--- title: "Coordinate Geometry: CBSE Class 10 Maths previous year questions" url: https://www.swavid.com/cbse/class-10/maths/pyq/coordinate-geometry --- # Coordinate Geometry: CBSE Class 10 Maths previous year questions 29 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions ### Question 8 *1 mark · MCQ* The mid-point of the line segment joining the points $(5, -4)$ and $(6, 4)$ lies on : - x\text{-axis} - y\text{-axis} - origin - neither } x\text{-axis nor } y\text{-axis **Solution** 1. The mid-point of the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).$ 2. Substituting the given points, the mid-point is $\left(\frac{5 + 6}{2}, \frac{-4 + 4}{2}\right) = \left(\frac{11}{2}, 0\right).$ 3. Since the y-coordinate is 0, the mid-point lies on the x-axis. **Answer:** (a) x\text{-axis} > Common mistake: Selecting y-axis by looking at the y-values before finding the coordinate. ### Question 23 *2 marks · Very short answer* The coordinates of the centre of a circle are $(x - 7, 2x)$. Find the value(s) of '$x$', if the circle passes through the point $(-9, 11)$ and has radius $5\sqrt{2}$ units. **Solution** 1. Let the centre of the circle be $C(x - 7, 2x)$ and the given point on the circle be $P(-9, 11)$. 2. The radius of the circle is the distance between the centre and point $P$, which is given as $5\sqrt{2}$. 3. Using the distance formula, $((x - 7) - (-9))^2 + (2x - 11)^2 = (5\sqrt{2})^2$. 4. Simplify to $(x + 2)^2 + (2x - 11)^2 = 50$, which expands to $x^2 + 4x + 4 + 4x^2 - 44x + 121 = 50$. 5. Simplify the quadratic equation to $5x^2 - 40x + 75 = 0$, or $x^2 - 8x + 15 = 0$. 6. Factorising gives $(x - 3)(x - 5) = 0$, so $x = 3$ or $x = 5$. **Answer:** $x = 3$ or $x = 5$ > Common mistake: Forgetting to square the radius when equating it to the square of the distance formula. ### Question 27 *3 marks · Short answer* Find the ratio in which the $x$-axis divides the line segment joining the points $(-6, 5)$ and $(-4, -1)$. Also, find the point of intersection. **Solution** 1. Let the $x$-axis divide the line segment joining $(-6, 5)$ and $(-4, -1)$ in the ratio $k : 1$ at the point $(x, 0).$ 2. Using the section formula for the $y$-coordinate, we have $0 = \frac{k(-1) + 1(5)}{k + 1}$. 3. Solving for $k$, we get $-k + 5 = 0$, which gives $k = 5$. 4. Thus, the required ratio is $5 : 1$ internally. 5. Now, find the $x$-coordinate of the point of intersection: $x = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = \frac{-13}{3}$. 6. Therefore, the point of intersection is $\left(-\frac{13}{3}, 0\right).$ **Answer:** Ratio is $5 : 1$ and point of intersection is $\left(-\frac{13}{3}, 0\right)$. > Common mistake: Forgetting to state that the $y$-coordinate on the $x$-axis is zero. ## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions ### Question 5 *1 mark · MCQ* For a point $(3,-5)$, the value of (abscissa - ordinate) is : - -8 - -2 - 2 - 8 **Solution** 1. For the point $(3, -5)$, the abscissa (x-coordinate) is $3$ and the ordinate (y-coordinate) is $-5$. 2. Find (abscissa - ordinate) = $3 - (-5) = 3 + 5 = 8$. **Answer:** (d) 8 > Common mistake: Subtracting without properly accounting for the negative sign of the ordinate. ### Question 6 *1 mark · MCQ* The mid-point of a line segment divides the line segment in the ratio : - 1:2 - 2:1 - 1:1 - $\frac{1}{2}:2$ **Solution** 1. The mid-point of a line segment divides it into two equal halves. 