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Page 1 Co-ordinate Geometry Practice Set 5.1 Q. 1. Find the distance between each of the following pairs of points. (1) A(2, 3), B(4, 1) (2) P(-5, 7), Q(-1, 3) (3) R(0, -3), S(0, 5/2) (4) L(5, -8), M(-7, -3) (5) T(-3, 6), R(9, -10) (6) , X(11, 4) Answer : The distance between points A(x1, y1) and B(x2, y2) is given by, 1. Given Points: A(2, 3) and B(4, 1) We can see that,x1 = 2 x2 = 4 y1 = 3 y2 = 1 Putting the values in the distance formula we get, d = ? d = ? d = v8 2. Given Points: P(-5, 7) and Q(-1, 3) we can see that,x1 = -5 x2 = -1 y1 = 7 y2 = 3 Putting these values in distance formula we get, d = d = v3 2 Page 2 Co-ordinate Geometry Practice Set 5.1 Q. 1. Find the distance between each of the following pairs of points. (1) A(2, 3), B(4, 1) (2) P(-5, 7), Q(-1, 3) (3) R(0, -3), S(0, 5/2) (4) L(5, -8), M(-7, -3) (5) T(-3, 6), R(9, -10) (6) , X(11, 4) Answer : The distance between points A(x1, y1) and B(x2, y2) is given by, 1. Given Points: A(2, 3) and B(4, 1) We can see that,x1 = 2 x2 = 4 y1 = 3 y2 = 1 Putting the values in the distance formula we get, d = ? d = ? d = v8 2. Given Points: P(-5, 7) and Q(-1, 3) we can see that,x1 = -5 x2 = -1 y1 = 7 y2 = 3 Putting these values in distance formula we get, d = d = v3 2 3. Given Points: R(0, -3), S(0, 5/2) we can see that,x1 = 0 x2 = 0 y1 = -3 y2 = 5/2 On putting these values in distance formula we get, d = d = d = 4. Given Points: L(5, -8), M(-7, -3) we can see that, x1 = 5 x2 = -7 y1 = -8 y2 = -3 On putting these values in distance formula we get, d = d = d = v1 6 9 = 13 5. Given Points: T(-3, 6), R(9, -10) we can see that, x1 = -3 x2 = 9 y1 = 6 y2 = -10 Page 3 Co-ordinate Geometry Practice Set 5.1 Q. 1. Find the distance between each of the following pairs of points. (1) A(2, 3), B(4, 1) (2) P(-5, 7), Q(-1, 3) (3) R(0, -3), S(0, 5/2) (4) L(5, -8), M(-7, -3) (5) T(-3, 6), R(9, -10) (6) , X(11, 4) Answer : The distance between points A(x1, y1) and B(x2, y2) is given by, 1. Given Points: A(2, 3) and B(4, 1) We can see that,x1 = 2 x2 = 4 y1 = 3 y2 = 1 Putting the values in the distance formula we get, d = ? d = ? d = v8 2. Given Points: P(-5, 7) and Q(-1, 3) we can see that,x1 = -5 x2 = -1 y1 = 7 y2 = 3 Putting these values in distance formula we get, d = d = v3 2 3. Given Points: R(0, -3), S(0, 5/2) we can see that,x1 = 0 x2 = 0 y1 = -3 y2 = 5/2 On putting these values in distance formula we get, d = d = d = 4. Given Points: L(5, -8), M(-7, -3) we can see that, x1 = 5 x2 = -7 y1 = -8 y2 = -3 On putting these values in distance formula we get, d = d = d = v1 6 9 = 13 5. Given Points: T(-3, 6), R(9, -10) we can see that, x1 = -3 x2 = 9 y1 = 6 y2 = -10 On putting these values in distance formula we get, d = d = d = 20 6. Given Points: W( ), X(11, 4) we can see that, x1 = -7/2 x2 = 11 y1 = 4 y2 = 4 On putting these values in distance formula we get, d = d = d = Q. 2. Determine whether the points are collinear. (1) A(1, -3), B(2, -5), C(-4, 7) (2) L(-2, 3), M(1, -3), N(5, 4) (3) R(0, 3), D(2, 1), S(3, -1) (4) P(-2, 3), Q(1, 2), R(4, 1) Answer : If Three points (a,b), (c,d), (e,f) are collinear then the area formed by the triangle by the three points is zero. (a,b) = (1,-3) Page 4 Co-ordinate Geometry Practice Set 5.1 Q. 1. Find the distance between