Samacheer Kalvi Class 9 Maths Solutions Chapter 5 Coordinate Geometry More Ques

I. Multiple Choice Questions.

Question 1. On which quadrant does the point (- 4, 3) lie?
(a) I
(b) II
(c) III
(d) IV
Answer: (b) II
In simple words: A point with a negative x-coordinate and a positive y-coordinate is always located in the second quadrant of the coordinate plane. This is because the x-axis extends negatively to the left and the y-axis extends positively upwards.

🎯 Exam Tip: Remember the signs of coordinates for each quadrant: (+,+), (-,+), (-,-), (+,-) for I, II, III, IV respectively.

Question 2. The point whose abscissa is 5 and lies on the x-axis is .......
(a) (-5, 0)
(b) (5, 5)
(c) (0, 5)
(d) (5, 0)
Answer: (d) (5, 0)
In simple words: The abscissa is the x-coordinate. When a point lies on the x-axis, its y-coordinate is always zero. Therefore, a point with x-coordinate 5 and y-coordinate 0 is (5, 0).

🎯 Exam Tip: Points on the x-axis always have a y-coordinate of 0, and points on the y-axis always have an x-coordinate of 0.

Question 3. A point which lies in the III quadrant is ......
(a) (5, 4)
(b) (5, -4)
(c) (-5, -4)
(d) (-5, 4)
Answer: (c) (-5, -4)
In simple words: In the third quadrant, both the x-coordinate and the y-coordinate are negative. This means both numbers in the pair will have a minus sign in front of them.

🎯 Exam Tip: Visualize the coordinate plane and the signs in each quadrant to quickly identify where a point lies.

Question 4. A point on the y-axis is ........
(a) (1, 1)
(b) (6, 0)
(c) (0, 6)
(d) (-1, -1)
Answer: (c) (0, 6)
In simple words: Any point that sits directly on the y-axis must have an x-coordinate of zero. This is because it does not move left or right from the center.

🎯 Exam Tip: Remember that "on the y-axis" means the point has no horizontal movement, so its x-value is zero.

Question 5. The distance between the points (4, -1) and the origin is ........
(a) \( \sqrt{24} \)
(b) \( \sqrt{37} \)
(c) \( \sqrt{26} \)
(d) \( \sqrt{17} \)
Answer: (d) \( \sqrt{17} \)
In simple words: The distance formula helps us find the length between two points. For the origin (0,0), it simplifies to finding the square root of the sum of the squares of the coordinates.

🎯 Exam Tip: When finding the distance from the origin (0, 0) to a point (x, y), you can use the simplified formula \( \sqrt{x^2 + y^2} \).

Question 6. The distance between the points (-1, 2) and (3, 2) is ........
(a) \( \sqrt{14} \)
(b) \( \sqrt{15} \)
(c) 4
(d) 0
Answer: (c) 4
In simple words: When two points have the same y-coordinate, they lie on a horizontal line. You can find the distance between them by simply finding the difference between their x-coordinates.

🎯 Exam Tip: If points share the same x-coordinate or y-coordinate, the distance calculation becomes much simpler, often just the absolute difference of the varying coordinate.

Question 7. The centre of a circle is (0, 0). One end point of a diameter is (5, -1), then the radius is ......
(a) \( \sqrt{24} \)
(b) \( \sqrt{37} \)
(c) \( \sqrt{26} \)
(d) \( \sqrt{17} \)
Answer: (c) \( \sqrt{26} \)
In simple words: The radius of a circle is the distance from its center to any point on its edge. Here, the center is the origin, and the given endpoint of the diameter is on the circle's edge.

🎯 Exam Tip: The radius is half the diameter. If you're given the center and one endpoint of the diameter, the distance between these two points is directly the radius.

Question 8. The point (0, -3) lies on
(a) + ve x-axis
(b) + ve y-axis
(c) - ve x-axis
(d) - ve y-axis
Answer: (d) - ve y-axis
In simple words: Since the x-coordinate is 0, the point is on the y-axis. Because the y-coordinate is -3 (a negative number), it lies on the negative part of the y-axis.

🎯 Exam Tip: A point with an x-coordinate of zero and a negative y-coordinate will always be found on the downward side of the y-axis.

Question 9. The point which is on y-axis with ordinate -5 is ........
(a) (0, -5)
(b) (-5, 0)
(c) (5, 0)
(d) (0, 5)
Answer: (a) (0, -5)
In simple words: The ordinate is the y-coordinate. If a point is on the y-axis, its x-coordinate must be zero. So, the point is (0, -5).

🎯 Exam Tip: Understand that "ordinate" refers to the y-coordinate and "abscissa" refers to the x-coordinate.

Question 10. The diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, -1) is ......
(a) 2
(b) 4
(c) \( \sqrt{2} \)
(d) 8
Answer: (a) 2
In simple words: To find the length of a diagonal, we can calculate the distance between two opposite vertices of the square. For example, the distance between (1, 0) and (-1, 0) represents the length of one side. The diagonal would be between (1,0) and (0, -1) or (1,0) and (-1,0). No, the diagonal would be between (1,0) and (-1,0) or (0,1) and (0,-1). Wait, these are not the points for a square from the origin. The problem describes a square centered at the origin, with its vertices on the axes. The points (1,0) and (-1,0) lie on the x-axis, and (0,1) and (0,-1) lie on the y-axis. The distance between (1,0) and (-1,0) is 2 units. The distance between (0,1) and (0,-1) is also 2 units. These represent the length of the diagonals in this special type of square that has its vertices on the axes.

🎯 Exam Tip: For a square with vertices on the axes and centered at the origin, the distance between opposite vertices (e.g., (a,0) and (-a,0) or (0,b) and (0,-b)) represents the diagonal length.

