This chapter covers the concepts of coordinate geometry, including finding distances between points and dividing line segments. Understanding these concepts is essential for solving geometry problems using algebra.
Coordinate Geometry - Practice Worksheet
Strengthen your foundation with key concepts and basic applications.
This worksheet covers essential long-answer questions to help you build confidence in Coordinate Geometry from Mathematics for Class X (Mathematics).
Basic comprehension exercises
Strengthen your understanding with fundamental questions about the chapter.
Questions
Explain the distance formula and derive it using the Pythagoras theorem.
The distance formula is used to find the distance between two points in a coordinate plane. If we have two points, P(x1, y1) and Q(x2, y2), the distance between them is given by √[(x2 - x1)² + (y2 - y1)²]. This formula is derived from the Pythagoras theorem. Consider a right-angled triangle formed by points P, Q, and R(x2, y1). The horizontal distance between P and R is (x2 - x1), and the vertical distance between R and Q is (y2 - y1). Applying Pythagoras theorem, PQ² = PR² + RQ², which leads to the distance formula. This formula is crucial in various real-life applications like navigation, architecture, and computer graphics.
Find the coordinates of the point which divides the line segment joining the points (2, -3) and (5, 6) in the ratio 1:2.
To find the coordinates of the point dividing the line segment joining (2, -3) and (5, 6) in the ratio 1:2, we use the section formula. The section formula states that if a point divides the line segment joining (x1, y1) and (x2, y2) in the ratio m:n, its coordinates are [(mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)]. Here, m = 1, n = 2, (x1, y1) = (2, -3), and (x2, y2) = (5, 6). Plugging these values into the formula, we get the coordinates as [(1*5 + 2*2)/(1 + 2), (1*6 + 2*(-3))/(1 + 2)] = [(5 + 4)/3, (6 - 6)/3] = (9/3, 0/3) = (3, 0). Thus, the required point is (3, 0).
Prove that the points (1, 7), (4, 2), (-1, -1), and (-4, 4) are the vertices of a square.
To prove that the points (1, 7), (4, 2), (-1, -1), and (-4, 4) form a square, we need to show that all sides are equal and the diagonals are also equal. First, calculate the distances between consecutive points: AB = √[(4-1)² + (2-7)²] = √[9 + 25] = √34, BC = √[(-1-4)² + (-1-2)²] = √[25 + 9] = √34, CD = √[(-4+1)² + (4+1)²] = √[9 + 25] = √34, DA = √[(1+4)² + (7-4)²] = √[25 + 9] = √34. Since all sides are equal, it's a rhombus. Now, check the diagonals: AC = √[(-1-1)² + (-1-7)²] = √[4 + 64] = √68, BD = √[(-4-4)² + (4-2)²] = √[64 + 4] = √68. Since diagonals are equal, it's a square.
Find the area of the triangle formed by the points (0, 0), (4, 0), and (0, 3).
The area of a triangle formed by the points (0, 0), (4, 0), and (0, 3) can be found using the formula for the area of a triangle given its vertices: (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Substituting the given points, we get (1/2) |0(0 - 3) + 4(3 - 0) + 0(0 - 0)| = (1/2) |0 + 12 + 0| = (1/2)*12 = 6 square units. Alternatively, since the points form a right-angled triangle with the right angle at (0, 0), the area can also be calculated as (1/2)*base*height = (1/2)*4*3 = 6 square units.
Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear.
To determine if the points (1, 5), (2, 3), and (-2, -11) are collinear, we can check if the area of the triangle formed by them is zero. The area formula is (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Substituting the points, we get (1/2) |1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3)| = (1/2) |1(14) + 2(-16) + (-2)(2)| = (1/2) |14 - 32 - 4| = (1/2)| -22 | = 11. Since the area is not zero, the points are not collinear. Alternatively, we can check the slopes of the lines joining the points. The slope between (1, 5) and (2, 3) is (3 - 5)/(2 - 1) = -2, and between (2, 3) and (-2, -11) is (-11 - 3)/(-2 - 2) = -14/-4 = 3.5. Since the slopes are not equal, the points are not collinear.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
A point on the x-axis has coordinates (x, 0). We need to find x such that the distance from (x, 0) to (2, -5) is equal to the distance from (x, 0) to (-2, 9). Using the distance formula, √[(2 - x)² + (-5 - 0)²] = √[(-2 - x)² + (9 - 0)²]. Squaring both sides, (2 - x)² + 25 = (-2 - x)² + 81. Expanding, 4 - 4x + x² + 25 = 4 + 4x + x² + 81. Simplifying, -4x + 29 = 4x + 85. Bringing like terms together, -8x = 56, so x = -7. Thus, the point is (-7, 0).
Find the coordinates of the centroid of the triangle whose vertices are (4, -1), (-2, -3), and (6, -5).
The centroid of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by the average of the coordinates: [(x1 + x2 + x3)/3, (y1 + y2 + y3)/3]. For the given points (4, -1), (-2, -3), and (6, -5), the centroid is [(4 + (-2) + 6)/3, (-1 + (-3) + (-5))/3] = [(8)/3, (-9)/3] = (8/3, -3). Thus, the centroid is at (8/3, -3).
