Samacheer Kalvi Class 9 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5.6

Question 1. If the y-coordinate of a point is zero, then the point always lies ........
(a) in the I quadrant
(b) in the II quadrant
(c) on x-axis
(d) on y-axis
Answer: (c) on x-axis
In simple words: When the y-coordinate of a point is zero, it means the point does not move up or down from the horizontal line. This horizontal line is known as the x-axis, so the point must always be on it.

🎯 Exam Tip: Remember that points on the x-axis always have a y-coordinate of zero (e.g., \( (x, 0) \)), and points on the y-axis always have an x-coordinate of zero (e.g., \( (0, y) \) ).

Question 2. The points (-5, 2) and (2, -5) lie in the ..........
(a) same quadrant
(b) II and III quadrant respectively
(c) II and IV quadrant respectively
(d) IV and II quadrant respectively
Answer: (c) II and IV quadrant respectively
In simple words: The first point \( (-5, 2) \) has a negative x and positive y, putting it in Quadrant II. The second point \( (2, -5) \) has a positive x and negative y, placing it in Quadrant IV.

🎯 Exam Tip: Quickly recall the sign conventions for coordinates in each quadrant: Quadrant I (+, +), Quadrant II (-, +), Quadrant III (-, -), Quadrant IV (+, -).

Question 3. On plotting the points O (0, 0), A (3, -4), B (3, 4) and C (0, 4) and joining OA, AB, BC and CO, which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Answer: (c) Trapezium
In simple words: When you connect these four points in order, you will form a shape that has one pair of parallel sides. This specific shape with only one pair of parallel sides is called a trapezium.

🎯 Exam Tip: To identify the figure correctly, either plot the points on graph paper or calculate the slopes of all four line segments to check for parallel and perpendicular lines, and calculate lengths to check for equal sides.

Question 4. If P (-1, 1), Q (3, -4), R (1, -1), S (-2, -3) and T (- 4, 4) are plotted on a graph paper, then the points in the fourth quadrant are ..........
(a) P and T
(b) Q and R
(c) only S
(d) P and Q
Answer: (b) Q and R
In simple words: The fourth quadrant is where the x-coordinate is positive and the y-coordinate is negative. Looking at the given points, only Q \( (3, -4) \) and R \( (1, -1) \) fit this description.

🎯 Exam Tip: The fourth quadrant contains points with positive x-values and negative y-values. Quickly checking the signs of the coordinates will help you identify the points belonging to this quadrant.

Question 5. The point whose ordinate is 4 and which lies on the y-axis is ..........
(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)
Answer: (b) (0, 4)
In simple words: 'Ordinate' means the y-coordinate. If a point is on the y-axis, its x-coordinate must be zero. So, a point with y-coordinate 4 on the y-axis is written as \( (0, 4) \).

🎯 Exam Tip: Remember that the x-coordinate is called the abscissa, and the y-coordinate is called the ordinate. A point on the y-axis always has an abscissa of zero.

Question 6. The distance between the two points (2, 3) and (1, 4) is ..........
(a) 2
(b) \( \sqrt{56} \)
(c) \( \sqrt{10} \)
(d) \( \sqrt{2} \)
Answer: (d) \( \sqrt{2} \)
In simple words: To find the distance, use the distance formula. Subtract the x-coordinates, square the result. Subtract the y-coordinates, square the result. Add these two squared numbers and then find the square root of their sum. This gives the straight-line distance between the points.

🎯 Exam Tip: The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Be careful with negative signs during subtraction.

Question 7. If the points A (2, 0), B (-6, 0), C (3, a – 3) lie on the x-axis then the value of a is
(a) 0
(b) 2
(c) 3
(d) -6
Answer: (c) 3
In simple words: If a point lies on the x-axis, its y-coordinate must be zero. For point C, the y-coordinate is \( a - 3 \). So, we set \( a - 3 \) equal to zero and solve for \( a \).

🎯 Exam Tip: A fundamental property of points on the x-axis is that their y-coordinate is always zero. This principle is key to solving problems involving points lying on the coordinate axes.

Question 8. If (x + 2, 4) = (5, y – 2), then the coordinates (x, y) are .........
(a) (7, 12)
(b) (6, 3)
(c) (3, 6)
(d) (2, 1)
Answer: (c) (3, 6)
In simple words: When two coordinate pairs are equal, their x-coordinates must be equal and their y-coordinates must be equal. So, we set \( x + 2 = 5 \) to find x, and \( 4 = y - 2 \) to find y.