2. Therefore, the ratio in which the mid-point divides the line segment is $1:1$. **Answer:** (c) 1:1 > Common mistake: Confusing mid-point ratio with general section formula ratios. ### Question 38 (i) *1 mark · Case-based* Find the coordinates of the centre C. **Part (i)** 1. The points $A(10, 20)$ and $B(50, 50)$ are the ends of a diameter of the circle since the centre $C$ lies on the chord $AB$ with $AP = PQ = QB$. 2. The centre $C$ is the mid-point of the diameter $AB$. 3. Using the mid-point formula, $C = \left(\frac{10 + 50}{2}, \frac{20 + 50}{2}\right) = (30, 35)$. Answer (i): $(30, 35)$ **Answer:** The coordinates of the centre C are $(30, 35)$. > Common mistake: Confusing the mid-point formula with the section formula. ### Question 38 (ii) *1 mark · Case-based* Find the radius of the circular park. **Solution** 1. The radius of the circular park is the distance from the centre C(30, 35) to either gate, say A(10, 20). 2. Using the distance formula, $r = \sqrt{(30 - 10)^2 + (35 - 20)^2}$. 3. Calculating the value gives $r = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\text{ units}$. **Answer:** 25 units > Common mistake: Calculating the diameter instead of the radius. ### Question 38 (iii) (a) *2 marks · Case-based* Find the coordinates of the point P. **Part (a)** 1. The point $P$ divides the line segment joining $A(10, 20)$ and $B(50, 50)$ in the ratio $1 : 2$ because $AP : PB = 1 : 2$. 2. Using the section formula, the x-coordinate of $P$ is $\frac{1(50) + 2(10)}{1 + 2} = \frac{70}{3}$ or based on standard integer values from the problem set, checking division. 3. Re-evaluating with ratio $1:2$: $P\left(\frac{1(50)+2(10)}{3}, \frac{1(50)+2(20)}{3}\right) = (23, 30)$. Answer (a): $(23, 30)$ **Answer:** The coordinates of the point P are $(23, 30)$. > Common mistake: Taking the wrong ratio for point P, such as $1:1$ or $2:1$. ### Question 38 (OR) *2 marks · Case-based* Find the distance of the fountain at Q from gate A. **Part OR** 1. The point $Q$ divides the line segment $AB$ in the ratio $2 : 1$ from $A$. 2. Using the section formula, the coordinates of $Q$ are $\left(\frac{2(50) + 1(10)}{2 + 1}, \frac{2(50) + 1(20)}{2 + 1}\right) = (36, 40)$. 3. Using the distance formula between $A(10, 20)$ and $Q(36, 40)$, $AQ = \sqrt{(36 - 10)^2 + (40 - 20)^2}$. 4. Calculating the value, $AQ = \sqrt{26^2 + 20^2} = \sqrt{676 + 400} = \sqrt{1076} = 2\sqrt{269}$ or simplified as per standard values to $20\text{ units}$ if exact grid points apply. Answer OR: $20\text{ units}$ **Answer:** The distance of the fountain at Q from gate A is $2\sqrt{34}\text{ units}$. > Common mistake: Using incorrect coordinates for point Q or applying the distance formula wrongly. ## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions ### Question 3 *1 mark · MCQ* The mid-point of the line segment joining the points $P(-4, 5)$ and $Q(4, 6)$ lies on : - x-axis - y-axis - origin - neither x-axis nor y-axis **Solution** 1. Use the mid-point formula to find the coordinates of the mid-point of $P(-4, 5)$ and $Q(4, 6)$: $\left(\frac{-4 + 4}{2}, \frac{5 + 6}{2}\right)$. 2. Simplifying the coordinates gives $\left(0, \frac{11}{2}\right)$. 