NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry — Free PDF

Overview of Chapter 7: Coordinate Geometry

Coordinate Geometry connects algebra with geometry by using coordinates to solve geometric problems. This chapter introduces three powerful formulas: the distance formula, the section formula, and the formula for area of a triangle given vertices.

The chapter has 4 exercises covering:
- Exercise 7.1: Distance formula — finding distances between two points, checking collinearity, and verifying shapes
- Exercise 7.2: Section formula and midpoint formula — finding the point that divides a segment in a given ratio
- Exercise 7.3: Area of a triangle using coordinates — computing areas and checking collinearity
- Exercise 7.4: Mixed problems combining all three formulas

This chapter typically carries 5-8 marks in the CBSE board exam and connects naturally with Triangles (Chapter 6) and Trigonometry (Chapter 8). The formulas are straightforward to apply, but careful handling of negative signs and fractions is essential for accuracy. Students who master these three formulas can score full marks in this chapter with consistent practice.

Key Concepts and Formulas

Distance Formula: The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is:

This formula is derived from the Pythagoras theorem. The horizontal distance $(x_2 - x_1)$ and the vertical distance $(y_2 - y_1)$ form the two legs of a right triangle, and the distance between the points is the hypotenuse.

Distance from the origin: The distance of point $P(x, y)$ from the origin $O(0, 0)$ is:

Section Formula: The point $P$ dividing the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m:n$ is:

Note that $m$ is the ratio towards the second point $B$, and $n$ is towards the first point $A$.

Midpoint Formula (special case when $m = n = 1$):

Area of a Triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$:

The absolute value ensures the area is always positive regardless of the order of vertices.

Collinearity: Three points are collinear (lie on the same straight line) if the area of the triangle formed by them is zero. This gives the condition:

Centroid of a triangle: The centroid (point where medians meet) of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is:

Exercise 7.1 — Distance Formula (Solved)

Problem 1: Find the distance between $(2, 3)$ and $(4, 1)$.

Solution:

Answer: $2\sqrt{2}$ units.

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Problem 2: Check if the points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are collinear.

Solution:
Let $A(1, 5)$, $B(2, 3)$, $C(-2, -11)$.

Check: $AB + BC = \sqrt{5} + \sqrt{212} \approx 2.24 + 14.56 = 16.80$, while $AC = \sqrt{265} \approx 16.28$.

Since $AB + BC \neq AC$, the points are not collinear.

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Problem 3: Find the point on the y-axis equidistant from $(-5, -2)$ and $(3, 2)$.

Solution:
Let the point on the y-axis be $P(0, y)$.

Squaring both sides:

Answer: The point is $(0, -2)$.

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Problem 4: Show that the points $(1, 7)$, $(4, 2)$, $(-1, -1)$, and $(-4, 4)$ are vertices of a square.

Solution:
Let $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$, $D(-4, 4)$.

Compute all four sides:

All sides are equal. Now check diagonals:

Diagonals are equal. Since all four sides are equal and both diagonals are equal, $ABCD$ is a square.

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Problem 5: Find the distance of the point $(6, -8)$ from the origin.

Solution:

Answer: $10$ units.

Exercise 7.2 — Section Formula (Solved)

Problem 1: Find the coordinates of the point dividing $A(-1, 7)$ and $B(4, -3)$ in the ratio $2:3$.

Solution:
Using the section formula with $m = 2$, $n = 3$:

Answer: The point is $(1, 3)$.

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Problem 2: Find the ratio in which the point $(-3, p)$ divides the segment joining $(-5, -4)$ and $(-2, 3)$. Also find $p$.

Solution:
Let the ratio be $k : 1$.

Using the x-coordinate:

The ratio is $2 : 1$.

Now find $p$ using the y-coordinate:

Answer: Ratio is $2:1$ and $p = \dfrac{2}{3}$.

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Problem 3: Find the midpoint of $A(4, -6)$ and $B(-2, 8)$.

Solution:

Answer: Midpoint is $(1, 1)$.

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Problem 4: Find the trisection points of the segment joining $A(2, -2)$ and $B(-7, 4)$.

