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Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Q1: The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q (2, -5) and R(-3, 6), find the coordinates of P. 
Ans:

Let the point P be (2k, k), Q(2,-5), R(-3, 6)
Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryPQ = PR …Given
PQ2 = PR2 …[Squaring both sides
(2k – 2)2 + (k + 5)2 = (2k + 3)2 + (k – 6)2 …Given
4k2 + 4 – 8k + k2 + 10k + 25 = 4k2 + 9 + 12k + k2 – 12k + 36
⇒ 2k + 29 = 45
⇒ 2k = 45 – 29
⇒ 2k = 16
⇒ k = 8
Hence coordinates of point P are (16, 8). 

Q2: Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B(2 + √3, 5) and C(2, 6).
Ans:

Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry
Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry
Since diagonal of a ||gm divides it into two equal areas.
Area of ABCD (||gm) = 2(Area of ∆ABC)
= 23 sq. units 

Q3: Find the coordinates of a point P, which lies on the line segment joining the points A(-2, -2) and B(2, -4) such that AP = 3/7AB.
Ans:

Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry
Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Q4: If P(2, 4) is equidistant from Q(7, 0) and R(x, 9), find the values of x. Also find the distance P.
Ans:

Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryPQ = PR …[Given]
PQ2 = PR2 … [Squaring both sides
∴ (7 – 2)2 + (0 – 4)2 = (x – 2)2 + (9 – 4)2
⇒ 25 + 16 = (x – 2)2 + 25
⇒ 16 = (x – 2)2
⇒ ±4 = x – 2 …[Taking sq. root of both sides
⇒ 2 ± 4 = x
⇒ x = 2 + 4 = 6 or x = 2 – 4 = -2 
Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Q5: Find the value of k, if the points P(5, 4), Q(7, k) and R(9, – 2) are collinear. 
Ans:

Given points are P(5, 4), Q(7, k) and R(9, -2).
x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 …[∵ Points are collinear
∴ 5 (k + 2) + 7 (- 2 – 4) + 9 (4 – k) = 0
5k + 10 – 14 – 28 + 36 – 9k = 0
4 = 4k 
∴ k = 1

Q6: If the points A(1, -2), B(2, 3), C(-3, 2) and D(-4, -3) are the vertices of parallelogram ABCD, then taking AB as the base, find the height of this parallelogram.
Ans:  
Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryClass 10 Maths Chapter 7 Question Answers - Coordinate Geometry
Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry
Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryClass 10 Maths Chapter 7 Question Answers - Coordinate Geometry
Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Q7: The three vertices of a parallelogram ABCD are A(3, 4), B(-1, -3) and C(-6, 2). Find the coordinates of vertex D and find the area of ABCD.
Ans:

Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryClass 10 Maths Chapter 7 Question Answers - Coordinate Geometry
Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Q8: If (3, 3), (6, y), (x, 7) and (5, 6) are the vertices of a parallelogram taken in order, find the values of x and y.
Ans:
Let A (3, 3), B (6, y), C (x, 7) and D (5, 6).
Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryClass 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Q9: Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, -3). Also, find the value of x.
Ans:

Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryClass 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Q10: Point P(x, 4) lies on the line segment joining the points A(-5, 8) and B(4, -10). Find the ratio in which point P divides the line segment AB. Also find the value of x.
Ans:

Class 10 Maths Chapter 7 Question Answers - Coordinate GeometryClass 10 Maths Chapter 7 Question Answers - Coordinate Geometry

The document Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

1. What are the basic concepts of Coordinate Geometry?
Ans. Coordinate Geometry, also known as Analytical Geometry, involves the study of geometric figures using a coordinate system. The basic concepts include points represented by ordered pairs (x, y), the Cartesian plane formed by two perpendicular axes (the x-axis and the y-axis), and the distance formula which helps calculate the distance between two points. Additionally, concepts such as slopes of lines, midpoints, and equations of lines are fundamental in this field.
2. How do you find the distance between two points in Coordinate Geometry?
Ans. The distance between two points A(x₁, y₁) and B(x₂, y₂) in the Cartesian plane can be calculated using the distance formula: \( d = \sqrt{(x₂ - x₁)² + (y₂ - y₁)²} \). This formula derives from the Pythagorean theorem and allows you to find the straight-line distance between the two points.
3. What is the slope of a line and how is it calculated in Coordinate Geometry?
Ans. The slope of a line measures its steepness and direction. It is calculated using the formula \( m = \frac{y₂ - y₁}{x₂ - x₁} \) where (x₁, y₁) and (x₂, y₂) are two distinct points on the line. A positive slope indicates the line rises from left to right, a negative slope indicates it falls, and a slope of zero indicates a horizontal line.
4. What are the different forms of the equation of a line in Coordinate Geometry?
Ans. In Coordinate Geometry, the equation of a line can be expressed in several forms: the slope-intercept form \( y = mx + b \) where m is the slope and b is the y-intercept; the point-slope form \( y - y₁ = m(x - x₁) \) which uses a known point (x₁, y₁) and the slope; and the standard form \( Ax + By = C \) where A, B, and C are constants. Each form is useful depending on the information available.
5. How can you determine if three points are collinear in Coordinate Geometry?
Ans. To determine if three points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃) are collinear, you can calculate the area of the triangle formed by these points. If the area is zero, the points are collinear. The area can be calculated using the formula: \( \text{Area} = \frac{1}{2} | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) | \). If the result is zero, then the points lie on the same straight line.
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