NCERT Solutions for Class 10 Mathematics Chapter 7: Coordinate Geometry (Complete Guide)
Coordinate Geometry serves as a vital mathematical bridge connecting pure algebraic analysis with visual Euclidean geometry on the two-dimensional Cartesian plane ($\mathbb{R}^2$). In the CBSE Class 10 curriculum, Chapter 7: Coordinate Geometry systematically formulates techniques to quantify spatial relationships, measure segment magnitudes, determine partition ratios, and verify geometric configurations algebraically without constructing synthetic figures.
Plaintext
+---------------------------------------------------------------------------------------------------+
| COORDINATE GEOMETRY ARCHITECTURE |
| |
| Cartesian Plane: Point P(x, y) |
| |
| +-----------------------------+-----------------------------+---------------------------------+ |
| | Distance Formula | Section Formula | Mid-Point Specialization | |
| | d = sqrt[(x2-x1)^2+(y2-y1)^2| P(x, y) = | M(x, y) = | |
| | Magnitude of straight line | ((m1*x2 + m2*x1)/(m1 + m2), | ((x1 + x2)/2, (y1 + y2)/2) | |
| | segment joining two points. | (m1*y2 + m2*y1)/(m1 + m2)) | Equipartition of line segment | |
| +-----------------------------+-----------------------------+---------------------------------+ |
+---------------------------------------------------------------------------------------------------+
The fundamental framework relies on the Distance Formula, derived directly from the Pythagorean theorem. For any two arbitrary points $P(x_1, y_1)$ and $Q(x_2, y_2)$, the shortest Euclidean distance is given by $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$. From this baseline, collinearity can be tested through triangle inequality boundary limits ($AB + BC = AC$), while quadrilaterals and triangles are classified via side lengths and diagonal equivalences.
The Section Formula enables the analytical determination of a point dividing a given line segment internally in a specified rational ratio $m_1 : m_2$. Setting $m_1 = m_2 = 1$ yields the Mid-Point Formula, which is widely used in diagonal bisector properties of parallelograms, rhombuses, and rectangles.
According to official CBSE marking schemes, full credit requires writing the general formula first, stating assigned coordinates $(x_1, y_1)$ and $(x_2, y_2)$ explicitly, substituting values step-by-step with proper sign handling, and stating the final answer with appropriate units (e.g., “units”).
| Geometrical Concept | Algebraic Formula | Standard Conditions & Constraints | Analytical Application |
| Distance Formula | $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$ | $d \ge 0$ always; $\sqrt{\Delta x^2 + \Delta y^2}$ | Length of segments, perimeter, collinearity testing. |
| Distance from Origin | $OP = \sqrt{x^2 + y^2}$ | $O(0, 0)$ is the reference frame origin | Absolute radial displacement of point $P(x, y)$. |
| Internal Section Formula | $\left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right)$ | $m_1, m_2 > 0$; ratio $k : 1 \implies \left(\frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1}\right)$ | Coordinates of dividing point on a line segment. |
| Mid-Point Formula | $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ | Special case where $m_1 : m_2 = 1 : 1$ | Bisectors, medians, centers of circles, diagonals. |
| Centroid of a Triangle | $G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$ | Concurrence of medians dividing in $2 : 1$ ratio | Center of mass / intersection of medians of $\triangle ABC$. |
Chapter-End Exercises – Exercise 7.1
Question 1. Find the distance between the following pairs of points:
(i) $(2, 3), (4, 1)$
(ii) $(-5, 7), (-1, 3)$
(iii) $(a, b), (-a, -b)$
(i) $(2, 3)$ and $(4, 1)$
Answer:
Given Data:
- Let $P(x_1, y_1) = (2, 3)$
- Let $Q(x_2, y_2) = (4, 1)$
Formula Required:
$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
Step-by-Step Calculation:
$$\begin{aligned} PQ &= \sqrt{(4 – 2)^2 + (1 – 3)^2} \\ &= \sqrt{(2)^2 + (-2)^2} \\ &= \sqrt{4 + 4} \\ &= \sqrt{8} \\ &= 2\sqrt{2}\text{ units} \end{aligned}$$
Final Statement:
The distance between the points is $2\sqrt{2}\text{ units}$.
(ii) $(-5, 7)$ and $(-1, 3)$
Answer:
Given Data:
- Let $P(x_1, y_1) = (-5, 7)$
- Let $Q(x_2, y_2) = (-1, 3)$
Formula Required:
$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
Step-by-Step Calculation:
$$\begin{aligned} PQ &= \sqrt{[-1 – (-5)]^2 + (3 – 7)^2} \\ &= \sqrt{(-1 + 5)^2 + (-4)^2} \\ &= \sqrt{(4)^2 + (-4)^2} \\ &= \sqrt{16 + 16} \\ &= \sqrt{32} \\ &= 4\sqrt{2}\text{ units} \end{aligned}$$
Final Statement:
The distance between the points is $4\sqrt{2}\text{ units}$.
