Coordinate Geometry

Coordinate Geometry

Coordinate Geometry carries 6 marks in every board paper: the 2026-27 sample paper and the 2025 and both 2026 papers all gave it exactly 6.

The marks usually come as one or two 1-mark MCQs (a distance, a mid-point, the fourth vertex of a parallelogram, or an Assertion-Reason item), a 2- or 3-mark question on the section formula (often "in what ratio does the x-axis or y-axis divide the segment"), and sometimes a 4-mark case study on a city map drawn on a grid. There has been no 5-mark question from this chapter. Only the distance formula and the section formula (internal division) are in the syllabus; the area of a triangle is not.

Where marks are usually lost:

• sign slips inside the distance formula, such as $(2 - (-3))$ written as $(2 - 3)$;
• putting the ratio the wrong way round in the section formula, or using $m_1 x_1 + m_2 x_2$ in place of $m_1 x_2 + m_2 x_1$;
• setting the wrong coordinate to zero: a point on the x-axis has y = 0;
• calling four equal sides a square without checking that the diagonals are equal.

Revise in 5 Minutes

Formulas (only these two, plus the mid-point, are in the syllabus)

• A point on the x-axis is $(x, 0)$; a point on the y-axis is $(0, y)$.
• The distance of $(x, y)$ from the x-axis is $|y|$, and from the y-axis is $|x|$.

Standard question types

• Ratio in which an axis divides $AB$: take the ratio $k : 1$, write the point, and put its y-coordinate $= 0$ (x-axis) or its x-coordinate $= 0$ (y-axis).
• Point on an axis equidistant from $A$ and $B$: take $(x, 0)$ or $(0, y)$ and put $PA^2 = PB^2$.
• Points of trisection: the ratios $1 : 2$ and $2 : 1$.
• $C$ on $AB$ produced with $AB : BC = m : n$: $B$ divides $AC$ in $m : n$.
• $AP = \frac{2}{5}AB$ means $AP : PB = 2 : 3$.
• Fourth vertex of a parallelogram: the diagonals bisect each other, so the mid-point of $AC$ = the mid-point of $BD$.
• Collinear by distances: the two shorter lengths add up to the longest.

Naming a quadrilateral from its lengths

Traps

• Square the whole difference: $(-3 - 2)^2 = 25$, not $-25$ or 1.
• Keep the ratio in the order it is given: $m_1$ goes with the second point.
• Area of a triangle from coordinates is not in the 2026-27 syllabus.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

The centre of a circle is $C(a, a + 1)$ and its radius is 5 units. If the circle passes through the point $P(6, 0)$, find the possible values of $a$ and the corresponding centres.

Answer.

1. $CP = 5$: $(a - 6)^2 + (a + 1)^2 = 25$, so $2a^2 - 10a + 12 = 0$, that is, $a^2 - 5a + 6 = 0$ — 1 mark
2. $a = 2$ or $a = 3$; the centre is $(2, 3)$ or $(3, 4)$ — 1 mark

Question 2 (2 marks)

Using the distance formula, show that the points $A(-1, -3)$, $B(1, 1)$ and $C(4, 7)$ are collinear.

Answer.

1. $AB = \sqrt{4 + 16} = 2\sqrt{5}$, $BC = \sqrt{9 + 36} = 3\sqrt{5}$, $AC = \sqrt{25 + 100} = 5\sqrt{5}$ — 1 mark
2. $AB + BC = 5\sqrt{5} = AC$, so $A$, $B$ and $C$ are collinear — 1 mark

Question 3 (2 marks)

$C$ is a point on $AB$ produced beyond $B$ such that $AB : BC = 1 : 2$. If $A$ is $(-2, 5)$ and $B$ is $(1, 3)$, find the coordinates of $C$.

Answer.

1. $B$ lies between $A$ and $C$ with $AB : BC = 1 : 2$, so $B$ divides $AC$ internally in $1 : 2$. Let $C = (x, y)$ — 1 mark
2. $\frac{x + 2 \times (-2)}{3} = 1$ and $\frac{y + 2 \times 5}{3} = 3$, so $C = (7, -1)$ — 1 mark

Question 4 (2 marks)

The vertices of $\triangle ABC$ are $A(5, 7)$, $B(-2, 3)$ and $C(6, -1)$. Find the length of the median $AD$.