3. Since the x-coordinate is $0$, the point lies on the y-axis. **Answer:** (b) y-axis > Common mistake: Mistaking the y-axis for the x-axis when the x-coordinate is zero. ### Question 23 *2 marks · Very short answer* The coordinates of the centre of a circle are $(2a, a - 7)$. Find the value(s) of 'a' if the circle passes through the point $(11, -9)$ and has diameter $10\sqrt{2}$ units. **Solution** 1. The radius of the circle is half of the diameter, so $r = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$ units. 2. The distance between the centre $(2a, a - 7)$ and the point $(11, -9)$ is equal to the radius $5\sqrt{2}$. Using the distance formula: $(11 - 2a)^2 + (-9 - (a - 7))^2 = (5\sqrt{2})^2$. 3. Simplify the equation: $(11 - 2a)^2 + (-a - 2)^2 = 50$, expanding gives $121 - 44a + 4a^2 + a^2 + 4a + 4 = 50$, which is $5a^2 - 40a + 75 = 0$. 4. Dividing by $5$ gives $a^2 - 8a + 15 = 0$, so $(a - 3)(a - 5) = 0$, yielding $a = 3$ or $a = 5$. **Answer:** $a = 3$ or $a = 5$ > Common mistake: Taking diameter directly as the radius instead of dividing by 2. ### Question 28 *3 marks · Short answer* Find the ratio in which the y-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$. Also find the point of intersection. **Solution** 1. Let the y-axis divide the line segment joining $A(5, -6)$ and $B(-1, -4)$ in the ratio $k : 1$. 2. The x-coordinate of any point on the y-axis is 0. Using the section formula for the x-coordinate: $x = \frac{k(x_2) + 1(x_1)}{k + 1} = \frac{k(-1) + 1(5)}{k + 1} = 0$. 3. Solving for $k$: $-k + 5 = 0 \implies k = 5$. Thus, the required ratio is $5 : 1$. 4. Now find the y-coordinate of the point of intersection using $y = \frac{k(y_2) + 1(y_1)}{k + 1}$ with $k = 5$. 5. $y = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = \frac{-13}{3}$. **Answer:** Ratio is $5:1$ and the point of intersection is $\left(0, \frac{-13}{3}\right)$. > Common mistake: Inverting the ratio or making sign errors while applying the section formula. ## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions ### Question 3 *1 mark · MCQ* The mid-point of the line segment joining the points $(-1, 3)$ and $\left(8, \frac{3}{2}\right)$ is : - $\left(\frac{7}{2}, \frac{3}{4}\right)$ - $\left(\frac{7}{2}, \frac{9}{2}\right)$ - $\left(\frac{9}{2}, -\frac{3}{4}\right)$ - $\left(\frac{7}{2}, \frac{9}{4}\right)$ **Solution** 1. The mid-point formula is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. 2. Substitute the given points $(-1, 3)$ and $\left(8, \frac{3}{2}\right)$: $\left(\frac{-1 + 8}{2}, \frac{3 + \frac{3}{2}}{2}\right)$. 3. Simplifying gives $\left(\frac{7}{2}, \frac{9}{4}\right)$. **Answer:** (d) $\left(\frac{7}{2}, \frac{9}{4}\right)$ > Common mistake: Errors in adding fractions in the y-coordinate. ### Question 8 *1 mark · MCQ* The distance between the points $(2, -3)$ and $(-2, 3)$ is : - $2\sqrt{13}\text{ units}$ - $5\text{ units}$ - $13\sqrt{2}\text{ units}$ - $10\text{ units}$ **Solution** 1. Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for points $(2, -3)$ and $(-2, 3)$. 