(iii) $(a, b)$ and $(-a, -b)$ [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
Given Data:
- Let $P(x_1, y_1) = (a, b)$
- Let $Q(x_2, y_2) = (-a, -b)$
Formula Required:
$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
Step-by-Step Calculation:
$$\begin{aligned} PQ &= \sqrt{(-a – a)^2 + (-b – b)^2} \\ &= \sqrt{(-2a)^2 + (-2b)^2} \\ &= \sqrt{4a^2 + 4b^2} \\ &= \sqrt{4(a^2 + b^2)} \\ &= 2\sqrt{a^2 + b^2}\text{ units} \end{aligned}$$
Final Statement:
The distance between the points is $2\sqrt{a^2 + b^2}\text{ units}$.
Question 2. Find the distance between the points $(0, 0)$ and $(36, 15)$. Can you now find the distance between the two towns A and B discussed in Section 7.2? [CBSE 2020]
Answer:
Given Data:
- Town A at origin $A(x_1, y_1) = (0, 0)$
- Town B at coordinates $B(x_2, y_2) = (36, 15)$
Formula Required:
$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
Step-by-Step Calculation:
$$\begin{aligned} AB &= \sqrt{(36 – 0)^2 + (15 – 0)^2} \\ &= \sqrt{36^2 + 15^2} \\ &= \sqrt{1296 + 225} \\ &= \sqrt{1521} \\ &= 39\text{ km} \end{aligned}$$
Final Statement:
The distance between the points is $39\text{ units}$. Yes, the actual physical distance between town A and town B is $39\text{ km}$.
Question 3. Determine if the points $(1, 5), (2, 3)$ and $(-2, -11)$ are collinear. [BOARD EXAM FAVORITE / CBSE 2018, 2022]
Answer:
Given Data:
Let the three given points be $A(1, 5)$, $B(2, 3)$, and $C(-2, -11)$.
Mathematical Principle of Collinearity:
Three points $A, B, C$ are collinear if the sum of the lengths of any two segments equals the length of the remaining segment (i.e., $AB + BC = AC$).
Step-by-Step Calculation:
- Distance $AB$:$$AB = \sqrt{(2 – 1)^2 + (3 – 5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$
- Distance $BC$:$$BC = \sqrt{(-2 – 2)^2 + (-11 – 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53}$$
- Distance $AC$:$$AC = \sqrt{(-2 – 1)^2 + (-11 – 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}$$
Checking the sum:
$$\begin{aligned} AB + BC &= \sqrt{5} + 2\sqrt{53} \neq \sqrt{265} = AC \\ AB + AC &\neq BC \\ BC + AC &\neq AB \end{aligned}$$
Final Statement:
Since the sum of no two segment lengths equals the third, the points $(1, 5), (2, 3)$, and $(-2, -11)$ are not collinear.
Question 4. Check whether $(5, -2), (6, 4)$ and $(7, -2)$ are the vertices of an isosceles triangle. [CBSE 2019, 2023]
Answer:
Given Data:
Let the vertices be $A(5, -2)$, $B(6, 4)$, and $C(7, -2)$.
Mathematical Principle:
A triangle is isosceles if any two of its three sides are equal in length.
Step-by-Step Calculation:
- Side $AB$:$$AB = \sqrt{(6 – 5)^2 + [4 – (-2)]^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}\text{ units}$$
- Side $BC$:$$BC = \sqrt{(7 – 6)^2 + (-2 – 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}\text{ units}$$
- Side $AC$:$$AC = \sqrt{(7 – 5)^2 + [-2 – (-2)]^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2\text{ units}$$
Since $AB = BC = \sqrt{37}\text{ units} \neq AC$:
Final Statement:
Yes, the points $A(5, -2), B(6, 4)$, and $C(7, -2)$ form an isosceles triangle.
Question 5. In a classroom, $4$ friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Answer:
Given Data (From Grid Coordinates):
- $A(3, 4)$
- $B(6, 7)$
- $C(9, 4)$
- $D(6, 1)$
Step-by-Step Calculation of Sides:
$$\begin{aligned} AB &= \sqrt{(6 – 3)^2 + (7 – 4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \\ BC &= \sqrt{(9 – 6)^2 + (4 – 7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \\ CD &= \sqrt{(6 – 9)^2 + (1 – 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \\ DA &= \sqrt{(3 – 6)^2 + (4 – 1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \end{aligned}$$
All four sides are equal: $AB = BC = CD = DA = 3\sqrt{2}\text{ units}$.
Step-by-Step Calculation of Diagonals:
$$\begin{aligned} AC &= \sqrt{(9 – 3)^2 + (4 – 4)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6\text{ units} \\ BD &= \sqrt{(6 – 6)^2 + (1 – 7)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6\text{ units} \end{aligned}$$
Both diagonals are equal: $AC = BD = 6\text{ units}$.
Final Statement:
Since all four sides and both diagonals are equal, quadrilateral $ABCD$ is a square. Therefore, Champa is correct.