each of the following pairs of points. (1) A(2, 3), B(4, 1) (2) P(-5, 7), Q(-1, 3) (3) R(0, -3), S(0, 5/2) (4) L(5, -8), M(-7, -3) (5) T(-3, 6), R(9, -10) (6) , X(11, 4) Answer : The distance between points A(x1, y1) and B(x2, y2) is given by, 1. Given Points: A(2, 3) and B(4, 1) We can see that,x1 = 2 x2 = 4 y1 = 3 y2 = 1 Putting the values in the distance formula we get, d = ? d = ? d = v8 2. Given Points: P(-5, 7) and Q(-1, 3) we can see that,x1 = -5 x2 = -1 y1 = 7 y2 = 3 Putting these values in distance formula we get, d = d = v3 2 3. Given Points: R(0, -3), S(0, 5/2) we can see that,x1 = 0 x2 = 0 y1 = -3 y2 = 5/2 On putting these values in distance formula we get, d = d = d = 4. Given Points: L(5, -8), M(-7, -3) we can see that, x1 = 5 x2 = -7 y1 = -8 y2 = -3 On putting these values in distance formula we get, d = d = d = v1 6 9 = 13 5. Given Points: T(-3, 6), R(9, -10) we can see that, x1 = -3 x2 = 9 y1 = 6 y2 = -10 On putting these values in distance formula we get, d = d = d = 20 6. Given Points: W( ), X(11, 4) we can see that, x1 = -7/2 x2 = 11 y1 = 4 y2 = 4 On putting these values in distance formula we get, d = d = d = Q. 2. Determine whether the points are collinear. (1) A(1, -3), B(2, -5), C(-4, 7) (2) L(-2, 3), M(1, -3), N(5, 4) (3) R(0, 3), D(2, 1), S(3, -1) (4) P(-2, 3), Q(1, 2), R(4, 1) Answer : If Three points (a,b), (c,d), (e,f) are collinear then the area formed by the triangle by the three points is zero. (a,b) = (1,-3) (c,d) = (2,-5) (e,f) = (-4,7) Hence the points are collinear. 2. (a,b) = (-2,3) (c,d) = (1,-3) (e,f) = (5,4) Area = Area = Hence the points are not collinear. 3. (a,b) = (0,3) (c,d) = (2,1) (e,f) = (3,-1) Area = Area = Hence the points are non collinear. 4. (a,b) = (-2,3) (c,d) = (1,2) (e,f) = (4,1) Page 5 Co-ordinate Geometry Practice Set 5.1 Q. 1. Find the distance between each of the following pairs of points. (1) A(2, 3), B(4, 1) (2) P(-5, 7), Q(-1, 3) (3) R(0, -3), S(0, 5/2) (4) L(5, -8), M(-7, -3) (5) T(-3, 6), R(9, -10) (6) , X(11, 4) Answer : The distance between points A(x1, y1) and B(x2, y2) is given by, 1. Given Points: A(2, 3) and B(4, 1) We can see that,x1 = 2 x2 = 4 y1 = 3 y2 = 1 Putting the values in the distance formula we get, d = ? d = ? d = v8 2. Given Points: P(-5, 7) and Q(-1, 3) we can see that,x1 = -5 x2 = -1 y1 = 7 y2 = 3 Putting these values in distance formula we get, d = d = v3 2 3. Given Points: R(0, -3), S(0, 5/2) we can see that,x1 = 0 x2 = 0 y1 = -3 y2 = 5/2 On putting these values in distance formula we get, d = d = d = 4. Given Points: L(5, -8), M(-7, -3) we can see that, x1 = 5 x2 = -7 y1 = -8 y2 = -3 On putting these values in distance formula we get, d = d = d = v1 6 9 = 13 5. Given Points: T(-3, 6), R(9, -10) we can see that, x1 = -3 x2 = 9 y1 = 6 y2 = -10 On putting these values in distance formula we get, d = d = d = 20 6. Given Points: W( ), X(11, 4) we can see that, x1 = -7/2 x2 = 11 y1 = 4 y2 = 4 On putting these values in distance formula we get, d = d = d = Q. 2. Determine whether the points are collinear. (1) A(1, -3), B(2, -5), C(-4, 7) (2) L(-2, 3), M(1, -3), N(5, 4) (3) R(0, 3), D(2, 1), S(3, -1) (4) P(-2, 3), Q(1, 2), R(4, 1) Answer : If Three points (a,b), (c,d), (e,f) are collinear then the area formed by the triangle by the three points is zero. (a,b) = (1,-3) (c,d) = (2,-5) (e,f) = (-4,7) Hence the points are collinear. 