Kaliyuga Ekalavya

NCERT Class X Chapter 7: Coordinate Geometry Question: Find a relation between \( x \) and \( y \) such that the point \( (x, y) \) is equidistant from the point \( (3, 6) \) and \( (–3, 4) \). Given: Let \( P(x, y) \) be a point equidistant from \( A(3, 6) \) and \( B(-3, 4) \). To Find: The relation between \( x \) and \( y \). Formula: Distance formula: $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let the distances from \( P(x, y) \) to \( A(3, 6) \) and \( B(-3, 4) \) be equal. $$ PA = PB $$ $$ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2} $$ Step 2: Square both sides to remove the square roots. $$ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 $$ Step 3: Expand both sides. $$ (x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16) $$ Step 4: Combine like terms and simplify. $$ x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 $$ $$ -6x - 12y ...

NCERT Class X Chapter 7: Coordinate Geometry Question: 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. Given: Points: Q(0, 1), P(5, -3), R(x, 6) To Find: 1. The values of \( x \) such that Q is equidistant from P and R. 2. The distances QR and PR. Formula: The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: $$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let the coordinates be Q(0, 1), P(5, -3), and R(x, 6). Since Q is equidistant from P and R, we have: $$ \text{QP} = \text{QR} $$ Step 2: Find QP using the distance formula: $$ \text{QP} = \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} $$ Step 3: Find QR using the distance formula: $$ \text{QR} = \sqrt{(x - 0)^2 + (6 - 1)^2} = \sqrt{x^2 + 25} $$ Step 4: Set QP = QR and solve for \( x \): $$ \sqrt{41} = \sqrt{x^2 + 25} $$ Squaring both ...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the values of \( y \) for which the distance between the points \( P(2, -3) \) and \( Q(10, y) \) is 10 units. Given: Points: \( P(2, -3) \), \( Q(10, y) \) Distance \( PQ = 10 \) units To Find: The values of \( y \) for which the distance between \( P \) and \( Q \) is 10 units. Formula: Distance formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Solution: Step 1: Assign the coordinates to the points. \[ P(x_1, y_1) = (2, -3), \quad Q(x_2, y_2) = (10, y) \] Step 2: Write the distance formula for \( PQ \) and substitute the values. \[ PQ = \sqrt{(10 - 2)^2 + (y - (-3))^2} \] \[ 10 = \sqrt{8^2 + (y + 3)^2} \] Step 3: Simplify inside the square root. \[ 10 = \sqrt{64 + (y + 3)^2} \] Step 4: Square both sides to remove the square root. \[ (10)^2 = 64 + (y + 3)^2 \] \[ 100 = 64 + (y + 3)^2 \] Step 5: Rearrange to solve for \( (y + 3)^2 \). \[ (y + 3)^2 = 100 - ...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). Given: Point A: \( (2, -5) \) Point B: \( (-2, 9) \) To Find: The point on the x-axis which is equidistant from \( (2, -5) \) and \( (-2, 9) \). Formula: Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let the required point on the x-axis be \( (x, 0) \). Step 2: Find the distance from \( (x, 0) \) to \( (2, -5) \) using the distance formula. $$ \text{Distance}_1 = \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - 2)^2 + 25} $$ Step 3: Find the distance from \( (x, 0) \) to \( (-2, 9) \) using the distance formula. $$ \text{Distance}_2 = \sqrt{(x + 2)^2 + (0 - 9)^2} = \sqrt{(x + 2)^2 + 81} $$ Step 4: Set the two distances equal since the point is equidistant from both points. $$ \sqrt{(x - 2)^2 +...