Question 6. 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) $A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)$ [BOARD EXAM FAVORITE]
Answer:
Calculate side lengths:
$$\begin{aligned} AB &= \sqrt{[1 – (-1)]^2 + [0 – (-2)]^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \\ BC &= \sqrt{(-1 – 1)^2 + (2 – 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \\ CD &= \sqrt{[-3 – (-1)]^2 + (0 – 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \\ DA &= \sqrt{[-1 – (-3)]^2 + (-2 – 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \end{aligned}$$
Calculate diagonal lengths:
$$\begin{aligned} AC &= \sqrt{[-1 – (-1)]^2 + [2 – (-2)]^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4 \\ BD &= \sqrt{(-3 – 1)^2 + (0 – 0)^2} = \sqrt{(-4)^2 + 0} = \sqrt{16} = 4 \end{aligned}$$
Since $AB = BC = CD = DA = 2\sqrt{2}$ and $AC = BD = 4$:
The quadrilateral is a square.
(ii) $A(-3, 5), B(3, 1), C(0, 3), D(-1, -4)$
Answer:
Calculate distances:
$$\begin{aligned} AB &= \sqrt{[3 – (-3)]^2 + (1 – 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \\ BC &= \sqrt{(0 – 3)^2 + (3 – 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \\ AC &= \sqrt{[0 – (-3)]^2 + (3 – 5)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \end{aligned}$$
Notice that $AC + BC = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = AB$.
This means points $A, C$, and $B$ are collinear (point $C$ lies on line segment $AB$).
Since three of the four points lie on the same straight line, they cannot form a closed four-sided polygon.
Therefore, no quadrilateral is formed.
(iii) $A(4, 5), B(7, 6), C(4, 3), D(1, 2)$
Answer:
Calculate side lengths:
$$\begin{aligned} AB &= \sqrt{(7 – 4)^2 + (6 – 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \\ BC &= \sqrt{(4 – 7)^2 + (3 – 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \\ CD &= \sqrt{(1 – 4)^2 + (2 – 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \\ DA &= \sqrt{(4 – 1)^2 + (5 – 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \end{aligned}$$
Here, opposite sides are equal: $AB = CD = \sqrt{10}$ and $BC = DA = 3\sqrt{2}$.
Calculate diagonal lengths:
$$\begin{aligned} AC &= \sqrt{(4 – 4)^2 + (3 – 5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2 \\ BD &= \sqrt{(1 – 7)^2 + (2 – 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \end{aligned}$$
Opposite sides are equal ($AB = CD$ and $BC = DA$), but diagonals are unequal ($AC \neq BD$).
Therefore, the quadrilateral is a parallelogram (and not a rectangle).
Question 7. Find the point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9)$. [BOARD EXAM FAVORITE / CBSE 2019, 2021, 2023]
Answer: Step 1: Variable Declaration Any point on the $x$-axis has its $y$-coordinate equal to zero. Let the required point on the $x$-axis be $P(x, 0)$. Let the given points be $A(2, -5)$ and $B(-2, 9)$.
Step 2: Equating Distances ($PA = PB \implies PA^2 = PB^2$)
$$\begin{aligned} PA^2 &= (x – 2)^2 + [0 – (-5)]^2 = (x – 2)^2 + 5^2 = x^2 – 4x + 4 + 25 = x^2 – 4x + 29 \\ PB^2 &= [x – (-2)]^2 + (0 – 9)^2 = (x + 2)^2 + (-9)^2 = x^2 + 4x + 4 + 81 = x^2 + 4x + 85 \end{aligned}$$
Since $P$ is equidistant from $A$ and $B$:
$$\begin{aligned} PA^2 &= PB^2 \\ x^2 – 4x + 29 &= x^2 + 4x + 85 \\ -4x – 4x &= 85 – 29 \\ -8x &= 56 \\ x &= \frac{56}{-8} = -7 \end{aligned}$$
Final Statement:
The required point on the $x$-axis is $(-7, 0)$.
Question 8. Find the values of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is $10\text{ units}$. [BOARD EXAM FAVORITE / CBSE 2018, 2020, 2022]
Answer:
Given Data:
- $P(x_1, y_1) = (2, -3)$
- $Q(x_2, y_2) = (10, y)$
- Distance $PQ = 10\text{ units}$
Step-by-Step Calculation:
Using the distance formula $PQ^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2$:
$$\begin{aligned} 10^2 &= (10 – 2)^2 + [y – (-3)]^2 \\ 100 &= 8^2 + (y + 3)^2 \\ 100 &= 64 + (y^2 + 6y + 9) \\ 100 &= y^2 + 6y + 73 \\ y^2 + 6y + 73 – 100 &= 0 \\ y^2 + 6y – 27 &= 0 \end{aligned}$$
Factorisation: Split $6y$ into $9y – 3y$:
$$\begin{aligned} y^2 + 9y – 3y – 27 &= 0 \\ y(y + 9) – 3(y + 9) &= 0 \\ (y + 9)(y – 3) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} y + 9 = 0 &\implies y = -9 \\ y – 3 = 0 &\implies y = 3 \end{aligned}$$
Final Statement:
The required values of $y$ are $3$ and $-9$.
