Worksheet
Coordinate Geometry explores the relationship between algebra and geometry through the use of coordinate systems to represent geometric shapes and solve problems.
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.
Draw a diagram with points P and Q and a right-angled triangle to visualize the distances.
Solution
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.
Recall the section formula and substitute the given values carefully.
Solution
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.
Calculate all four sides and both diagonals to verify the properties of a square.
Solution
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).
Plot the points to visualize the triangle and identify the base and height.
Solution
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.
Use the area method or slope method to check collinearity.
Solution
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).
Use the distance formula and set the distances equal to each other.
Solution
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).
Recall the formula for the centroid of a triangle as the average of its vertices' coordinates.
Solution
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).
Use the section formula and solve for the ratio k:1.
Solution
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).
Divide the quadrilateral into two triangles and calculate their areas separately.
Solution
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).
Find the perpendicular bisectors of two sides of the triangle and their intersection point.
Solution
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).
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.
Remember the distance formula is derived from the Pythagorean theorem. Plotting points accurately is key to verification.
Solution
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.
A square has equal sides and equal diagonals. Use the distance formula for each side and diagonal.
Solution
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.
The section formula for internal division is (m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2).
Solution
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.
For collinearity, the sum of distances between two pairs should equal the distance between the first and last point.
Solution
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.
For division by y-axis, set x-coordinate of the dividing point to 0 and solve for the ratio.
Solution
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.
Use the formula: 1/2 |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|.
Solution
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).
Trisection points divide the segment into three equal parts, so use section formula for ratios 1:2 and 2:1.
Solution
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.
Use the property that diagonals of a parallelogram bisect each other to find p.
Solution
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).
The center of the circle is the midpoint of the diameter AB.
Solution
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.
Area of a rhombus is half the product of its diagonals. Calculate diagonals using distance formula between opposite vertices.
Solution
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.
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.
Calculate the distances AB, BC, and AC using the distance formula. Check if the sum of the squares of the two shorter sides equals the square of the longest side.
Solution
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.
Remember that for external division, the ratio is taken as k:1 where k is greater than 1. Think about how this concept is used in physics or engineering.
Solution
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?
The y-coordinate of P is 0. Use this to set up an equation using the section formula.
Solution
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.
Let the ratio be k:1. Use the section formula to find k. Then, calculate the distances PA and PB to verify.
Solution
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.
The mid-points of the diagonals AC and BD should be the same. Use this to find p.
Solution
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.
Use the shoelace formula for each triangle and add the areas. Consider how this method might not work for self-intersecting polygons.
Solution
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 mid-point of the diameter is the center. Use the mid-point formula in reverse to find the other end.
Solution
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.
The point is on the y-axis, so its x-coordinate is 0. Set the distances from (0, y) to both given points equal and solve for y.
Solution
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.
All sides of a square are equal, and the diagonals are equal. Adjacent sides are perpendicular, so their slopes are negative reciprocals.
Solution
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.
The set of points equidistant from two given points is the perpendicular bisector of the line segment joining them. Find the equation of this line.
Solution
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.
Explore the world of Polynomials, understanding their types, degrees, and operations to solve algebraic expressions and equations effectively.
Explore the methods to solve a pair of linear equations in two variables, including graphical, substitution, elimination, and cross-multiplication techniques.
Explore the world of quadratic equations, learning to solve them using various methods like factoring, completing the square, and the quadratic formula.
A chapter that explores sequences where each term after the first is obtained by adding a constant difference, focusing on their properties, nth term, and sum formulas.
Explore the properties, types, and theorems related to triangles, including congruence and similarity, to solve geometric problems effectively.