Solution:
The trisection points $P$ and $Q$ divide $AB$ in ratios $1:2$ and $2:1$ respectively.

For $P$ (ratio $1:2$):

For $Q$ (ratio $2:1$):

Answer: The trisection points are $(-1, 0)$ and $(-4, 2)$.

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Problem 5: The three vertices of a parallelogram are $(3, -1)$, $(-2, -1)$, and $(1, 3)$. Find the fourth vertex.

Solution:
Let $A(3, -1)$, $B(-2, -1)$, $C(1, 3)$, and $D(x, y)$.

In a parallelogram, diagonals bisect each other, so the midpoint of $AC$ equals the midpoint of $BD$.

Midpoint of $AC = \left(\dfrac{3+1}{2},\; \dfrac{-1+3}{2}\right) = (2, 1)$

Midpoint of $BD = \left(\dfrac{-2+x}{2},\; \dfrac{-1+y}{2}\right) = (2, 1)$

Answer: The fourth vertex is $(6, 3)$.

Exercise 7.3 — Area of a Triangle (Solved)

Problem 1: Find the area of the triangle with vertices $(2, 3)$, $(-1, 0)$, $(2, -4)$.

Solution:

Answer: Area $= 10.5$ sq units.

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Problem 2: Find the value of $k$ if $(7, -2)$, $(5, 1)$, $(3, k)$ are collinear.

Solution:
For collinear points, area $= 0$:

Answer: $k = 4$.

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Problem 3: Find the area of the quadrilateral with vertices $(−4, −2)$, $(−3, −5)$, $(3, −2)$, $(2, 3)$ taken in order.

Solution:
Divide the quadrilateral into two triangles by joining $(-4, -2)$ and $(3, -2)$.

Triangle 1: $(-4, -2)$, $(-3, -5)$, $(3, -2)$.

Triangle 2: $(-4, -2)$, $(3, -2)$, $(2, 3)$.

Total area $= \dfrac{21}{2} + \dfrac{35}{2} = \dfrac{56}{2} = 28$ sq units.

Answer: Area of the quadrilateral $= 28$ sq units.

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Problem 4: Find the area of the triangle formed by the midpoints of the sides of a triangle with vertices $(1, -1)$, $(-4, 6)$, $(-3, -5)$.

Solution:
Midpoint of side joining $(1,-1)$ and $(-4,6)$: $M_1 = \left(\dfrac{1-4}{2},\; \dfrac{-1+6}{2}\right) = \left(\dfrac{-3}{2},\; \dfrac{5}{2}\right)$

Midpoint of side joining $(-4,6)$ and $(-3,-5)$: $M_2 = \left(\dfrac{-4-3}{2},\; \dfrac{6-5}{2}\right) = \left(\dfrac{-7}{2},\; \dfrac{1}{2}\right)$

Midpoint of side joining $(-3,-5)$ and $(1,-1)$: $M_3 = \left(\dfrac{-3+1}{2},\; \dfrac{-5-1}{2}\right) = (-1, -3)$

Area of triangle $M_1 M_2 M_3$:

Answer: Area $= 6$ sq units. (This is $\dfrac{1}{4}$ of the area of the original triangle, which is a general result.)

Worked Examples — Additional Practice

Example 1: Prove that the points $(a, 0)$, $(0, b)$, and $(1, 1)$ are collinear if $\dfrac{1}{a} + \dfrac{1}{b} = 1$.

Solution:
For collinearity, area of triangle $= 0$:

This matches the given condition, so the points are collinear.

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Example 2: If $(1, 2)$, $(4, y)$, $(x, 6)$, and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

Solution:
Diagonals of a parallelogram bisect each other.

Midpoint of diagonal joining $(1,2)$ and $(x,6) =$ midpoint of diagonal joining $(4,y)$ and $(3,5)$.

Answer: $x = 6$, $y = 3$.

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Example 3: Find the centroid of the triangle with vertices $(-2, 3)$, $(4, -1)$, and $(1, 7)$.

Solution:

Answer: Centroid is $(1, 3)$.

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Example 4: Find the ratio in which the line $2x + y - 4 = 0$ divides the segment joining $A(2, -2)$ and $B(3, 7)$.