Question 9. 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 $PR$. [BOARD EXAM FAVORITE / CBSE 2020, 2023]
Answer: Step 1: Finding the Value of $x$ Given that $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$, we have $QP = QR \implies QP^2 = QR^2$.
$$\begin{aligned} QP^2 &= (5 – 0)^2 + (-3 – 1)^2 = 5^2 + (-4)^2 = 25 + 16 = 41 \\ QR^2 &= (x – 0)^2 + (6 – 1)^2 = x^2 + 5^2 = x^2 + 25 \end{aligned}$$
Equating the squared distances:
$$\begin{aligned} x^2 + 25 &= 41 \\ x^2 &= 41 – 25 = 16 \\ x &= \pm 4 \end{aligned}$$
Thus, point $R$ is either $R_1(4, 6)$ or $R_2(-4, 6)$.
Step 2: Calculating Distances $QR$ and $PR$
For $x = \pm 4$:
$$QR = \sqrt{(\pm 4 – 0)^2 + (6 – 1)^2} = \sqrt{16 + 25} = \sqrt{41}\text{ units}$$
For $x = 4$, calculating $PR$:
$$PR = \sqrt{(4 – 5)^2 + [6 – (-3)]^2} = \sqrt{(-1)^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82}\text{ units}$$
For $x = -4$, calculating $PR$:
$$PR = \sqrt{(-4 – 5)^2 + [6 – (-3)]^2} = \sqrt{(-9)^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}\text{ units}$$
Final Statement:
The values of $x$ are $\pm 4$, $QR = \sqrt{41}\text{ units}$, and $PR$ is $\sqrt{82}\text{ units}$ or $9\sqrt{2}\text{ units}$.
Question 10. Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the point $(3, 6)$ and $(-3, 4)$. [BOARD EXAM FAVORITE / CBSE 2019, 2022]
Answer:
Given Data:
- Let point $P(x, y)$ be equidistant from $A(3, 6)$ and $B(-3, 4)$.
Step-by-Step Calculation:
Since $PA = PB \implies PA^2 = PB^2$:
$$\begin{aligned} (x – 3)^2 + (y – 6)^2 &= [x – (-3)]^2 + (y – 4)^2 \\ (x – 3)^2 + (y – 6)^2 &= (x + 3)^2 + (y – 4)^2 \end{aligned}$$
Expanding both sides:
$$x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16$$
Canceling $x^2$ and $y^2$ from both sides:
$$\begin{aligned} -6x – 12y + 45 &= 6x – 8y + 25 \\ -6x – 6x – 12y + 8y + 45 – 25 &= 0 \\ -12x – 4y + 20 &= 0 \end{aligned}$$
Divide the entire equation by $-4$:
$$3x + y – 5 = 0$$
Final Statement:
The required relation between $x$ and $y$ is $3x + y – 5 = 0$ (or $3x + y = 5$).
Chapter-End Exercises – Exercise 7.2
Question 1. Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$. [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
Given Data:
- $(x_1, y_1) = (-1, 7)$
- $(x_2, y_2) = (4, -3)$
- Partition ratio $m_1 : m_2 = 2 : 3$
Formula Required:
$$P(x, y) = \left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right)$$
Step-by-Step Calculation:
$$\begin{aligned} x &= \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 – 3}{5} = \frac{5}{5} = 1 \\ y &= \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3 \end{aligned}$$
Final Statement:
The coordinates of the required dividing point are $(1, 3)$.
Question 2. Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$. [BOARD EXAM FAVORITE / CBSE 2018, 2020, 2022]
Answer:
Given Data:
Let the line segment join $A(4, -1)$ and $B(-2, -3)$.
Trisection divides the line segment into three equal parts using two points $P$ and $Q$:
- $P$ divides $AB$ internally in the ratio $1 : 2$.
- $Q$ divides $AB$ internally in the ratio $2 : 1$ (or is the midpoint of $PB$).
Step 1: Coordinates of Point $P$ (Ratio $1 : 2$):
$$\begin{aligned} x_P &= \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2 \\ y_P &= \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 – 2}{3} = -\frac{5}{3} \end{aligned}$$
So, $P = \left(2, -\frac{5}{3}\right)$.
Step 2: Coordinates of Point $Q$ (Mid-Point of $PB$):
$$\begin{aligned} x_Q &= \frac{x_P + x_B}{2} = \frac{2 + (-2)}{2} = \frac{0}{2} = 0 \\ y_Q &= \frac{y_P + y_B}{2} = \frac{-\frac{5}{3} + (-3)}{2} = \frac{-\frac{14}{3}}{2} = -\frac{7}{3} \end{aligned}$$
So, $Q = \left(0, -\frac{7}{3}\right)$.
Final Statement:
The coordinates of the points of trisection are $\left(2, -\frac{5}{3}\right)$ and $\left(0, -\frac{7}{3}\right)$.
Question 3. To conduct Sports Day activities, in your rectangular shaped school ground $ABCD$, lines have been drawn with chalk powder at a distance of $1\text{ m}$ each. $100$ flower pots have been placed at a distance of $1\text{ m}$ from each other along $AD$, as shown in Fig. 7.12. Niharika runs $\frac{1}{4}\text{th}$ the distance $AD$ on the $2\text{nd}$ line and posts a green flag. Preet runs $\frac{1}{5}\text{th}$ the distance $AD$ on the $8\text{th}$ line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? [CBSE 2019, 2020]
Answer:
Step 1: Determining Flag Coordinates
- Total distance $AD = 100 \times 1\text{ m} = 100\text{ m}$.