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
To find the ratio in which (-1, 6) divides the line segment joining (-3, 10) and (6, -8), we use the section formula. Let the ratio be k:1. Then, -1 = [6k + (-3)*1]/(k + 1). Solving, -1(k + 1) = 6k - 3 → -k - 1 = 6k - 3 → -7k = -2 → k = 2/7. Thus, the ratio is 2:7. We can verify the y-coordinate: 6 = [-8*(2/7) + 10*1]/(2/7 + 1) = [-16/7 + 10]/(9/7) = [(-16 + 70)/7]/(9/7) = (54/7)/(9/7) = 6, which matches. Therefore, the ratio is 2:7.
Find the area of the quadrilateral whose vertices, taken in order, are (-3, 2), (5, 4), (7, -6), and (-5, -4).
To find the area of the quadrilateral with vertices (-3, 2), (5, 4), (7, -6), and (-5, -4), we can divide it into two triangles and sum their areas. Using the area formula for a triangle given vertices, the area of triangle ABC (A(-3, 2), B(5, 4), C(7, -6)) is (1/2) |-3(4 - (-6)) + 5(-6 - 2) + 7(2 - 4)| = (1/2) |-3(10) + 5(-8) + 7(-2)| = (1/2) |-30 - 40 - 14| = (1/2)(84) = 42. The area of triangle ACD (A(-3, 2), C(7, -6), D(-5, -4)) is (1/2) |-3(-6 - (-4)) + 7(-4 - 2) + (-5)(2 - (-6))| = (1/2) |-3(-2) + 7(-6) + (-5)(8)| = (1/2) |6 - 42 - 40| = (1/2)(76) = 38. Total area = 42 + 38 = 80 square units.
Find the coordinates of the point which is equidistant from the three points (3, 0), (0, 3), and (-3, 0).
The point equidistant from (3, 0), (0, 3), and (-3, 0) is the circumcenter of the triangle formed by these points. First, find the perpendicular bisectors of at least two sides. The midpoint of (3, 0) and (0, 3) is [(3 + 0)/2, (0 + 3)/2] = (1.5, 1.5). The slope of the line joining them is (3 - 0)/(0 - 3) = -1, so the slope of the perpendicular bisector is 1. Its equation is y - 1.5 = 1(x - 1.5) → y = x. The midpoint of (3, 0) and (-3, 0) is (0, 0). The line joining them is horizontal, so the perpendicular bisector is vertical: x = 0. The intersection of y = x and x = 0 is (0, 0). Thus, the equidistant point is (0, 0).
Question 1 of 10
Explain the distance formula and derive it using the Pythagoras theorem.
Coordinate Geometry - Mastery Worksheet
Advance your understanding through integrative and tricky questions.
This worksheet challenges you with deeper, multi-concept long-answer questions from Coordinate Geometry to prepare for higher-weightage questions in Class X Mathematics.
Intermediate analysis exercises
Deepen your understanding with analytical questions about themes and characters.
Questions
Find the distance between the points A(2, 3) and B(4, 1) using the distance formula. Also, verify your answer by plotting these points on a graph paper and measuring the distance between them.
Using the distance formula, the distance between A(2, 3) and B(4, 1) is √[(4-2)² + (1-3)²] = √[4 + 4] = √8 = 2√2 units. On graph paper, plotting these points and measuring the distance with a ruler should approximately match this calculation, confirming the accuracy of the distance formula.
Show that the points (1, 7), (4, 2), (-1, -1), and (-4, 4) form a square. Calculate the lengths of all sides and diagonals to support your answer.
Calculate the distances between consecutive points to find all sides equal (e.g., AB = BC = CD = DA = √34) and diagonals equal (AC = BD = √68). Since all sides are equal and diagonals are equal, the figure is a square.
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.
Using the section formula, the coordinates are [(3*8 + 1*4)/(3+1), (3*5 + 1*(-3))/(3+1)] = (7, 3).
Determine if the points (1, 5), (2, 3), and (-2, -11) are collinear using the distance formula.
Calculate distances AB, BC, and AC. If AB + BC = AC, then points are collinear. AB = √5, BC = √221, AC = √260. Since √5 + √221 ≈ √260, the points are collinear.
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.
The y-axis divides the line in the ratio 5:1. The point of intersection is (0, -13/3).
Calculate the area of the triangle formed by the points (3, 0), (4, 5), and (-1, 4) using the formula for area of a triangle given vertices.
Area = 1/2 |3(5-4) + 4(4-0) + (-1)(0-5)| = 1/2 |3 + 16 + 5| = 12 square units.
Find the coordinates of the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4).
First trisection point (P) divides AB in 1:2 ratio: [(-7+4)/3, (4-4)/3] = (-1, 0). Second trisection point (Q) divides AB in 2:1 ratio: [(-14+2)/3, (8-2)/3] = (-4, 2).
Prove that the points A(6, 1), B(8, 2), C(9, 4), and D(p, 3) are the vertices of a parallelogram. Find the value of p.