🎯 Exam Tip: For two ordered pairs to be equal, their corresponding components must be equal. This means \( (a, b) = (c, d) \implies a = c \) and \( b = d \). This is a basic property of coordinate geometry.

Question 9. If Q1, Q2, Q3, Q4 are the quadrants in a Cartesian plane then \( Q_2 \cap Q_3 \) is .........
(a) \( Q_1 \cup Q_2 \)
(C) Null set
(d) Negative x-axis
Answer: (c) Null set
In simple words: Quadrant 2 has points with negative x and positive y. Quadrant 3 has points with negative x and negative y. Since there's no point that can have both a positive and negative y-coordinate at the same time, there's no overlap between these two quadrants, so their intersection is an empty set.

🎯 Exam Tip: Quadrants are distinct regions. They only share boundaries (the axes), but the points on the axes themselves are not considered part of any quadrant. Therefore, the intersection of any two different quadrants is always an empty set.

Question 10. The distance between the point (5, -1 ) and the origin is ..........
(a) \( \sqrt{24} \)
(b) \( \sqrt{37} \)
(c) \( \sqrt{26} \)
(d) \( \sqrt{17} \)
Answer: (c) \( \sqrt{26} \)
In simple words: The origin is the point \( (0, 0) \). To find the distance between \( (5, -1) \) and \( (0, 0) \), we use the distance formula. This simplifies to squaring each coordinate, adding them, and taking the square root.

🎯 Exam Tip: The distance of a point \( (x, y) \) from the origin \( (0, 0) \) is a special case of the distance formula, simplified to \( \sqrt{x^2 + y^2} \). This is a quick way to find the length of the hypotenuse if you imagine a right triangle formed by the point, the origin, and the x-axis or y-axis.

Question 11. The coordinates of the point C dividing the line segment joining the points P (2, 4) and Q (5, 7) internally in the ratio 2 : 1 is ........
(a) \( (\frac{7}{2}, \frac{11}{2}) \)
(b) (3, 5)
(c) (4, 4)
(d) (4, 6)
Answer: (d) (4, 6)
In simple words: We use the section formula to find the coordinates of point C. If a point divides a line segment internally in a ratio \( m:n \), we apply the formula for the x-coordinate and y-coordinate separately. Here, \( m=2 \) and \( n=1 \).

🎯 Exam Tip: The section formula for internal division is \( \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \). Remember to assign \( (x_1, y_1) \) and \( (x_2, y_2) \) correctly and substitute the ratio \( m:n \) carefully.

Question 12. If p \( (\frac{a}{3}, \frac{b}{2}) \) is the mid-point of the line segment joining A (-4, 3) and B (-2, 4) then (a, b) is .........
(a) (-9, 7)
(b) \( (-3, \frac{7}{2}) \)
(c) (9, -7)
(d) \( (3, -\frac{7}{2}) \)
Answer: (a) (-9, 7)
In simple words: The midpoint's coordinates are found by averaging the x-coordinates and averaging the y-coordinates of the two endpoints. Once we have the midpoint's x and y values, we can set them equal to \( \frac{a}{3} \) and \( \frac{b}{2} \) respectively to find \( a \) and \( b \).

🎯 Exam Tip: The midpoint formula is \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \). Be sure to correctly set up the equations for \( \frac{a}{3} \) and \( \frac{b}{2} \) with the calculated midpoint values.

Question 13. The point Q (1, 6) divide the line segment joining the points P (2, 7) and R(-2, 3) .........
(a) 1:2
(6) 2:1
(c) 1:3
(d) 3:1
Answer: (c) 1:3
In simple words: We need to find the ratio in which point Q divides the line segment PR. We can use the section formula and set the known coordinates of Q equal to the section formula expression. Solving for \( m/n \) will give us the ratio. We can use either the x-coordinates or the y-coordinates for this calculation.

🎯 Exam Tip: When finding the ratio of division, you can choose to work with either the x-coordinates or the y-coordinates. Both calculations should yield the same ratio. Ensure consistent labeling of \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x, y) \).

Question 14. If one end of a diameter of a circle is (3, 4) and the coordinates of its centre is (-3, 2), then the coordinate of the other end of the diameter is ........
(a) (0, -3)
(b) (0, 9)
(c) (3, 0)
(d) (-9, 0)
Answer: (d) (-9, 0)
In simple words: The center of a circle is the midpoint of its diameter. We know one end of the diameter and the center. We can use the midpoint formula backwards to find the coordinates of the other end of the diameter. This involves setting up equations for the x and y coordinates.