- Niharika’s Green Flag ($G$): On $2\text{nd}$ line, $y_G = \frac{1}{4} \times 100 = 25 \implies G(2, 25)$.
- Preet’s Red Flag ($R$): On $8\text{th}$ line, $y_R = \frac{1}{5} \times 100 = 20 \implies R(8, 20)$.
Step 2: Distance Between the Flags ($GR$):
$$\begin{aligned} GR &= \sqrt{(8 – 2)^2 + (20 – 25)^2} \\ &= \sqrt{6^2 + (-5)^2} \\ &= \sqrt{36 + 25} \\ &= \sqrt{61}\text{ m} \end{aligned}$$
Step 3: Position of Blue Flag Posted by Rashmi (Mid-Point $M$):
$$\begin{aligned} x_M &= \frac{2 + 8}{2} = \frac{10}{2} = 5 \\ y_M &= \frac{25 + 20}{2} = \frac{45}{2} = 22.5 \end{aligned}$$
Final Statement:
The distance between both flags is $\sqrt{61}\text{ m}$. Rashmi should post her blue flag on the $5\text{th}$ line at a distance of $22.5\text{ m}$.
Question 4. Find the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$. [BOARD EXAM FAVORITE / CBSE 2019, 2021, 2023]
Answer:
Given Data:
- $A(x_1, y_1) = (-3, 10)$
- $B(x_2, y_2) = (6, -8)$
- Dividing point $P(x, y) = (-1, 6)$
Step-by-Step Calculation:
Let the required internal division ratio be $k : 1$.
Applying the section formula for the $x$-coordinate:
$$\begin{aligned} x &= \frac{k x_2 + 1(x_1)}{k + 1} \\ -1 &= \frac{k(6) + 1(-3)}{k + 1} \\ -(k + 1) &= 6k – 3 \\ -k – 1 &= 6k – 3 \\ -k – 6k &= -3 + 1 \\ -7k &= -2 \\ k &= \frac{2}{7} \end{aligned}$$
Verifying with the $y$-coordinate:
$$y = \frac{2(-8) + 7(10)}{2 + 7} = \frac{-16 + 70}{9} = \frac{54}{9} = 6 \quad \text{[Consistent]}$$
Final Statement:
The point $(-1, 6)$ divides the line segment in the ratio $2 : 7$.
Question 5. 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. [BOARD EXAM FAVORITE / CBSE 2020, 2022]
Answer:
Step 1: Finding the Ratio
Any point on the $x$-axis has $y = 0$. Let the point be $P(x, 0)$ and the ratio be $k : 1$.
Applying the section formula for $y$:
$$\begin{aligned} y &= \frac{k y_2 + 1(y_1)}{k + 1} \\ 0 &= \frac{k(5) + 1(-5)}{k + 1} \\ 5k – 5 &= 0 \\ 5k &= 5 \implies k = 1 \end{aligned}$$
Thus, the $x$-axis divides $AB$ in the ratio $1 : 1$ (the point of division is the midpoint).
Step 2: Finding the Coordinates of Division Point $P$
Substitute $k = 1$ to calculate $x$:
$$x = \frac{1(-4) + 1(1)}{1 + 1} = \frac{-4 + 1}{2} = -\frac{3}{2}$$
Final Statement:
The ratio is $1 : 1$, and the coordinates of the point of division are $\left(-\frac{3}{2}, 0\right)$.
Question 6. If $(1, 2), (4, y), (x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$. [BOARD EXAM FAVORITE / CBSE 2018, 2020, 2023]
Answer:
Given Data:
Let the ordered vertices of parallelogram $ABCD$ be $A(1, 2)$, $B(4, y)$, $C(x, 6)$, and $D(3, 5)$.
Mathematical Principle:
The diagonals of a parallelogram bisect each other, meaning the midpoint of diagonal $AC$ coincides with the midpoint of diagonal $BD$.
Step 1: Equating Midpoint Coordinates:
$$\text{Midpoint of } AC = \left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) = \left(\frac{1 + x}{2}, 4\right)$$
$$\text{Midpoint of } BD = \left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right) = \left(\frac{7}{2}, \frac{y + 5}{2}\right)$$
Step 2: Solving for $x$ and $y$:
Equating $x$-coordinates:
$$\begin{aligned} \frac{1 + x}{2} &= \frac{7}{2} \\ 1 + x &= 7 \implies x = 6 \end{aligned}$$
Equating $y$-coordinates:
$$\begin{aligned} \frac{y + 5}{2} &= 4 \\ y + 5 &= 8 \implies y = 3 \end{aligned}$$
Final Statement:
The values are $x = 6$ and $y = 3$.
Question 7. Find the coordinates of a point A, where $AB$ is the diameter of a circle whose centre is $(2, -3)$ and B is $(1, 4)$. [BOARD EXAM FAVORITE / CBSE 2019, 2021]
Answer:
Given Data:
- Let the coordinates of point $A$ be $(x, y)$.
