9) = 5 units.
Distance Formula
Pythagoras theorem written in coordinates.
Take only the positive root. The order of the points does not matter.
- P(4, 6), Q(6, 8): PQ = β(4 + 4) = 2β2.
- P(6, 4), Q(β5, β3): PQ = β(121 + 49) = β170.
- Towns (0, 0) and (36, 15): β1521 = 39 km.
Distance from the Origin
When one point is O(0, 0), the formula shrinks.
Distance from the x-axis is |y|; distance from the y-axis is |x|.
- (β6, 8) is β(36 + 64) = 10 units from O.
- (a, b) and (βa, βb) are 2β(a2 + b2) apart.
- (3, β4) is 4 units from the x-axis and 3 from the y-axis.
Collinear or Triangle
Compare the three distances.
Simplify surds first: 3β2 + 2β2 = 5β2 is exact.
- Right triangle: longest2 = sum of squares of the other two.
- P(3, 2), Q(β2, β3), R(2, 3): 50 + 2 = 52, so β P = 90Β°.
- (5, β2), (6, 4), (7, β2): two sides β37, isosceles.
Naming the Shape
Find four sides and both diagonals, vertices in order.
Four equal sides alone prove only a rhombus.
- Square: all sides equal, diagonals equal.
- Rhombus: all sides equal, diagonals unequal.
- (β3, 5), (3, 1), (0, 3), (β1, β4): three collinear, no quadrilateral.
Equidistant Points
Set the squared distances equal and the squares cancel.
The result is the perpendicular bisector of AB.
- Equidistant from (7, 1) and (3, 5): x β y = 2.
- On the y-axis from (6, 5) and (β4, 3): (0, 9).
- On the x-axis from (2, β5) and (β2, 9): (β7, 0).
Section Formula
The point P with PA : PB = m1 : m2, internally.
Cross-multiply: m1 with B, m2 with A. Proved by AA similarity.
- (4, β3) and (8, 5) in 3 : 1 gives (7, 3).
- (β1, 7) and (4, β3) in 2 : 3 gives (1, 3).
- Relay tower, 1 : 2 from (0, 0) to (36, 15): (12, 5).
The k : 1 Method
Let the ratio be k : 1 so only one unknown remains.
Solve with one coordinate, check with the other.
- (β4, 6) on A(β6, 10), B(3, β8): k = 2/7, ratio 2 : 7.
- y-axis on (5, β6), (β1, β4): x = 0 gives 5 : 1, point (0, β13/3).
- x-axis on (1, β5), (β4, 5): y = 0 gives 1 : 1, point (β3/2, 0).
Mid-point and Trisection
Ratio 1 : 1 gives the average of the coordinates.
Trisection uses 1 : 2 and 2 : 1; four equal parts use mid-points twice.
- Trisect A(2, β2), B(β7, 4): (β1, 0) and (β4, 2).
- Centre (2, β3), end B(1, 4): other end A(3, β10).
- A(β2, 2), B(2, 8) in four parts: (β1, 7/2), (0, 5), (1, 13/2).
Diagonals Bisect Each Other
Mid-point of AC equals mid-point of BD.
Take the vertices in the given order before pairing diagonals.
- A(6, 1), B(8, 2), C(9, 4), D(p, 3): p = 7.
- (1, 2), (4, y), (x, 6), (3, 5): x = 6, y = 3.
- Rhombus (3, 0), (4, 5), (β1, 4), (β2, β1): area Β½ Γ 4β2 Γ 6β2 = 24.
π Practice Coordinate Geometry - 10 board questions
CBSE previous-year and competency-based Β· with answers & explanations
Start βClose β
Why A: Step 1 (distance formula, one point on each axis): dΒ² = (a cos ΞΈ + b sin ΞΈ)Β² + (a sin ΞΈ β b cos ΞΈ)Β². Step 2 (expand): aΒ² cosΒ²ΞΈ + 2ab sin ΞΈ cos ΞΈ + bΒ² sinΒ²ΞΈ + aΒ² sinΒ²ΞΈ β 2ab sin ΞΈ cos ΞΈ + bΒ² cosΒ²ΞΈ; the middle terms cancel. Step 3 (sinΒ²ΞΈ + cosΒ²ΞΈ = 1): dΒ² = aΒ²(cosΒ²ΞΈ + sinΒ²ΞΈ) + bΒ²(sinΒ²ΞΈ + cosΒ²ΞΈ) = aΒ² + bΒ², so d = β(aΒ² + bΒ²).