2. (a,b) = (-2,3) (c,d) = (1,-3) (e,f) = (5,4) Area = Area = Hence the points are not collinear. 3. (a,b) = (0,3) (c,d) = (2,1) (e,f) = (3,-1) Area = Area = Hence the points are non collinear. 4. (a,b) = (-2,3) (c,d) = (1,2) (e,f) = (4,1) Area = Area = Q. 3. Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4). Answer : A point in the x = axis is of the form (a,0) Distance d between two points(a,b) and (c,d)is given by Distance between (-3,4) and (a,0) = D = D Distance between (1,-4) and (a,0) D = D = As the two points are equidistant from the point (a.0) = Squaring both sides, we get (1-a) 2 + 16 = (3 + a) 2 + 16 1 + a 2 -2a = 9 + a 2 + 6a 8a = -8 a = -1 Hence the point is (-1,0)Read More
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FAQs on Textbook Solutions: Co-ordinate Geometry - Mathematics Class 10 (Maharashtra SSC Board)
| 1. What is the basic concept of coordinate geometry? |  |
Ans. Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. It involves the use of algebraic equations to describe geometric figures and their properties. In this system, points are represented by ordered pairs (x, y) in a two-dimensional space, where 'x' indicates the horizontal position and 'y' indicates the vertical position. This approach allows for the analysis of shapes, lines, distances, and angles mathematically.
| 2. How do you determine the distance between two points in a coordinate plane? | |
Ans. The distance between two points (x1, y1) and (x2, y2) in a coordinate plane can be calculated using the distance formula: Distance = √[(x2 - x1)² + (y2 - y1)²]. This formula is derived from the Pythagorean theorem and provides a straightforward way to measure the length of the line segment connecting the two points.
| 3. What is the midpoint formula in coordinate geometry? | |
Ans. The midpoint formula is used to find the midpoint of a line segment connecting two points. If the points are (x1, y1) and (x2, y2), the coordinates of the midpoint (M) can be calculated using the formula: M = ((x1 + x2)/2, (y1 + y2)/2). This gives the average of the x-coordinates and the average of the y-coordinates, resulting in the point that is exactly halfway between the two given points.
| 4. Can you explain the slope of a line and how it is calculated? | |
Ans. The slope of a line is a measure of its steepness and direction. It is calculated as the ratio of the vertical change to the horizontal change between two points on the line. If you have two points (x1, y1) and (x2, y2), the slope (m) can be calculated using the formula: m = (y2 - y1) / (x2 - x1). A positive slope indicates the line rises as it moves from left to right, while a negative slope indicates it falls.
| 5. What are the different types of lines in coordinate geometry? | |
Ans. In coordinate geometry, lines can be classified based on their slopes. They can be horizontal lines (slope = 0), vertical lines (undefined slope), and oblique lines (slope ≠ 0). Additionally, lines can be parallel (having the same slope) or perpendicular (the product of their slopes equals -1). Understanding these classifications helps in analyzing the relationships between different lines on the coordinate plane.
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