- Center of the circle $C = (2, -3)$.
- Point $B = (1, 4)$.
Mathematical Principle:
The center of a circle is the midpoint of its diameter $AB$.
Step-by-Step Calculation:
$$\begin{aligned} x_C &= \frac{x_A + x_B}{2} \implies 2 = \frac{x + 1}{2} \implies 4 = x + 1 \implies x = 3 \\ y_C &= \frac{y_A + y_B}{2} \implies -3 = \frac{y + 4}{2} \implies -6 = y + 4 \implies y = -10 \end{aligned}$$
Final Statement:
The coordinates of point A are $(3, -10)$.
Question 8. If A and B are $(-2, -2)$ and $(2, -4)$, respectively, find the coordinates of P such that $AP = \frac{3}{7}AB$ and P lies on the line segment $AB$. [BOARD EXAM FAVORITE / CBSE 2019, 2022, 2023]
Answer:
Step 1: Determining the Partition Ratio
We are given $AP = \frac{3}{7}AB$.
$$\begin{aligned} PB &= AB – AP = AB – \frac{3}{7}AB = \frac{4}{7}AB \\ \frac{AP}{PB} &= \frac{\frac{3}{7}AB}{\frac{4}{7}AB} = \frac{3}{4} \end{aligned}$$
Thus, point $P$ divides segment $AB$ internally in the ratio $m_1 : m_2 = 3 : 4$.
Step 2: Section Formula Calculation:
$$\begin{aligned} x_P &= \frac{3(2) + 4(-2)}{3 + 4} = \frac{6 – 8}{7} = -\frac{2}{7} \\ y_P &= \frac{3(-4) + 4(-2)}{3 + 4} = \frac{-12 – 8}{7} = -\frac{20}{7} \end{aligned}$$
Final Statement:
The coordinates of point P are $\left(-\frac{2}{7}, -\frac{20}{7}\right)$.
Question 9. Find the coordinates of the points which divide the line segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts. [BOARD EXAM FAVORITE / CBSE 2020]
Answer:
Given Data:
Let the three interior division points be $P, Q, R$.
- $Q$ is the midpoint of $AB$.
- $P$ is the midpoint of $AQ$.
- $R$ is the midpoint of $QB$.
Step 1: Midpoint $Q$ of $AB$:
$$Q = \left(\frac{-2 + 2}{2}, \frac{2 + 8}{2}\right) = (0, 5)$$
Step 2: Midpoint $P$ of $AQ$ (between $(-2, 2)$ and $(0, 5)$):
$$P = \left(\frac{-2 + 0}{2}, \frac{2 + 5}{2}\right) = \left(-1, \frac{7}{2}\right)$$
Step 3: Midpoint $R$ of $QB$ (between $(0, 5)$ and $(2, 8)$):
$$R = \left(\frac{0 + 2}{2}, \frac{5 + 8}{2}\right) = \left(1, \frac{13}{2}\right)$$
Final Statement:
The three points dividing $AB$ into four equal parts are $\left(-1, \frac{7}{2}\right), (0, 5)$, and $\left(1, \frac{13}{2}\right)$.
Question 10. 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 $= \frac{1}{2}(\text{product of its diagonals})$]. [CBSE 2020, 2023]
Answer:
Given Data:
Let the vertices in order be $A(3, 0)$, $B(4, 5)$, $C(-1, 4)$, and $D(-2, -1)$.
Step 1: Length of Diagonal $AC$ ($d_1$):
$$d_1 = AC = \sqrt{(-1 – 3)^2 + (4 – 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\text{ units}$$
Step 2: Length of Diagonal $BD$ ($d_2$):
$$d_2 = BD = \sqrt{(-2 – 4)^2 + (-1 – 5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\text{ units}$$
Step 3: Area Calculation:
$$\begin{aligned} \text{Area of Rhombus} &= \frac{1}{2} \times d_1 \times d_2 \\ &= \frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2}) \\ &= \frac{1}{2} \times 24 \times 2 \\ &= 24\text{ sq units} \end{aligned}$$
Final Statement:
The area of the rhombus is $24\text{ sq units}$.
15 High-Yield Board Exam FAQs
FAQ 1 (Centroid of a Triangle). Find the coordinates of the centroid of a triangle whose vertices are $A(3, -5), B(-7, 4),$ and $C(10, -2)$.
Answer:
$$\begin{aligned} x_G &= \frac{x_1 + x_2 + x_3}{3} = \frac{3 + (-7) + 10}{3} = \frac{6}{3} = 2 \\ y_G &= \frac{y_1 + y_2 + y_3}{3} = \frac{-5 + 4 + (-2)}{3} = \frac{-3}{3} = -1 \end{aligned}$$
The centroid is $(2, -1)$.
FAQ 2 (HOTS / Third Vertex from Centroid). Two vertices of $\triangle ABC$ are $A(1, -6)$ and $B(-5, 2)$. If the centroid is $G(-2, 1)$, find the coordinates of vertex $C$.