Answer.

1. $D$ is the mid-point of $BC$: $D = \left(\frac{-2 + 6}{2}, \frac{3 - 1}{2}\right) = (2, 1)$ — 1 mark
2. $AD = \sqrt{(5 - 2)^2 + (7 - 1)^2} = \sqrt{45} = 3\sqrt{5}$ units — 1 mark

Question 5 (3 marks)

The line segment joining $A(-3, a)$ and $B(6, 8)$ meets the y-axis at $P(0, 4)$. Find the ratio $AP : PB$ and the value of $a$.

Answer.

1. Let $AP : PB = k : 1$; the x-coordinate of $P$ gives $\frac{6k - 3}{k + 1} = 0$ — 1 mark
2. $k = \frac{1}{2}$, so $AP : PB = 1 : 2$ — 1 mark
3. y-coordinate: $\frac{1 \times 8 + 2 \times a}{3} = 4$, so $a = 2$ — 1 mark

Question 6 (3 marks)

$A(-1, 3)$ and $B(9, -2)$ are two points, and $P$ is a point on the line segment $AB$ such that $PB = \frac{3}{5}AB$. Find the coordinates of $P$, and verify your answer using the distance formula.

Answer.

1. $PB = \frac{3}{5}AB$ gives $AP = \frac{2}{5}AB$, so $AP : PB = 2 : 3$ — 1 mark
2. $P = \left(\frac{2 \times 9 + 3 \times (-1)}{5}, \frac{2 \times (-2) + 3 \times 3}{5}\right) = (3, 1)$ — 1 mark
3. $AP = \sqrt{16 + 4} = 2\sqrt{5}$, $PB = \sqrt{36 + 9} = 3\sqrt{5}$, so $AP : PB = 2 : 3$; verified — 1 mark

Question 7 (3 marks)

Using the section formula, show that the points $A(-4, -1)$, $P(2, 3)$ and $B(8, 5)$ are not collinear.

Answer.

Model answer:

Suppose $A$, $P$ and $B$ were collinear. Then $P$ would divide $AB$ in some ratio $k : 1$, and both coordinates of $P$ would come from the section formula with the same $k$.

x-coordinate: $\frac{8k + (-4)}{k + 1} = 2$ gives $8k - 4 = 2k + 2$, so $k = 1$.

With $k = 1$, the y-coordinate would be $\frac{1 \times 5 + 1 \times (-1)}{2} = 2$.

But the y-coordinate of $P$ is 3. The two coordinates do not give the same ratio, so no such $k$ exists. Hence $A$, $P$ and $B$ are not collinear. (The point $(2, 2)$, the mid-point of $AB$, is on the line; $P(2, 3)$ is 1 unit above it.)

Marking scheme:

1. If they were collinear, $P$ would divide $AB$ in some ratio $k : 1$; its x-coordinate gives $\frac{8k - 4}{k + 1} = 2$ — 1 mark
2. $8k - 4 = 2k + 2$, so $k = 1$; then the y-coordinate would be $\frac{5 + (-1)}{2} = 2$ — 1 mark
3. But the y-coordinate of $P$ is 3, not 2; so $P$ is not on the line $AB$, and the points are not collinear — 1 mark

Question 8 (3 marks)

The points $A(2, 1)$, $B(6, 3)$ and $C(k, 7)$ are the vertices of a triangle which is right-angled at $B$. Find the value of $k$.

Answer.

1. $AB^2 = 16 + 4 = 20$, $BC^2 = (k - 6)^2 + 16$, $AC^2 = (k - 2)^2 + 36$ — 1 mark
2. Right angle at $B$: $AB^2 + BC^2 = AC^2$, so $20 + k^2 - 12k + 52 = k^2 - 4k + 40$ — 1 mark
3. $-8k = -32$, so $k = 4$ — 1 mark

Question 9 (3 marks)

The points $A(-3, 2)$, $B(1, 0)$ and $C(4, 3)$ are three vertices of a parallelogram $ABCD$, taken in order. Find the fourth vertex $D$. Is $ABCD$ a rectangle? Give a reason.

