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Let P, Q be points of trisection of AB, with P closer to A and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, –2).
Trisection means dividing a segment into three equal lengths. First, find the total distance: x increases by 12 (16 minus 4) and y decreases by 9 (–2 minus 7). Dividing these by three gives steps of 4 for x and –3 for y. Starting from A (4, 7) and adding one step gives P (8, 4). Adding another stepRead more
Trisection means dividing a segment into three equal lengths. First, find the total distance: x increases by 12 (16 minus 4) and y decreases by 9 (–2 minus 7). Dividing these by three gives steps of 4 for x and –3 for y. Starting from A (4, 7) and adding one step gives P (8, 4). Adding another step to P gives Q (12, 1). These two points, P and Q, trisect the segment AB.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See less(i) Given the points A (1, – 8), B (– 4, 7) and C (–7, – 4), show that they lie on a circle K whose center is the origin O (0, 0). What is the radius of circle K? (ii) Given the points D (–5, 6) and E (0, 9), check whether D and E lie within the circle, on the circle or outside the circle K.
(i) By calculating the distance from the origin O (0, 0) to points A, B and C using the formula (x squared plus y squared), we find all equal 65. Thus, they lie on circle K with a radius of the square root of 65. (ii) For point D (–5, 6), the sum is 61, which is less than 65, placing it inside the cRead more
(i) By calculating the distance from the origin O (0, 0) to points A, B and C using the formula (x squared plus y squared), we find all equal 65. Thus, they lie on circle K with a radius of the square root of 65. (ii) For point D (–5, 6), the sum is 61, which is less than 65, placing it inside the circle. For point E (0, 9), the sum is 81, placing it outside the circle.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessThe midpoints of the sides of triangle ABC are the points D, E and F. Given that the coordinates of D, E and F are (5, 1), (6, 5) and (0, 3), respectively, find the coordinates of A, B and C.
In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposRead more
In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposite F), we calculate C = D + E – F, resulting in (11, 3). These are the three vertices.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessA city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction. (i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines. (ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find: (a) how many street intersections can be referred to as (4, 3). (b) how many street intersections can be referred to as (3, 4).
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th NortRead more
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th North-South street meets the 3rd East-West street. (b) Likewise, there is only one unique intersection referred to as (3, 4). These represent two different physical locations in the city.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessA computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180Read more
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180 pixels. Since the distance between the centers is less than the combined radii, the two circular icons must overlap or intersect.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
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