Answer:
Let $C = (x, y)$.
$$\begin{aligned} -2 &= \frac{1 – 5 + x}{3} \implies -6 = -4 + x \implies x = -2 \\ 1 &= \frac{-6 + 2 + y}{3} \implies 3 = -4 + y \implies y = 7 \end{aligned}$$
The coordinates of vertex C are $(-2, 7)$.
FAQ 3 (Perpendicular Bisector Relation). Find the equation of the perpendicular bisector of the line segment joining $A(2, 3)$ and $B(6, -5)$.
Answer:
Any point $P(x, y)$ on the perpendicular bisector satisfies $PA^2 = PB^2$:
$$\begin{aligned} (x – 2)^2 + (y – 3)^2 &= (x – 6)^2 + [y – (-5)]^2 \\ x^2 – 4x + 4 + y^2 – 6y + 9 &= x^2 – 12x + 36 + y^2 + 10y + 25 \\ -4x – 6y + 13 &= -12x + 10y + 61 \\ 8x – 16y – 48 &= 0 \implies x – 2y – 6 = 0 \end{aligned}$$
The required relation is $x – 2y – 6 = 0$.
FAQ 4 (HOTS / Circumcenter Calculation). Find the circumcenter of the triangle formed by $A(5, 1), B(-3, -7),$ and $C(7, -1)$.
Answer:
Let the circumcenter be $O(x, y)$, which is equidistant from all three vertices ($OA^2 = OB^2 = OC^2$).
Equating $OA^2 = OC^2$:
$$\begin{aligned} (x – 5)^2 + (y – 1)^2 &= (x – 7)^2 + (y + 1)^2 \\ -10x – 2y + 26 &= -14x + 2y + 50 \implies 4x – 4y = 24 \implies x – y = 6 \quad \text{— (1)} \end{aligned}$$
Equating $OA^2 = OB^2$:
$$\begin{aligned} (x – 5)^2 + (y – 1)^2 &= (x + 3)^2 + (y + 7)^2 \\ -10x – 2y + 26 &= 6x + 14y + 58 \implies -16x – 16y = 32 \implies x + y = -2 \quad \text{— (2)} \end{aligned}$$
Adding (1) and (2) yields $2x = 4 \implies x = 2$.
From (2), $2 + y = -2 \implies y = -4$.
The circumcenter is $(2, -4)$.
FAQ 5 ($y$-axis Partition Ratio). In what ratio does the $y$-axis divide the line segment joining $(-4, 5)$ and $(3, -7)$?
Answer:
Any point on the $y$-axis has $x = 0$. Let the ratio be $k : 1$.
$$0 = \frac{k(3) + 1(-4)}{k + 1} \implies 3k – 4 = 0 \implies k = \frac{4}{3}$$
The $y$-axis divides the segment in the ratio $4 : 3$.
FAQ 6 (Assertion-Reasoning).
- Assertion (A): The point $(0, 4)$ lies on the $y$-axis.
- Reason (R): Any point on the $y$-axis has an $x$-coordinate of $0$.Answer:Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
FAQ 7 (Right Triangle Verification). Show that the points $A(1, 1), B(-1, -1),$ and $C(-\sqrt{3}, \sqrt{3})$ form an equilateral triangle.
Answer:
$$\begin{aligned} AB^2 &= (-1 – 1)^2 + (-1 – 1)^2 = (-2)^2 + (-2)^2 = 4 + 4 = 8 \\ BC^2 &= (-\sqrt{3} + 1)^2 + (\sqrt{3} + 1)^2 = (3 – 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) = 8 \\ CA^2 &= (1 + \sqrt{3})^2 + (1 – \sqrt{3})^2 = (1 + 2\sqrt{3} + 3) + (1 – 2\sqrt{3} + 3) = 8 \end{aligned}$$
Since $AB = BC = CA = \sqrt{8} = 2\sqrt{2}\text{ units}$, $\triangle ABC$ is equilateral.
FAQ 8 (Collinearity Value of $k$). For what value of $k$ are the points $A(2, 3), B(4, k),$ and $C(6, -3)$ collinear?
Answer:
For collinearity, the slope of $AB$ equals the slope of $BC$:
$$\frac{k – 3}{4 – 2} = \frac{-3 – k}{6 – 4} \implies \frac{k – 3}{2} = \frac{-3 – k}{2} \implies k – 3 = -3 – k \implies 2k = 0 \implies k = 0$$
The value of $k$ is $0$.
FAQ 9 (Equidistant Point on $y$-axis). Find a point on the $y$-axis equidistant from $A(6, 5)$ and $B(-4, 3)$.
Answer:
Let $P(0, y)$. Then $PA^2 = PB^2$:
$$\begin{aligned} (0 – 6)^2 + (y – 5)^2 &= [0 – (-4)]^2 + (y – 3)^2 \\ 36 + y^2 – 10y + 25 &= 16 + y^2 – 6y + 9 \\ -10y + 61 &= -6y + 25 \\ -4y &= -36 \implies y = 9 \end{aligned}$$
The point on the $y$-axis is $(0, 9)$.
