Kaliyuga Ekalavya

NCERT Class X Chapter 7: Coordinate Geometry Question: If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. Given: The vertices of a parallelogram are A(1, 2), B(4, y), C(x, 6), and D(3, 5), taken in order. To Find: The values of \( x \) and \( y \). Formula: Midpoint formula: If the endpoints are \( (x_1, y_1) \) and \( (x_2, y_2) \), then the midpoint is $$ \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) $$ In a parallelogram, the diagonals bisect each other, so their midpoints are equal. Solution: Step 1: Let the vertices be \( A(1,2) \), \( B(4,y) \), \( C(x,6) \), \( D(3,5) \) in order. In a parallelogram, the diagonals bisect each other. So, the midpoints of \( AC \) and \( BD \) are equal. Step 2: Find the midpoint of \( AC \): $$ \text{Midpoint of } AC = \left( \frac{1 + x}{2},\ \frac{2 + 6}{2} \right ) = \left( \frac{1 + x}{2},\ 4 \right ) $$ Step 3: Find the midpoint of \( BD ...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) taken in order. Given: The vertices of the rhombus are: A(3, 0) B(4, 5) C(–1, 4) D(–2, –1) To Find: The area of the rhombus. Formula: Area of a rhombus = \( \dfrac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of the diagonals. Also, the area of a quadrilateral with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \) is: \[ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \] Solution: Step 1: Assign the vertices in order as A(3, 0), B(4, 5), C(–1, 4), D(–2, –1). Step 2: Use the formula for the area of a quadrilateral given its vertices: $$ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| $$ Step 3: Substitute the coordinate...

NCERT Class X Chapter 7: Coordinate Geometry Question: If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = (3/7)AB and P lies on the line segment AB. Given: A = (–2, –2) B = (2, –4) AP = \( \dfrac{3}{7} \) AB P lies on the line segment AB To Find: Coordinates of P Formula: Section formula: If a point P divides the line segment joining A(\(x_1, y_1\)) and B(\(x_2, y_2\)) in the ratio \(m:n\), then \[ P = \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right) \] Solution: Step 1: Let the coordinates of P be \( (x, y) \). Step 2: Since \( AP = \dfrac{3}{7} AB \), P divides AB in the ratio 3:4 (because AP:PB = 3:4). Step 3: Using the section formula, substitute the values: \[ x = \frac{3 \times 2 + 4 \times (-2)}{3 + 4} \] \[ y = \frac{3 \times (-4) + 4 \times (-2)}{3 + 4} \] Step 4: Simplify the numerators and denominators: \[ x = \frac{6 + (-8)}{7} =...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the coordinates of the points which divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts. Given: Points \( A(-2,\,2) \) and \( B(2,\,8) \). To Find: The coordinates of the points that divide the line segment \( AB \) into four equal parts. Formula: Section formula: The coordinates of the point \( P \) dividing the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \) are: $$ P = \left( \frac{mx_2 + nx_1}{m+n}, \; \frac{my_2 + ny_1}{m+n} \right) $$ Solution: Step 1: Let the points dividing \( AB \) into four equal parts be \( P \), \( Q \), and \( R \). These points divide \( AB \) at 1/4, 1/2, and 3/4 of the way from \( A \) to \( B \). Step 2: The required points divide \( AB \) in the ratios 1:3, 1:1, and 3:1 respectively. Let us calculate the coordinates for each. Step 3: For the point \( P_1 \) dividing \( AB \) in the ratio \( 1:3 ...

NCERT Class X Chapter 7: Coordinate Geometry Question: 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). Given: The centre of the circle is at \( O(2, -3) \). One end of the diameter is \( B(1, 4) \). To Find: The coordinates of point \( A \), the other end of the diameter \( AB \). Formula: Midpoint formula: If the endpoints of a line segment are \( (x_1, y_1) \) and \( (x_2, y_2) \), then the midpoint is given by: $$ \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) $$ Solution: Step 1: Let the coordinates of point \( A \) be \( (x, y) \). Step 2: Since \( O \) is the midpoint of \( AB \), using the midpoint formula: $$ \left( \frac{x + 1}{2},\ \frac{y + 4}{2} \right) = (2,\ -3) $$ Step 3: Equate the \( x \)-coordinates and solve for \( x \): $$ \frac{x + 1}{2} = 2 \\ x + 1 = 4 \\ x = 3 $$ Step 4: Equate the \( y \)-coordinates and solve fo...