NCERT Class X Chapter 7: Coordinate Geometry Question: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2) Given: Points: \( A(4, 5),\ B(7, 6),\ C(4, 3),\ D(1, 2) \) To Find: The type of quadrilateral formed by the given points. Formula: Distance formula: \[ \text{Distance between } (x_1, y_1) \text{ and } (x_2, y_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Solution: Step 1: Assign the points as \( A(4,5) \), \( B(7,6) \), \( C(4,3) \), \( D(1,2) \). Step 2: Find the length of \( AB \): \[ AB = \sqrt{(7-4)^2 + (6-5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \] Step 3: Find the length of \( BC \): \[ BC = \sqrt{(4-7)^2 + (3-6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \] Step 4: Find the length of \( CD \): \[ CD = \sqrt{(1-4)^2 + (2-3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \] Step 5: Find the length of \( DA \): \...

NCERT Class X Chapter 7: Coordinate Geometry Question: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, –4) Given: Points: \( A(-3, 5) \), \( B(3, 1) \), \( C(0, 3) \), \( D(-1, -4) \) To Find: The type of quadrilateral formed by the given points. Formula: Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \): $$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Slope of the line joining \( (x_1, y_1) \) and \( (x_2, y_2) \): $$ m = \frac{y_2 - y_1}{x_2 - x_1} $$ Solution: Step 1: Assign the points as follows: \( A(-3, 5) \), \( B(3, 1) \), \( C(0, 3) \), \( D(-1, -4) \) Step 2: Find the lengths of all sides using the distance formula. Step 3: Calculate \( AB \): $$ AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} $$ Step 4: Calculate \( BC \): $$ BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 ...

NCERT Class X Chapter 7: Coordinate Geometry Question: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–1, –2), (1, 0), (–1, 2), (–3, 0) Given: Points: \( A(-1, -2) \), \( B(1, 0) \), \( C(-1, 2) \), \( D(-3, 0) \) To Find: The type of quadrilateral formed by the given points. Formula: Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \): $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Find the length of side \( AB \). $$ AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$ Step 2: Find the length of side \( BC \). $$ BC = \sqrt{((-1) - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$ Step 3: Find the length of side \( CD \). $$ CD = \sqrt{((-3) - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$ Step 4: Find the length of side \( DA \). ...

NCERT Class X Chapter 7: Coordinate Geometry Question: In a classroom, 4 friends are seated at the points A, B, C and D as shown in the figure. 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. Given: The coordinates of the points are: A(1, 1) B(4, 1) C(4, 5) D(1, 5) To Find: Using the distance formula, determine whether ABCD is a square. Formula: Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Find the length of side AB. $$ AB = \sqrt{(4-1)^2 + (1-1)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 $$ Step 2: Find the length of side BC. $$ BC = \sqrt{(4-4)^2 + (5-1)^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4 $$ Step 3: Find the length of side CD. $$ CD = \sqrt{(1-4)^2 + (5-5)^2} ...

NCERT Class X Chapter 7: Coordinate Geometry Question: Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle. Given: Points: \( A(5, -2) \), \( B(6, 4) \), \( C(7, -2) \). To Find: Whether triangle \( ABC \) is an isosceles triangle. Formula: Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \): $$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Find the length of side \( AB \). $$ AB = \sqrt{(6 - 5)^2 + [4 - (-2)]^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} $$ Step 2: Find the length of side \( BC \). $$ BC = \sqrt{(7 - 6)^2 + [(-2) - 4]^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37} $$ Step 3: Find the length of side \( CA \). $$ CA = \sqrt{(5 - 7)^2 + [(-2) - (-2)]^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4 + 0} = 2 $$ Step 4: Compare the side lengths. We have \( AB = \sqrt{37} \), \( BC = \sqrt{37} \), and \( CA = 2 \). Since two sides are equal, the triangle is isosceles. ...