In a parallelogram, midpoints of diagonals coincide. Midpoint of AC = (7.5, 2.5), midpoint of BD = ((8+p)/2, 2.5). Setting equal: (8+p)/2 = 7.5 → p = 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose center is (2, -3) and B is (1, 4).
Center is midpoint of AB. Let A be (x, y). Then (x+1)/2 = 2 → x = 3; (y+4)/2 = -3 → y = -10. So, A is (3, -10).
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4), and (-2, -1) taken in order.
First, find lengths of diagonals using distance formula: d1 = distance between (3,0) and (-1,4) = √32; d2 = distance between (4,5) and (-2,-1) = √72. Area = 1/2 * d1 * d2 = 1/2 * √32 * √72 = 24 square units.
Question 1 of 10
Find the distance between the points A(2, 3) and B(4, 1) using the distance formula. Also, verify your answer by plotting these points on a graph paper and measuring the distance between them.
Coordinate Geometry - Challenge Worksheet
Push your limits with complex, exam-level long-form questions.
The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Coordinate Geometry in Class X.
Advanced critical thinking
Test your mastery with complex questions that require critical analysis and reflection.
Questions
Prove that the points (3, 0), (6, 4), and (-1, 3) form a right-angled triangle. Also, find the area of the triangle.
To prove the points form a right-angled triangle, calculate the distances between all pairs of points and verify Pythagoras' theorem. The area can be found using the formula for the area of a triangle with coordinates.
Find the coordinates of the point which divides the line segment joining the points (2, -2) and (-7, 4) in the ratio 2:1 externally. Discuss the significance of external division in coordinate geometry.
Use the section formula for external division to find the coordinates. Discuss how external division differs from internal division and its applications in real-life scenarios.
A point P divides the line segment joining A(1, -5) and B(-4, 5) in the ratio k:1. Find the value of k if P lies on the x-axis. What does this imply about the position of P?
Since P lies on the x-axis, its y-coordinate is 0. Use the section formula to set up an equation for the y-coordinate and solve for k. Discuss the geometric interpretation of P's position.
Determine the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by the point (-1, 6). Verify your answer using the distance formula.
Use the section formula to find the ratio. Verify by calculating the distances from the dividing point to the endpoints and comparing the ratios.
If the points A(6, 1), B(8, 2), C(9, 4), and D(p, 3) are the vertices of a parallelogram, find the value of p. Explain the properties of a parallelogram used in this problem.
In a parallelogram, the diagonals bisect each other. Use the mid-point formula to find p. Discuss the properties of parallelograms that are relevant here.
Find the area of the quadrilateral formed by the points (3, 0), (4, 5), (-1, 4), and (-2, -1) taken in order. Discuss the method used and its limitations.
Divide the quadrilateral into two triangles and use the area formula for triangles with coordinates. Discuss the limitations of this method for more complex shapes.
A circle has its center at (2, -3) and one end of a diameter at (1, 4). Find the coordinates of the other end of the diameter. Explain the geometric principle used.
The center of the circle is the mid-point of the diameter. Use the mid-point formula to find the other end. Discuss the properties of circles and diameters.
Find the coordinates of the point on the y-axis which is equidistant from the points (6, 5) and (-4, 3). Discuss the significance of equidistant points in coordinate geometry.
Let the point be (0, y). Use the distance formula to set up an equation and solve for y. Discuss the concept of equidistant points and their applications.
Show that the points (1, 7), (4, 2), (-1, -1), and (-4, 4) are the vertices of a square. Verify using both distance and slope formulas.
Calculate all side lengths and diagonals to show they are equal, and verify that adjacent sides are perpendicular using slopes. Discuss the properties of squares.
Find the relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). Interpret this relation geometrically.
Use the distance formula to set up an equation and simplify to find the relation. Discuss how this represents the perpendicular bisector of the line segment joining the two points.
Question 1 of 10
Prove that the points (3, 0), (6, 4), and (-1, 3) form a right-angled triangle. Also, find the area of the triangle.
This chapter discusses polynomials, their degrees, and classifications such as linear, quadratic, and cubic. Understanding polynomials is essential for solving various mathematical problems.
This chapter focuses on solving pairs of linear equations with two variables and their real-life applications.
This chapter explores quadratic equations, highlighting their forms and significance in real-world applications.
This chapter introduces arithmetic progressions, which are sequences of numbers generated by adding a fixed value to the previous term. Understanding these patterns is crucial for solving real-life mathematical problems.
This chapter focuses on the properties of triangles, specifically their similarity and how it can be applied in various real-world contexts.
This chapter focuses on the foundational concepts of trigonometry, particularly the relationships between the angles and sides of right triangles.
This chapter explores how trigonometry is applied in real-life situations, particularly in measuring heights and distances.
This chapter explores the properties of circles, particularly focusing on tangents and their relationship with radii and secants.
This chapter focuses on sectors and segments of circles, essential concepts in geometry. Understanding these helps in solving real-life problems related to areas and measurements.
This chapter explores how to find the surface areas and volumes of various solids, including combinations of basic shapes like cubes, cones, cylinders, and spheres, essential for real-world applications.