FAQ 10 (Median Length). Find the length of the median through vertex $A$ of the triangle with vertices $A(0, 0), B(6, 0),$ and $C(0, 8)$.
Answer:
The median through $A$ meets the midpoint $D$ of $BC$:
$$D = \left(\frac{6 + 0}{2}, \frac{0 + 8}{2}\right) = (3, 4)$$
Length of median $AD$:
$$AD = \sqrt{(3 – 0)^2 + (4 – 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ units}$$
The length of the median is $5\text{ units}$.
FAQ 11 (Line Ratio Division). Find the ratio in which the line $2x + y – 4 = 0$ divides the segment joining $A(2, -2)$ and $B(3, 7)$.
Answer:
Let the line divide $AB$ in the ratio $k : 1$. The division point is:
$$P\left(\frac{3k + 2}{k + 1}, \frac{7k – 2}{k + 1}\right)$$
Since $P$ lies on $2x + y – 4 = 0$:
$$\begin{aligned} 2\left(\frac{3k + 2}{k + 1}\right) + \left(\frac{7k – 2}{k + 1}\right) – 4 &= 0 \\ (6k + 4) + (7k – 2) – 4(k + 1) &= 0 \\ 13k + 2 – 4k – 4 &= 0 \\ 9k – 2 &= 0 \implies k = \frac{2}{9} \end{aligned}$$
The ratio is $2 : 9$.
FAQ 12 (Opposite Vertices of a Square). If $(0, -1)$ and $(0, 3)$ are two opposite vertices of a square, find the other two vertices.
Answer:
Let opposite vertices be $A(0, -1)$ and $C(0, 3)$. The center $M$ is the midpoint of $AC$:
$$M = \left(0, \frac{-1 + 3}{2}\right) = (0, 1)$$
Diagonal length $AC = 3 – (-1) = 4\text{ units}$, so semi-diagonal length is $2\text{ units}$.
Since $AC$ lies on the $y$-axis, the perpendicular diagonal $BD$ of length $4$ lies along the horizontal line $y = 1$, extending $2\text{ units}$ left and right from $M(0, 1)$.
The remaining vertices are $(2, 1)$ and $(-2, 1)$.
FAQ 13 (Distance Identity). Show that the distance of the point $(a\cos\theta, a\sin\theta)$ from the origin is independent of $\theta$.
Answer:
$$OP = \sqrt{(a\cos\theta – 0)^2 + (a\sin\theta – 0)^2} = \sqrt{a^2\cos^2\theta + a^2\sin^2\theta} = \sqrt{a^2(\cos^2\theta + \sin^2\theta)} = \sqrt{a^2(1)} = |a|$$
The distance is $|a|$, which is independent of $\theta$.
FAQ 14 (Case-Study: GPS Tower Placement). A town has three communication towers at $A(1, 1)$, $B(5, 4)$, and $C(9, 7)$. Determine if an engineer can connect them along a straight fiber-optic route.
Answer:
Check collinearity via distance:
$$\begin{aligned} AB &= \sqrt{(5 – 1)^2 + (4 – 1)^2} = \sqrt{16 + 9} = 5 \\ BC &= \sqrt{(9 – 5)^2 + (7 – 4)^2} = \sqrt{16 + 9} = 5 \\ AC &= \sqrt{(9 – 1)^2 + (7 – 1)^2} = \sqrt{64 + 36} = 10 \end{aligned}$$
Since $AB + BC = 5 + 5 = 10 = AC$, the points are collinear.
Yes, they can be connected along a single straight route.
FAQ 15 (Midpoint Coordinate Geometry). If the coordinates of the midpoints of the sides of a triangle are $(1, 1), (2, -3),$ and $(3, 4)$, find the coordinates of its vertices.
Answer:
Let vertices be $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$.
$$\begin{aligned} x_1 + x_2 &= 2, \quad x_2 + x_3 = 4, \quad x_3 + x_1 = 6 \implies x_1 + x_2 + x_3 = 6 \\ x_3 &= 6 – 2 = 4, \quad x_1 = 6 – 4 = 2, \quad x_2 = 6 – 6 = 0 \end{aligned}$$
$$\begin{aligned} y_1 + y_2 &= 2, \quad y_2 + y_3 = -6, \quad y_3 + y_1 = 8 \implies y_1 + y_2 + y_3 = 2 \\ y_3 &= 2 – 2 = 0, \quad y_1 = 2 – (-6) = 8, \quad y_2 = 2 – 8 = -6 \end{aligned}$$
The vertices are $(2, 8), (0, -6),$ and $(4, 0)$.
To secure full marks on Coordinate Geometry in CBSE examinations, structure every numerical solution methodically. Always state the given coordinate pairs explicitly before calculation. Write the complete algebraic formula first, substitute coordinates carefully within square brackets to prevent sign errors, and label all intermediate distances with their correct geometric segments. For geometric proofs (such as verifying parallelograms, squares, or isosceles triangles), state the specific algebraic properties being tested (such as equality of opposite sides alongside diagonal comparisons). For ratio problems, use the $k : 1$ formulation to simplify linear algebra, verify the solution using the second coordinate, and write the final answers clearly with “units” or “sq units”.