NCERT Class X Chapter 7: Coordinate Geometry Question: Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear. Given: Points: \( A(1, 5) \), \( B(2, 3) \), \( C(-2, -11) \) To Find: Are the points \( A \), \( B \), and \( C \) collinear? Formula: Area of triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \): $$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$ If the area is 0, the points are collinear. Solution: Step 1: Assign coordinates to the points. $$ A(x_1, y_1) = (1, 5),\quad B(x_2, y_2) = (2, 3),\quad C(x_3, y_3) = (-2, -11) $$ Step 2: Substitute the values in the area formula. $$ \text{Area} = \frac{1}{2} \left| 1(3 - (-11)) + 2((-11) - 5) + (-2)(5 - 3) \right| $$ Step 3: Simplify the expressions inside the modulus. $$ = \frac{1}{2} \left| 1 \times 14 + 2 \times (-16) + (-2) \times 2 \right| $$ Step 4: Calculate each term. $$ = \frac{1}{2} \left| 14 - 32...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B, where town B is located 36 km east and 15 km north of the town A? Given: Point A: (0, 0) Point B: (36, 15) Town B is 36 km east and 15 km north of town A To Find: The distance between the points (0, 0) and (36, 15) The distance between town A and town B Formula: Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \): $$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let the points be \( A(0, 0) \) and \( B(36, 15) \). Step 2: Apply the distance formula: $$ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Here, \( x_1 = 0,\, y_1 = 0,\, x_2 = 36,\, y_2 = 15 \). Step 3: Substitute the values into the formula: $$ AB = \sqrt{(36 - 0)^2 + (15 - 0)^2} $$ Step 4: Simplify the squares: $$ AB = \sqrt{3...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the distance between the points \((a, b)\) and \((-a, -b)\). Given: Two points: \((a, b)\) and \((-a, -b)\) To Find: The distance between the points \((a, b)\) and \((-a, -b)\). Formula: Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let \((x_1, y_1) = (a, b)\) and \((x_2, y_2) = (-a, -b)\). Step 2: Substitute the values into the distance formula: $$ \sqrt{((-a) - a)^2 + ((-b) - b)^2} $$ Step 3: Simplify the expressions inside the square root: $$ \sqrt{(-2a)^2 + (-2b)^2} $$ Step 4: Calculate the squares: $$ \sqrt{4a^2 + 4b^2} $$ Step 5: Factor out 4 from the square root: $$ \sqrt{4(a^2 + b^2)} $$ Step 6: Take the square root of 4: $$ 2\sqrt{a^2 + b^2} $$ Result: The distance between the points \((a, b)\) and \((-a, -b)\) is: $$ 2\sqrt{a^2 + b^2} $$ Next question solution: ...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the distance between the points (–5, 7) and (–1, 3). Given: Points \(A(-5, 7)\) and \(B(-1, 3)\). To Find: The distance between points \(A\) and \(B\). Formula: Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Assign the coordinates. $$ x_1 = -5,\quad y_1 = 7,\quad x_2 = -1,\quad y_2 = 3 $$ Step 2: Substitute the values into the distance formula. $$ AB = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} $$ Step 3: Simplify the expressions inside the square root. $$ AB = \sqrt{(4)^2 + (-4)^2} $$ Step 4: Calculate the squares and add. $$ AB = \sqrt{16 + 16} $$ Step 5: Add the values inside the root. $$ AB = \sqrt{32} $$ Step 6: Express 32 as \(16 \times 2\) and simplify. $$ AB = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2} $$ Result: The...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the distance between the points (2, 3) and (4, 1). Given: Points: \((x_1, y_1) = (2, 3)\), \((x_2, y_2) = (4, 1)\) To Find: The distance between the two given points. Formula: The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: $$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Substitute the given values into the distance formula. $$ d = \sqrt{(4 - 2)^2 + (1 - 3)^2} $$ Step 2: Calculate the differences. $$ d = \sqrt{(2)^2 + (-2)^2} $$ Step 3: Square the differences. $$ d = \sqrt{4 + 4} $$ Step 4: Add the results inside the square root. $$ d = \sqrt{8} $$ Step 5: Simplify the square root. $$ d = \sqrt{4 \times 2} = 2\sqrt{2} $$ Result: The distance between the points (2, 3) and (4, 1) is \(2\sqrt{2}\). Next question solution: NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Qu...