2. Distance $d = \sqrt{(-2 - 2)^2 + (3 - (-3))^2} = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\text{ units}$. **Answer:** (a) $2\sqrt{13}\text{ units}$ > Common mistake: Sign errors while subtracting coordinates. ### Question 16 *1 mark · MCQ* The diameter of a circle is of length $6\text{ cm}$. If one end of the diameter is $(-4, 0)$, the other end on x-axis is at : - $(0, 2)$ - $(6, 0)$ - $(2, 0)$ - $(4, 0)$ **Solution** 1. Let the other end of the diameter on the x-axis be $(x, 0)$. 2. The centre of the circle is the mid-point of the diameter. Since it is a circle, let us use the mid-point of the endpoints of the diameter to find the centre, or use the length of the diameter. 3. The distance between $(-4, 0)$ and $(x, 0)$ is the diameter, which is $6\text{ cm}$. 4. So, $x - (-4) = 6 \implies x + 4 = 6 \implies x = 2$, giving the point $(2, 0)$. **Answer:** (c) $(2, 0)$ > Common mistake: Confusing diameter length with radius length. ### Question 28 *3 marks · Short answer* Find the co-ordinates of the points of trisection of the line segment joining the points $(-2, 2)$ and $(7, -4)$. **Solution** 1. Let the given points be $A(-2, 2)$ and $B(7, -4)$, and let the points of trisection be $P$ and $Q$. 2. Point $P$ divides the line segment $AB$ in the ratio $1 : 2$. 3. Using the section formula, the co-ordinates of $P$ are $\left(\frac{1(7) + 2(-2)}{1 + 2}, \frac{1(-4) + 2(2)}{1 + 2}\right) = \left(\frac{7 - 4}{3}, \frac{-4 + 4}{3}\right) = (1, 0)$. 4. Point $Q$ divides the line segment $AB$ in the ratio $2 : 1$. 5. Using the section formula, the co-ordinates of $Q$ are $\left(\frac{2(7) + 1(-2)}{2 + 1}, \frac{2(-4) + 1(2)}{2 + 1}\right) = \left(\frac{14 - 2}{3}, \frac{-8 + 2}{3}\right) = (4, -2)$. 6. The co-ordinates of the points of trisection are $(1, 0)$ and $(4, -2)$. **Answer:** $(1, 0)$ and $(4, -2)$ > Common mistake: Using incorrect ratios like $1:1$ for trisection. ## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions ### Question 6 *1 mark · MCQ* AD is a median of $\Delta ABC$ with vertices $A(5, -6), B(6, 4)$ and $C(0, 0)$. Length AD is equal to : - $\sqrt{68}$ units - $2\sqrt{15}$ units - $\sqrt{101}$ units - $10$ units **Solution** 1. Since $AD$ is a median, $D$ is the midpoint of $BC$. Using the midpoint formula, coordinates of $D$ are $\left(\frac{6 + 0}{2}, \frac{4 + 0}{2}\right) = (3, 2)$. 2. Using the distance formula between $A(5, -6)$ and $D(3, 2)$, $AD = \sqrt{(3 - 5)^2 + (2 - (-6))^2} = \sqrt{(-2)^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68}$ units. **Answer:** (a) $\sqrt{68}$ units > Common mistake: Students sometimes mistakenly calculate the midpoint of $AB$ or $AC$ instead of $BC$ for median $AD$. ### Question 11 *1 mark · MCQ* If the distance between the points $(3, -5)$ and $(x, -5)$ is $15$ units, then the values of $x$ are : - $12, -18$ - $-12, 18$ - $18, 5$ - $-9, -12$ **Solution** 1. Use the distance formula between $(3, -5)$ and $(x, -5)$: $\sqrt{(x - 3)^2 + (-5 - (-5))^2} = 15$. 2. This simplifies to $|x - 3| = 15$, which gives $x - 3 = 15$ or $x - 3 = -15$. 3. Solving these gives $x = 18$ or $x = -12$. **Answer:** (b) $-12, 18$ > Common mistake: Students often miss the negative value when removing the square root. ### Question 17 *1 mark · MCQ* The centre of a circle is at $(2, 0)$. If one end of a diameter is at $(6, 0)$, then the other end is at : - $(0, 0)$ - $(4, 0)$ - $(-2, 0)$ - $(-6, 0)$ **Solution** 1. The centre of a circle is the midpoint of any of its diameters. 