Why not B: aΒ² β bΒ² has a sign slip and also skips the square root.
Why not C: β(aΒ² β bΒ²) comes from a sign slip while expanding the second bracket.
Why not D: aΒ² + bΒ² is dΒ², the square of the distance.
Remember: after the distance formula, look for pairs that make sinΒ²ΞΈ + cosΒ²ΞΈ = 1.
Why A: Step 1 (section formula in the ratio k : 1): the dividing point has x = (10k + (β4))/(k + 1). Step 2 (point on the y-axis has x = 0): 10k β 4 = 0, so k = 2/5 and the ratio is 2/5 : 1 = 2 : 5.
Why not B: 1 : 2 is not what the condition x = 0 gives.
Why not C: 2 : 1 is not what the condition x = 0 gives.
Why not D: 5 : 2 is the correct ratio read from Q’s end, the reverse of PQ order.
Remember: the y-axis cuts where x = 0, so the ratio is |x of P| : |x of Q| = 4 : 10 = 2 : 5.
Why A: Step 1 (mid-point formula): ((5 + 6)/2, (β4 + 4)/2) = (11/2, 0). Step 2 (points on the x-axis have y = 0): the y-coordinate is 0, so the mid-point lies on the x-axis.
Why not B: points on the y-axis have x = 0; here x = 11/2.
Why not C: the origin needs both coordinates 0, but x = 11/2.
Why not D: that comes from subtracting instead of adding, which gives (1/2, 4).
Remember: y = 0 means on the x-axis, x = 0 means on the y-axis.
Why B: Name the points P(β5, 0), Q(5, 0), R(0, 4). Step 1 (distance formula): PQ = 10, PR = β(25 + 16) = β41, QR = β(25 + 16) = β41. Step 2 (compare sides): exactly two sides are equal, so the triangle is isosceles. Step 3 (Pythagoras check): 41 + 41 = 82 β 100, so there is no right angle.
Why not A: the squares of the two shorter sides add to 82, not to 100.
Why not C: PQ = 10 is not equal to β41.
Why not D: two sides are equal, so it is not scalene.
Remember: R is on the y-axis, the perpendicular bisector of PQ, so it is equidistant from P and Q.
Why D: Step 1 (distance formula): diameter = β[(5 β 5)Β² + (2 β (β2))Β²] = β16 = 4. Step 2 (radius is half the diameter): radius = 4/2 = 2.
Why not A: a length is never negative, so Β± 2 is not a length.
Why not B: Β± 4 is the diameter, and with a negative sign that no length can have.
Why not C: 4 is the diameter, not the radius.
Remember: a distance comes from a square root and is always non-negative; radius is half the diameter.
Why B: Step 1 (vertices in order): in AOBC, A and B are opposite corners, so AB is a diagonal. Step 2 (distance formula squared): ABΒ² = (4 β 0)Β² + (0 β 2)Β² = 16 + 4 = 20.
Why not A: 36 = (4 + 2)Β² adds the sides before squaring.
Why not C: 16 is only the square of side OB.
Why not D: 4 is only the square of side OA.
Remember: diagonalΒ² of a rectangle = lengthΒ² + breadthΒ².
Why C: Take mβ = 2 (next to P) and mβ = 1. Step 1 (section formula, x): (2 Γ 5 + 1 Γ (β1))/3 = 9/3 = 3. Step 2 (section formula, y): (2 Γ 2 + 1 Γ 5)/3 = 9/3 = 3. So the point is (3, 3).
Why not A: (3, β3) has the wrong sign on y; 2 Γ 2 + 1 Γ 5 = 9 is positive.
Why not B: (5, 5) mixes Q’s x-coordinate with P’s y-coordinate instead of using the formula.
Why not D: (5, 1) matches no correct working of the formula in either ratio order.
Remember: a point dividing PQ in 2 : 1 is closer to Q, so it should lie two-thirds of the way from P.
Why B: Step 1 (mid-point formula): ((β4 + 4)/2, (5 + 6)/2) = (0, 11/2). Step 2 (points on the y-axis have x = 0): the x-coordinate is 0, so the mid-point lies on the y-axis.
Why not A: points on the x-axis have y = 0; here the y-coordinate is 11/2.