NCERT Class X Chapter 7: Coordinate Geometry Question: 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. Given: Points A(1, –5) and B(–4, 5). To Find: The ratio in which the x-axis divides the line segment AB. The coordinates of the point of division. Formula: Section formula: If a point \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then \[ x = \frac{mx_2 + nx_1}{m + n}, \quad y = \frac{my_2 + ny_1}{m + n} \] Solution: Step 1: Let the x-axis divide AB at point \( P(x, 0) \) in the ratio \( k:1 \). Step 2: Using the section formula, the coordinates of \( P \) are: \[ x = \frac{k \cdot (-4) + 1 \cdot 1}{k + 1} = \frac{-4k + 1}{k + 1} \] \[ y = \frac{k \cdot 5 + 1 \cdot (-5)}{k + 1} = \frac{5k - 5}{k + 1} \] Step 3: Since the point lies on the x-axis, its y-c...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6). Given: Point A: \( (-3, 10) \) Point B: \( (6, -8) \) Point P: \( (-1, 6) \), which divides AB To Find: The ratio in which point P divides the line segment AB. Formula: Section formula: If a point \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then \[ x = \frac{mx_2 + nx_1}{m + n}, \quad y = \frac{my_2 + ny_1}{m + n} \] Solution: Step 1: Let the required ratio be \( m:n \). Using the section formula, the coordinates of the dividing point \( P \) are: \[ P = \left( \frac{m \cdot 6 + n \cdot (-3)}{m+n},\ \frac{m \cdot (-8) + n \cdot 10}{m+n} \right) \] Step 2: Since \( P = (-1, 6) \), equate the x-coordinates: \[ -1 = \frac{6m - 3n}{m+n} \] Step 3: Cross-multiply and solve for \( m \) and \( n \): \[ -1(m+n) = 6m - 3...

NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.2 Question 3 Question: To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Figure. Niharika runs (1/4)th the distance AD on the 2nd line and (1/5)th the distance AD on the eighth 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? Given: Distance between lines = 1 m Number of flower pots along AD = 100 Distance between flower pots = 1 m Total length of AD = 100 m Niharika posts a red flag at \( \frac{1}{4} \)th of AD on the 2nd line Niharika posts another red flag at \( \frac{1}{5} \)th of AD on the 8th line To Find: The distance between the two red flags The p...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3). Given: The endpoints of the line segment are: A(4, -1) and B(-2, -3) To Find: The coordinates of the points that divide the line segment AB into three equal parts (i.e., the points of trisection). Formula: If a point divides the line joining \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m:n\), then its coordinates are: $$ \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right) $$ Solution: Step 1: Let the points of trisection be \(P\) and \(Q\). \(P\) divides AB in the ratio \(1:2\) (i.e., \(AP:PB = 1:2\)). \(Q\) divides AB in the ratio \(2:1\) (i.e., \(AQ:QB = 2:1\)). Step 2: Find the coordinates of \(P\) (dividing AB in \(1:2\)). $$ P = \left( \frac{1 \times (-2) + 2 \times 4}{1+2},\ \frac{1 \times (-3) + 2 \times (-1)}{1+2} \right) $$ Step 3: Calculate the x-coordinate and y-c...

NCERT Class X Chapter 7: Coordinate Geometry Question: Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. Given: Points: \( A = (-1, 7) \), \( B = (4, -3) \) Ratio: \( 2 : 3 \) To Find: The coordinates of the point dividing \( AB \) in the ratio \( 2 : 3 \). Formula: Section formula: If a point \( P \) divides the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m : n \), then the coordinates of \( P \) are: $$ x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n} $$ Solution: Step 1: Let the required point be \( P(x, y) \). Identify the values to substitute: \( (x_1, y_1) = (-1, 7) \), \( (x_2, y_2) = (4, -3) \), \( m = 2, \ n = 3 \) Step 2: Apply the section formula for the x-coordinate: $$ x = \frac{m x_2 + n x_1}{m + n} = \frac{2 \times 4 + 3 \times (-1)}{2 + 3} $$ Step 3: Simplify the x-coordinate: $$ x = \frac{8 + (-3)}{5} = \frac{5}{5} = 1 $...