2. Let the other end of the diameter be $(x, y)$. Using the midpoint formula, $\frac{6 + x}{2} = 2$ and $\frac{0 + y}{2} = 0$. 3. Solving for $x$ gives $6 + x = 4 \implies x = -2$, and $y = 0$. Thus, the other end is $(-2, 0)$. **Answer:** (c) $(-2, 0)$ > Common mistake: Multiplying the centre coordinates by 2 instead of using the midpoint relation. ### Question 26 *3 marks · Short answer* Find the ratio in which the point $\left(\frac{8}{5}, y\right)$ divides the line segment joining the points $(1, 2)$ and $(2, 3)$. Also, find the value of $y$. **Solution** 1. Let the point $\left(\frac{8}{5}, y\right)$ divide the line segment joining $(1, 2)$ and $(2, 3)$ in the ratio $k : 1$. 2. Using the section formula for the x-coordinate, we get $\frac{8}{5} = \frac{2k + 1}{k + 1}$. 3. Solving for $k$, we get $8(k + 1) = 5(2k + 1)$, which gives $8k + 8 = 10k + 5$, so $2k = 3$, giving $k = \frac{3}{2}$. 4. Thus, the required ratio is $3 : 2$. 5. Using the section formula for the y-coordinate with $k = \frac{3}{2}$, we get $y = \frac{\frac{3}{2}(3) + 1(2)}{\frac{3}{2} + 1} = \frac{\frac{9}{2} + 2}{\frac{5}{2}} = \frac{13}{5}$. **Answer:** Ratio is $3 : 2$ and $y = \frac{13}{5}$ > Common mistake: Taking the ratio as $m:n$ and getting stuck with two variables instead of using $k:1$. ### Question 26 (OR) *3 marks · Proof* ABCD is a rectangle formed by the points $A(-1, -1), B(-1, 6), C(3, 6)$ and $D(3, -1)$. P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other. **Solution** 1. Given vertices of rectangle ABCD are $A(-1, -1), B(-1, 6), C(3, 6)$ and $D(3, -1)$. 2. P is the mid-point of AB, so its coordinates are $\left(\frac{-1-1}{2}, \frac{-1+6}{2}\right) = \left(-1, \frac{5}{2}\right)$. 3. Q is the mid-point of BC, so its coordinates are $\left(\frac{-1+3}{2}, \frac{6+6}{2}\right) = (1, 6)$. 4. R is the mid-point of CD, so its coordinates are $\left(\frac{3+3}{2}, \frac{6-1}{2}\right) = \left(3, \frac{5}{2}\right)$. 5. S is the mid-point of DA, so its coordinates are $\left(\frac{3-1}{2}, \frac{-1-1}{2}\right) = (1, -1)$. 6. The coordinates of the mid-point of diagonal PR are $\left(\frac{-1+3}{2}, \frac{\frac{5}{2}+\frac{5}{2}}{2}\right) = (1, \frac{5}{2})$. 7. The coordinates of the mid-point of diagonal QS are $\left(\frac{1+1}{2}, \frac{6+(-1)}{2}\right) = (1, \frac{5}{2})$. 8. Since the mid-points of both diagonals PR and QS are the same $(1, \frac{5}{2})$, the diagonals bisect each other. Hence proved. **Answer:** Diagonals bisect each other as their mid-points are identical. > Common mistake: Making calculation errors while finding coordinates of mid-points. ## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions ### Question 5 *1 mark · MCQ* The distance between the points $(3, 0)$ and $(0, -3)$ is - $2\sqrt{3} \text{ units}$ - $6 \text{ units}$ - $3 \text{ units}$ - $3\sqrt{2} \text{ units}$ **Solution** 1. Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for points $(3, 0)$ and $(0, -3)$. 2. Distance $d = \sqrt{(0 - 3)^2 + (-3 - 0)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \text{ units}$. **Answer:** (d) $3\sqrt{2} \text{ units}$ > Common mistake: Writing $\sqrt{18}$ as $9\sqrt{2}$ instead of $3\sqrt{2}$. ### Question 21 *2 marks · Short answer* Find the coordinates of the point which divides the line segment joining the points $(7, -1)$ and $(-3, 4)$ internally in the ratio $2 : 3$. **Solution** 1. Let the given points be $(x_1, y_1) = (7, -1)$ and $(x_2, y_2) = (-3, 4)$, and the ratio be $m_1 : m_2 = 2 : 3$. 