Solution:
Let the ratio be $k:1$. The dividing point is:

Since $P$ lies on $2x + y - 4 = 0$:

Answer: The ratio is $2:9$.

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Example 5: Find the coordinates of the point that is $\dfrac{3}{4}$ of the way from $A(2, -5)$ to $B(6, 3)$.

Solution:
The point divides $AB$ in ratio $3:1$.

Answer: $(5, 1)$.

Common Mistakes to Avoid

Mistake 1: Sign errors in the distance formula.
The most common error is dropping a negative sign. For example, finding the distance between $(3, -4)$ and $(-2, 1)$: the calculation should be $(3-(-2))^2 = (5)^2 = 25$, not $(3-2)^2 = 1$. Always use brackets around negative coordinates.

Mistake 2: Confusing $m$ and $n$ in the section formula.
In the formula $P = \left(\dfrac{mx_2 + nx_1}{m+n}, \dfrac{my_2 + ny_1}{m+n}\right)$, the ratio $m:n$ means $m$ is the part from $A$ towards $B$. So if $P$ divides $AB$ in ratio $2:3$, then $m = 2$ (closer to $A$) and $n = 3$ (closer to $B$). Swapping $m$ and $n$ gives a completely different point.

Mistake 3: Forgetting the absolute value in the area formula.
The area formula can give a negative number depending on the order of vertices. Always take the absolute value. Area is always positive.

Mistake 4: Incorrect simplification of square roots.
$\sqrt{8}$ should be simplified to $2\sqrt{2}$, not left as $\sqrt{8}$. CBSE examiners expect simplified surds. Similarly, $\sqrt{50} = 5\sqrt{2}$ and $\sqrt{72} = 6\sqrt{2}$.

Mistake 5: Assuming collinear when you have not verified.
To prove collinearity, you must either show that the area of the triangle is zero, or show that $AB + BC = AC$ (where $AC$ is the largest distance). Simply showing that three points have a pattern is not sufficient.

Exam Tips for Coordinate Geometry

Practice Questions with Answers

Q1. Find the distance between $(-3, 4)$ and $(5, -2)$.

Answer: $d = \sqrt{(5-(-3))^2 + (-2-4)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ units.

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Q2. Find the midpoint of the line segment joining $(7, -3)$ and $(-1, 5)$.

Answer: $M = \left(\dfrac{7-1}{2}, \dfrac{-3+5}{2}\right) = (3, 1)$.

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Q3. Show that the points $(1, 1)$, $(3, -1)$, and $(5, -3)$ are collinear.

Answer: Area $= \dfrac{1}{2}|1(-1+3) + 3(-3-1) + 5(1+1)| = \dfrac{1}{2}|2 - 12 + 10| = \dfrac{1}{2}|0| = 0$. Since area $= 0$, the points are collinear.

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Q4. Find the ratio in which the x-axis divides the segment joining $(2, -3)$ and $(5, 6)$.

Answer: Let the ratio be $k:1$. The y-coordinate of the dividing point is $0$ (on x-axis): $\dfrac{6k - 3}{k+1} = 0 \implies k = \dfrac{1}{2}$. Ratio is $1:2$.

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Q5. Find the area of the triangle with vertices $(0, 0)$, $(4, 0)$, and $(0, 3)$.

Answer: $\text{Area} = \dfrac{1}{2}|0(0-3) + 4(3-0) + 0(0-0)| = \dfrac{1}{2}|12| = 6$ sq units. (This is a right triangle with base $4$ and height $3$, so area $= \dfrac{1}{2} \times 4 \times 3 = 6$. Verified.)

Key Takeaways

- The distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ is derived from the Pythagoras theorem.
- The section formula gives the coordinates of a point dividing a segment in ratio $m:n$.
- The midpoint formula is a special case of the section formula with $m = n = 1$.
- The area formula for a triangle uses coordinates and always requires taking the absolute value.
- Three points are collinear if and only if the area of the triangle they form is zero.
- To classify quadrilaterals (square, rhombus, rectangle, parallelogram), compute side lengths and diagonals using the distance formula.
- Always simplify surds (e.g., $\sqrt{8} = 2\sqrt{2}$) and fractions in your final answer.

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