Why not C: the origin needs both coordinates 0, but y = 11/2.
Why not D: that comes from subtracting instead of adding, which gives (β4, β1/2).
Remember: x = 0 means on the y-axis, y = 0 means on the x-axis.
Why B: Step 1 (a point on the x-axis has y = 0): putting y = 0 gives x = a, so the line meets the x-axis at (a, 0); putting x = 0 gives y = b, so it meets the y-axis at (0, b). Step 2 (area of a right triangle): the triangle has vertices O(0, 0), (a, 0) and (0, b), with legs a and b along the axes and the right angle at O, so its area is (1/2) Γ a Γ b = (1/2)ab.
Why not A: ab is the area of the rectangle on those two legs, not of the triangle.
Why not C: (1/4)ab halves the area twice.
Why not D: 2ab doubles the rectangle instead of halving it.
Remember: x/a + y/b = 1 cuts the x-axis at a and the y-axis at b, and the triangle with the axes has area half of ab.
Why B: Step 1 (meaning of ordinate): the y-coordinate tells how far the point is above or below the x-axis. Step 2 (distance is non-negative): the y-coordinate is 7, so the point is |7| = 7 units from the x-axis.
Why not A: β1 is the abscissa, which measures distance from the y-axis, and a distance cannot be negative anyway.
Why not C: 6 comes from 7 β 1, mixing the two coordinates, which has no geometric meaning here.
Why not D: β50 = β(1 + 49) is the distance from the origin, not from the x-axis.
Remember: distance from the x-axis is |y|, distance from the y-axis is |x|.
Chapter Navigation
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Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
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Frequently Asked Questions
The distance between P(xβ, yβ) and Q(xβ, yβ) is β[(xβ β xβ)Β² + (yβ β yβ)Β²]. Drop perpendiculars from P and Q to the x-axis and draw a horizontal line from P. This makes a right triangle with legs xβ β xβ and yβ β yβ and hypotenuse PQ, so Pythagoras theorem gives the formula. Only the positive root is taken because distance is never negative.
Find all three distances between the points. If the sum of the two smaller distances equals the largest one, the points lie on one line. For A(3, 1), B(6, 4) and C(8, 6), AB = 3β2, BC = 2β2 and AC = 5β2, and 3β2 + 2β2 = 5β2, so they are collinear. Simplify surds first so the addition is exact.
Find all four sides and both diagonals. A rhombus has four equal sides; a square also has equal diagonals. So show all sides equal and AC = BD. Another route is four equal sides plus one right angle, found by the converse of Pythagoras theorem, for example ADΒ² + DCΒ² = ACΒ². Also check that no three vertices are collinear.
If P divides the join of A(xβ, yβ) and B(xβ, yβ) internally in the ratio mβ : mβ, meaning PA : PB = mβ : mβ, then P = ((mβxβ + mβxβ)/(mβ + mβ), (mβyβ + mβyβ)/(mβ + mβ)). The ratio parts cross-multiply: mβ goes with B and mβ with A. It is proved using AA similarity of two right triangles.
Take the ratio as k : 1, so the dividing point is ((kxβ + xβ)/(k + 1), (kyβ + yβ)/(k + 1)). Equate one coordinate to the known value and solve for k. For the x-axis, set the y-coordinate to 0; for the y-axis, set the x-coordinate to 0. Then check the other coordinate with the same k.
The diagonals of a parallelogram bisect each other, so the mid-point of AC equals the mid-point of BD when the vertices are taken in order. For A(6, 1), B(8, 2), C(9, 4) and D(p, 3), the mid-point of AC is (15/2, 5/2) and of BD is ((8 + p)/2, 5/2), which gives p = 7.
Let the point be P(x, y) and write PAΒ² = PBΒ² so the square roots disappear. The xΒ² and yΒ² terms cancel, leaving a linear relation, which is the perpendicular bisector of AB. If the point must lie on an axis, use (x, 0) or (0, y) instead. For A(6, 5) and B(β4, 3), the point on the y-axis is (0, 9).
The mid-point divides a segment in the ratio 1 : 1, so putting mβ = mβ = 1 in the section formula gives ((xβ + xβ)/2, (yβ + yβ)/2). It is used to find the centre of a circle from the ends of a diameter, to find a missing vertex of a parallelogram because its diagonals bisect each other, and to split a segment into four equal parts.