Answer.

1. The diagonals bisect each other: mid-point of $AC$ $= \left(\frac{1}{2}, \frac{5}{2}\right)$ = mid-point of $BD$ — 1 mark
2. $\frac{1 + x}{2} = \frac{1}{2}$ and $\frac{0 + y}{2} = \frac{5}{2}$, so $D = (0, 5)$ — 1 mark
3. $AC = \sqrt{49 + 1} = \sqrt{50}$ and $BD = \sqrt{1 + 25} = \sqrt{26}$; the diagonals are not equal, so it is not a rectangle — 1 mark

Question 10 (3 marks)

The vertices of $\triangle ABC$ are $A(1, 5)$, $B(-3, -1)$ and $C(7, 1)$. $D$ and $E$ are the mid-points of $AB$ and $AC$. Find the coordinates of $D$ and $E$, and verify that $DE = \frac{1}{2}BC$.

Answer.

1. $D = \left(\frac{1 - 3}{2}, \frac{5 - 1}{2}\right) = (-1, 2)$ and $E = \left(\frac{1 + 7}{2}, \frac{5 + 1}{2}\right) = (4, 3)$ — 1 mark
2. $DE = \sqrt{25 + 1} = \sqrt{26}$ — 1 mark
3. $BC = \sqrt{100 + 4} = \sqrt{104} = 2\sqrt{26}$, so $DE = \frac{1}{2}BC$; verified — 1 mark

Long Answer and Case-Based Questions

Question 11 (4 marks)

For Makar Sankranti, Aarav draws the design of a kite on graph paper, taking the centre of the sheet as the origin. The four corners of the kite are $A(0, 6)$, $B(-3, 2)$, $C(0, -4)$ and $D(3, 2)$, and its two sticks lie along $AC$ and $BD$ (see the figure).

(i) Find the length $AB$. (1 mark)

Answer.

1. $AB = \sqrt{(-3 - 0)^2 + (2 - 6)^2} = \sqrt{9 + 16} = 5$ units — 1 mark

(ii) Find the mid-point of $BD$, and show that it lies on the stick $AC$. (1 mark)

Answer.

1. Mid-point of $BD$ $= (0, 2)$; $A$ and $C$ both have x-coordinate 0, so $AC$ lies on the y-axis and $(0, 2)$ is on it, between $A$ and $C$ — 1 mark

(iii) Show that $AB = AD$ and $CB = CD$. (2 marks)

Answer.

1. $AD = \sqrt{9 + 16} = 5 = AB$ — 1 mark
2. $CB = \sqrt{9 + 36} = 3\sqrt{5}$ and $CD = \sqrt{9 + 36} = 3\sqrt{5}$, so $CB = CD$ — 1 mark

OR

(iii) In what ratio does the stick $BD$ divide the stick $AC$? (2 marks)

Answer.

1. $BD$ meets $AC$ at $M(0, 2)$; let $AM : MC = k : 1$, then $\frac{-4k + 6}{k + 1} = 2$ — 1 mark
2. $k = \frac{2}{3}$, so the ratio is $2 : 3$ (check: $AM = 4$, $MC = 6$) — 1 mark

Question 12 (4 marks)

A new metro line runs straight from station $A(-7, -2)$ to station $B(5, 7)$ on a city map drawn on a coordinate grid, where 1 unit = 1 km. A hospital is at $H(-4, 6)$ (see the figure).

(i) How long is the metro line? (1 mark)

Answer.

1. $AB = \sqrt{12^2 + 9^2} = \sqrt{225} = 15$ km — 1 mark

(ii) Two more stations $P$ and $Q$ divide $AB$ into three equal parts, with $P$ nearer to $A$. Find their coordinates. (1 mark)

Answer.

1. $P$ divides $AB$ in $1 : 2$: $P = \left(\frac{5 - 14}{3}, \frac{7 - 4}{3}\right) = (-3, 1)$; $Q$ is the mid-point of $PB$: $Q = (1, 4)$ — 1 mark

(iii) A river flows along the y-axis. In what ratio does the river divide $AB$, and at which point does the metro line cross the river? (2 marks)

Answer.

