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 $...