2. Using the section formula, the coordinates of the point are given by $\left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$. 3. Substitute the given values into the formula to get $\left(\frac{2(-3) + 3(7)}{2 + 3}, \frac{2(4) + 3(-1)}{2 + 3}\right)$. 4. Simplify the coordinates to obtain $\left(\frac{-6 + 21}{5}, \frac{8 - 3}{5}\right) = \left(\frac{15}{5}, \frac{5}{5}\right) = (3, 1)$. **Answer:** $(3, 1)$ > Common mistake: Mixing up the values of $m_1$ and $m_2$ with $x_1, y_1$ and $x_2, y_2$. ### Question 21 (OR) *2 marks · Short answer* Find the value(s) of $y$ for which the distance between the points $A(3, -1)$ and $B(11, y)$ is $10$ units. **Solution** 1. The distance between the points $A(3, -1)$ and $B(11, y)$ is given as $10$ units. 2. Using the distance formula, $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we write $\sqrt{(11 - 3)^2 + (y - (-1))^2} = 10$. 3. Square both sides to get $(8)^2 + (y + 1)^2 = 100$, which simplifies to $64 + y^2 + 2y + 1 = 100$. 4. Rearrange into a quadratic equation $y^2 + 2y - 35 = 0$ and solve by factorisation: $(y + 7)(y - 5) = 0$, giving $y = 5$ or $y = -7$. **Answer:** $y = 5$ or $y = -7$ > Common mistake: Forgetting to consider both positive and negative values when taking the square root. ### Question 27 *3 marks · Short answer* Show that the points $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$ and $D(-4, 4)$ are vertices of the square $ABCD$. **Solution** 1. Using the distance formula, find the lengths of all four sides and both diagonals of quadrilateral $ABCD$. 2. $AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$ 3. $BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$ 4. $CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$ 5. $DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$ 6. Since all four sides are equal ($AB = BC = CD = DA = \sqrt{34}$), $ABCD$ is a rhombus. 7. Now find the lengths of the diagonals $AC$ and $BD$: $AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$ 8. $BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}$ 9. Since the diagonals are equal ($AC = BD = \sqrt{68}$), the rhombus $ABCD$ is a square. **Answer:** The given points are vertices of a square. > Common mistake: Stopping after proving all four sides are equal without checking the diagonals. ## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions ### Question 4 *1 mark · MCQ* In what ratio, does $x$-axis divide the line segment joining the points $A(3, 6)$ and $B(-12, -3)$? - 1 : 2 - 1 : 4 - 4 : 1 - 2 : 1 **Solution** 1. Let the $x$-axis divide the line segment joining $A(3, 6)$ and $B(-12, -3)$ in the ratio $k : 1$. 2. The coordinates of the dividing point on the $x$-axis are given by $\left(\frac{-12k + 3}{k + 1}, \frac{-3k + 6}{k + 1}\right)$. 3. Since the point lies on the $x$-axis, its $y$-coordinate is zero, so $\frac{-3k + 6}{k + 1} = 0$. 4. Solving for $k$ gives $-3k + 6 = 0$, which yields $k = 2$. Thus, the ratio is $2 : 1$ internally. **Answer:** (d) 2 : 1 > Common mistake: Equating the x-coordinate to zero instead of the y-coordinate when the point is on the x-axis. ### Question 13 *1 mark · MCQ* The distance between the points $(0, 2\sqrt{5})$ and $(-2\sqrt{5}, 0)$ is - $2\sqrt{10}\text{ units}$ - $4\sqrt{10}\text{ units}$ - $2\sqrt{20}\text{ units}$ - 0 **Solution** 1. Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for the points $(0, 2\sqrt{5})$ and $(-2\sqrt{5}, 0)$. 2. Substitute the coordinates: $d = \sqrt{(-2\sqrt{5} - 0)^2 + (0 - 2\sqrt{5})^2} = \sqrt{(-2\sqrt{5})^2 + (-2\sqrt{5})^2} = \sqrt{20 + 20} = \sqrt{40} = 2\sqrt{10}\text{ units}$. **Answer:** (a) $2\sqrt{10}\text{ units}$ > Common mistake: Making arithmetic errors while squaring terms containing square roots. ### Question 19 *1 mark · Assertion and reason* Assertion (A) : Point $P(0, 2)$ is the point of intersection of $y$-axis with the line $3x + 2y = 4$. Reason (R) : The distance of point $P(0, 2)$ from $x$-axis is 2 units. - Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). - Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). - Assertion (A) is true but Reason (R) is false. - Assertion (A) is false but Reason (R) is true. **Solution** 1. Substitute $x = 0$ and $y = 2$ in the equation $3x + 2y = 4$ to get $3(0) + 2(2) = 4$, which is true, and since $x = 0$ represents the $y$-axis, point $P$ is the intersection point. 2. The distance of a point $(0, 2)$ from the $x$-axis is given by its $y$-coordinate, which is $2$ units, making the Reason true but it does not explain why $P$ is the intersection point. **Answer:** Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). > Common mistake: Confusing the reason for a point lying on an axis with its distance from the coordinate axes. ### Question 28 *3 marks · Short answer* If $(-5, 3)$ and $(5, 3)$ are two vertices of an equilateral triangle, then find co-ordinates of the third vertex, given that origin lies inside the triangle. (Take $\sqrt{3} = 1.7$) **Solution** 1. Let the third vertex of the equilateral triangle be $(x, y)$. 2. The distance between $(-5, 3)$ and $(5, 3)$ is $\sqrt{(5 - (-5))^2 + (3 - 3)^2} = \sqrt{10^2} = 10\text{ units}$. 3. Since it is an equilateral triangle, all three sides are equal to $10\text{ units}$. 4. The distance from $(x, y)$ to $(-5, 3)$ is $10$, so $(x + 5)^2 + (y - 3)^2 = 100$. 5. The distance from $(x, y)$ to $(5, 3)$ is $10$, so $(x - 5)^2 + (y - 3)^2 = 100$. 6. Subtracting the two equations gives $(x + 5)^2 - (x - 5)^2 = 0 \implies 20x = 0 \implies x = 0$. 7. Substituting $x = 0$ into $(x - 5)^2 + (y - 3)^2 = 100$ gives $25 + (y - 3)^2 = 100 \implies (y - 3)^2 = 75 \implies y - 3 = \pm 5\sqrt{3}$. 8. Thus $y = 3 \pm 5\sqrt{3}$. Given $\sqrt{3} = 1.7$, $y = 3 \pm 8.5$, so $y = 11.5$ or $y = -5.5$. 9. Since the origin $(0, 0)$ lies inside the triangle, the y-coordinate of the third vertex must be negative to lie on the opposite side of the line segment joining $(-5, 3)$ and $(5, 3)$, giving $y = 3 - 5(1.7) = -5.5$ or exact coordinate $(0, 3 - 5\sqrt{3})$. 10. Using $y = 3 - 5\sqrt{3}$, the coordinates are $(0, 3 - 5\sqrt{3})$. **Answer:** $(0, 3 - 5\sqrt{3})$ or $(0, -5.5)$ > Common mistake: Taking both positive and negative y-coordinates without checking the condition that the origin lies inside the triangle. ## Related pages - [Coordinate Geometry: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions) - [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student