NCERT Class X Chapter 7: Coordinate Geometry Example 10 Question: If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p. Given: Points A(6, 1), B(8, 2), C(9, 4), and D(p, 3) are the vertices of a parallelogram ABCD. To Find: The value of \( p \). Formula: Midpoint formula: The midpoint of a line segment with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is: $$ \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) $$ Solution: Step 1: In a parallelogram, the diagonals bisect each other. Thus, the midpoints of diagonals \( AC \) and \( BD \) must be the same. Step 2: Find the midpoint of diagonal \( AC \), where \( A(6, 1) \) and \( C(9, 4) \): $$ \text{Midpoint of } AC = \left( \frac{6 + 9}{2},\ \frac{1 + 4}{2} \right) = \left( \frac{15}{2},\ \frac{5}{2} \right) $$ Step 3: Find the midpoint of diagonal \( BD \), where \( B(8, 2) \) and \( D(p, 3) \): $$ \t...

NCERT Class X Chapter 7: Coordinate Geometry Example 9 Question: 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. Given: Points \( A(5, -6) \) and \( B(-1, -4) \). To Find: The ratio in which the y-axis divides the line segment AB. The coordinates of the point of intersection. Formula: Section formula: The coordinates of the point dividing the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio \( m:n \) are: $$ \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right) $$ Solution: Step 1: Let the y-axis divide the line segment AB at point P in the ratio \( k:1 \). Step 2: The coordinates of P using the section formula are: $$ P = \left( \frac{k \cdot (-1) + 1 \cdot 5}{k+1},\ \frac{k \cdot (-4) + 1 \cdot (-6)}{k+1} \right) $$ Step 3: Since P lies on the y-axis, its x-coordinate is 0. $$ \frac{k \cdot (-1) + 1 \cdot 5}{k+1}...

NCERT Class X Chapter 7: Coordinate Geometry Example 8 Question: Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, –2) and B(–7, 4). Given: Points \( A(2, -2) \) and \( B(-7, 4) \). To Find: Coordinates of the points of trisection of the line segment \( AB \). Formula: Section formula: If a point \( P \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then the coordinates of \( P \) are: \[ P = \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right) \] Solution: Step 1: Let the points of trisection be \( P \) and \( Q \). Point \( P \) divides \( AB \) in the ratio \( 1:2 \), and point \( Q \) divides \( AB \) in the ratio \( 2:1 \). Step 2: Find the coordinates of \( P \) (dividing \( AB \) in \( 1:2 \)). \[ \begin{align*} x_P &= \frac{1 \times (-7) + 2 \times 2}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}...

NCERT Class X Chapter 7: Coordinate Geometry Example 7 Question: In what ratio does the point (–4, 6) divide the line segment joining the points A(–6, 10) and B(3, –8)? Given: A = (–6, 10), B = (3, –8), and P = (–4, 6) divides AB. To Find: The ratio in which the point P divides the line segment AB. Formula: Section formula: If a point \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then \[ x = \frac{mx_2 + nx_1}{m + n}, \quad y = \frac{my_2 + ny_1}{m + n} \] Solution: Step 1: Let the required ratio be \( m:n \). Using the section formula for the x-coordinate: \[ -4 = \frac{m \times 3 + n \times (-6)}{m + n} \] Step 2: Cross multiply and simplify the equation for x-coordinate. \[ -4(m + n) = 3m - 6n \] \[ -4m - 4n = 3m - 6n \] \[ -4m - 4n - 3m + 6n = 0 \] \[ -7m + 2n = 0 \] \[ 2n = 7m \] Step 3: Express the ratio \( \frac{m}{n} \) from the above equation. \[ 2n = 7m \implies \fr...

NCERT Class X Chapter 7: Coordinate Geometry Example 6 Question: Find the coordinates of the point which divides the line segment joining the points (4, –3) and (8, 5) in the ratio 3 : 1 internally. Given: Point A: (4, –3) Point B: (8, 5) The point divides AB in the ratio 3:1 internally. To Find: The coordinates of the point that divides the line segment AB in the ratio 3:1 internally. Formula: If a point \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \) internally, then: $$ x = \frac{mx_2 + nx_1}{m + n}, \qquad y = \frac{my_2 + ny_1}{m + n} $$ Solution: Step 1: Assign values to the variables. Let \( A(x_1, y_1) = (4, -3) \), \( B(x_2, y_2) = (8, 5) \), and the ratio \( m:n = 3:1 \). Step 2: Write the section formula for \( x \)-coordinate. $$ x = \frac{m x_2 + n x_1}{m + n} $$ Substitute the values: $$ x = \frac{3 \times 8 + 1 \times 4}{3 + 1} $$ ...

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 - ...