🎯 Exam Tip: Visualize the problem: the center is exactly halfway between the two ends of the diameter. If you have two of these three points, you can always find the third using the midpoint formula: \( x_{centre} = \frac{x_1 + x_2}{2} \) and \( y_{centre} = \frac{y_1 + y_2}{2} \).

Question 15. The ratio in which the x-axis divides the line segment joining the points A \( (a_1, b_1) \) and B \( (a_2, b_2) \) is ..........
(b) \( -b_1 : b_2 \)
(c) \( a_1 : a_2 \)
(d) \( -a_1 : a_2 \)
Answer: (b) \( -b_1 : b_2 \)
In simple words: When the x-axis divides a line segment, the dividing point will have a y-coordinate of zero. Using the section formula for the y-coordinate and setting it to zero allows us to find the ratio \( m:n \). The result will be \( -b_1 : b_2 \).

🎯 Exam Tip: A key fact for this type of problem is that any point on the x-axis has a y-coordinate of 0. Apply the section formula for the y-coordinate and solve for the ratio \( m:n \).

Question 16. The ratio in which the x-axis divides the line segment joining the points (6, 4) and (1, -7) is .........
(a) 2:3
(b) 3:4
(c) 4:7
(d) 4:3
Answer: (c) 4:7
In simple words: Since the x-axis divides the line segment, the y-coordinate of the division point is zero. We use the section formula for the y-coordinate, setting it to 0 and then solve for the ratio \( m:n \). This ratio will be 4:7.

🎯 Exam Tip: When the x-axis divides a line segment, it means the y-coordinate of the point of division is 0. If the y-axis divides a line segment, the x-coordinate of the point of division is 0.

Question 17. If the coordinates of the mid-points of the sides AB, BC and CA of a triangle are (3, 4), (1, 1) and (2, -3) respectively, then the vertices A and B of the triangle are .........
(a) (3, 2), (2, 4)
(b) (4, 0), (2, 8)
(c) (3,4), (2,0)
(d) (4, 3), (2, 4)
Answer: (b) (4, 0), (2, 8)
In simple words: We are given the midpoints of the triangle's sides. We can set up a system of equations using the midpoint formula for each side. By adding and subtracting these equations, we can find the coordinates of the vertices A, B, and C.

🎯 Exam Tip: Let the vertices be \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \). Set up three equations for the x-coordinates and three for the y-coordinates using the midpoint formula. Summing the equations and then subtracting individual pairs will efficiently yield the vertex coordinates.

Question 18. The mid-point of the line joining (-a, 2b) and (-3a, -4b) is ........
(a) (2a, 3b)
(b) (-2a, -b)
(c) (2a, b)
(d) (-2a, -3b)
Answer: (b) (-2a, -b)
In simple words: To find the midpoint, we add the x-coordinates and divide by 2, and then add the y-coordinates and divide by 2. This mathematical operation helps us find the exact center point between any two given points.

🎯 Exam Tip: The midpoint formula is \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \). Remember to be careful with algebraic signs when adding the coordinates.

Question 19. In what ratio does the y-axis divides the line joining the points (-5, 1) and (2, 3) internally ..........
(a) 1:3
(b) 2:5
(c) 3:1
(d) 5:2
Answer: (d) 5:2
In simple words: When the y-axis divides a line segment, the dividing point will have an x-coordinate of zero. We can use the section formula for the x-coordinate, set it to zero, and then solve for the ratio \( m:n \). This helps to determine the exact proportion in which the line is split.

🎯 Exam Tip: For a point on the y-axis, the x-coordinate is always zero. Use the x-coordinate part of the section formula: \( 0 = \frac{mx_2 + nx_1}{m+n} \), and solve for \( m/n \).

Question 20. If (1, -2), (3, 6), (x, 10) and (3, 2) are the vertices of the parallelogram taken in order, then the value of x is .........
(b) 5
(c) 4
(d) 3
Answer: (b) 5
In simple words: In a parallelogram, the diagonals bisect each other, meaning their midpoints are the same. We can find the midpoint of one diagonal (AC) and the midpoint of the other diagonal (BD) and then set their coordinates equal to each other to solve for \( x \). This property is crucial for understanding parallelogram geometry.

🎯 Exam Tip: The key property for this question is that the midpoints of the diagonals of a parallelogram coincide. Equating the midpoint coordinates of AC and BD provides two equations, one for x and one for y, which can be solved easily.