1. Let the ratio be $k : 1$; on the y-axis, $\frac{5k - 7}{k + 1} = 0$, so $k = \frac{7}{5}$: ratio $7 : 5$ — 1 mark
2. Point $= \left(0, \frac{7 \times 7 + 5 \times (-2)}{12}\right) = \left(0, \frac{13}{4}\right)$ — 1 mark

OR

(iii) Which of the stations $P$ and $Q$ is nearer to the hospital $H$? Justify your answer. (2 marks)

Answer.

1. $HP = \sqrt{1 + 25} = \sqrt{26}$ km and $HQ = \sqrt{25 + 4} = \sqrt{29}$ km — 1 mark
2. $\sqrt{26} < \sqrt{29}$, so station $P$ is nearer — 1 mark

Question 13 (4 marks)

A triangular park in Chandigarh is drawn on a coordinate grid, where 1 unit = 10 m. Its three gates are at the corners $A(2, 9)$, $B(-4, 0)$ and $C(8, 0)$. A fountain $D$ is to be built at the mid-point of the side $BC$, and a straight path will join gate $A$ to the fountain.

(i) Find the length of the side $BC$ in metres. (1 mark)

Answer.

1. $BC = 8 - (-4) = 12$ units $= 120$ m — 1 mark

(ii) Find the coordinates of the fountain $D$. (1 mark)

Answer.

1. $D = \left(\frac{-4 + 8}{2}, 0\right) = (2, 0)$ — 1 mark

(iii) A lamp post $L$ is fixed on the path $AD$ such that $AL : LD = 2 : 1$. Find the coordinates of $L$ and its distance from gate $A$ in metres. (2 marks)

Answer.

1. $L = \left(\frac{2 \times 2 + 1 \times 2}{3}, \frac{2 \times 0 + 1 \times 9}{3}\right) = (2, 3)$ — 1 mark
2. $AL = 9 - 3 = 6$ units $= 60$ m — 1 mark

OR

(iii) Show that the gate $A$ is equidistant from the gates $B$ and $C$, and find this distance. (2 marks)

Answer.

1. $AB = \sqrt{36 + 81} = \sqrt{117}$ and $AC = \sqrt{36 + 81} = \sqrt{117}$, so $AB = AC$ — 1 mark
2. $\sqrt{117} = 3\sqrt{13}$ units, that is, $30\sqrt{13}$ m — 1 mark

Question 14 (4 marks)

On Yoga Day, four students stand on the school ground at the points $A(1, 2)$, $B(5, 5)$, $C(9, 2)$ and $D(5, -1)$ of a coordinate grid marked on it, where 1 unit = 1 m. Ropes are tied along $AB$, $BC$, $CD$ and $DA$ (see the figure).

(i) Find the length of the rope $AB$. (1 mark)

Answer.

1. $AB = \sqrt{16 + 9} = 5$ m — 1 mark

(ii) Show that the diagonals $AC$ and $BD$ bisect each other. (1 mark)

Answer.

1. Mid-point of $AC$ $= (5, 2)$ and mid-point of $BD$ $= (5, 2)$: the same point, so they bisect each other — 1 mark

(iii) The coach says that $ABCD$ is a square. Is he right? Justify your answer. (2 marks)

Answer.

1. $AB = BC = CD = DA = 5$ m (each is $\sqrt{16 + 9}$) — 1 mark
2. But $AC = 8$ m and $BD = 6$ m are not equal, so $ABCD$ is a rhombus, not a square; he is not right — 1 mark

OR

(iii) A water bottle is kept at the point $T$ on $AC$ such that $AT : TC = 1 : 3$. Find the coordinates of $T$ and the distance of the student at $B$ from the bottle. (2 marks)

Answer.

1. $T = \left(\frac{1 \times 9 + 3 \times 1}{4}, \frac{1 \times 2 + 3 \times 2}{4}\right) = (3, 2)$ — 1 mark
2. $BT = \sqrt{(5 - 3)^2 + (5 - 2)^2} = \sqrt{13}$ m — 1 mark