Question 11. The distance between the points (-2, 2) and (3, 2) is ........
(a) 10 units
(b) 5 units
(c) \( 5\sqrt{3} \) units
(d) 20 units
Answer: (b) 5 units
In simple words: Since both points have the same y-coordinate (2), the line segment connecting them is horizontal. We simply find the absolute difference between their x-coordinates to get the distance.

🎯 Exam Tip: When calculating the distance between two points, always check if they share an x or y coordinate first, as it simplifies the calculation to just the difference in the other coordinate.

Question 12. The midpoint of the line joining the points (1, -1) and (-5, 3) is .........
(a) (2, 1)
(b) (2, -1)
(c) (-2, -1)
(d) (-2, 1)
Answer: (d) (-2, 1)
In simple words: To find the midpoint of a line segment, we average the x-coordinates and average the y-coordinates separately. This gives us the exact center point.

🎯 Exam Tip: Remember the midpoint formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).

Question 13. If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then the third vertex is ......
(a) (-2, 2)
(b) (2, -2)
(c) (-2, -2)
(d) (2, 2)
Answer: (b) (2, -2)
In simple words: The centroid of a triangle is the average of its three vertices' coordinates. If you know the centroid and two vertices, you can work backward to find the missing third vertex.

🎯 Exam Tip: The centroid formula is \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \). Use this to set up equations and solve for the unknown coordinates.

Question 14. The X-axis divides the line segment joining the points (6, 4) and (1, -7) is ........
(a) 1:2
(b) 2:3
(c) 4:7
(d) 7:4
Answer: (c) 4:7
In simple words: When a line segment is divided by the X-axis, the y-coordinate of the division point is zero. We use the section formula to find the ratio in which the X-axis divides the segment. The point of division is (x, 0).

🎯 Exam Tip: To find the ratio (m:n) in which the x-axis divides a segment, set the y-coordinate in the section formula to zero and solve for m/n.

Question 15. The centroid of a triangle (3, -5), (-7, 4) and (10, -2) is .......
(a) (2, -1)
(b) (2, 1)
(c) (-2, 1)
(d) (1, -2)
Answer: (a) (2, -1)
In simple words: To find the centroid, we add up all the x-coordinates and divide by 3, and do the same for all the y-coordinates. This gives us the average position of the vertices.

🎯 Exam Tip: Ensure you sum all three x-coordinates correctly and all three y-coordinates correctly before dividing by 3 to avoid calculation errors.

II. Answer the Following Questions.

Question 1. Show that the given points (1, 1), (5, 4), (-2, 5) are the vertices of an isosceles right angled triangle.
Answer: Let the points be A(1, 1), B(5, 4), and C(-2, 5).
We calculate the square of the distance between each pair of points using the distance formula \( d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \).
Length of AB: \( AB^2 = (5-1)^2 + (4-1)^2 = 4^2 + 3^2 = 16 + 9 = 25 \)
\( AB = \sqrt{25} = 5 \)
Length of BC: \( BC^2 = (-2-5)^2 + (5-4)^2 = (-7)^2 + 1^2 = 49 + 1 = 50 \)
\( BC = \sqrt{50} \)
Length of AC: \( AC^2 = (-2-1)^2 + (5-1)^2 = (-3)^2 + 4^2 = 9 + 16 = 25 \)
\( AC = \sqrt{25} = 5 \)
Since AB = AC = 5, the triangle is isosceles. This means two of its sides are equal in length.
Now, we check for a right angle using the Pythagorean theorem: \( BC^2 = AB^2 + AC^2 \)
\( 50 = 25 + 25 \)
\( 50 = 50 \)
Since the sum of the squares of two sides equals the square of the third side, triangle ABC is a right-angled triangle, with the right angle at A.
Therefore, the given points form an isosceles right-angled triangle.
In simple words: First, we find how long each side of the triangle is. We see that two sides are the same length, so it's an isosceles triangle. Then, we check if the longest side squared is equal to the sum of the other two sides squared. If it is, then the triangle also has a right angle, making it an isosceles right-angled triangle.

🎯 Exam Tip: To prove a triangle is isosceles, show two sides are equal. To prove it's right-angled, show \( a^2 + b^2 = c^2 \).

Question 2. Show that the point (3, -2), (3, 2), (-1, 2) and (-1, -2) taken in order are the vertices of a square.
Answer: Let the given points be A(3, -2), B(3, 2), C(-1, 2), and D(-1, -2).
We calculate the square of the length of each side:
Side AB: \( AB^2 = (3-3)^2 + (2-(-2))^2 = 0^2 + (4)^2 = 16 \)
\( AB = \sqrt{16} = 4 \)
Side BC: \( BC^2 = (-1-3)^2 + (2-2)^2 = (-4)^2 + 0^2 = 16 \)
\( BC = \sqrt{16} = 4 \)
Side CD: \( CD^2 = (-1-(-1))^2 + (-2-2)^2 = 0^2 + (-4)^2 = 16 \)
\( CD = \sqrt{16} = 4 \)
Side DA: \( DA^2 = (3-(-1))^2 + (-2-(-2))^2 = (4)^2 + 0^2 = 16 \)
\( DA = \sqrt{16} = 4 \)
Since AB = BC = CD = DA = 4, all four sides are equal, which means it could be a rhombus or a square.
Next, we calculate the length of the diagonals:
Diagonal AC: \( AC^2 = (-1-3)^2 + (2-(-2))^2 = (-4)^2 + (4)^2 = 16 + 16 = 32 \)
\( AC = \sqrt{32} \)
Diagonal BD: \( BD^2 = (-1-3)^2 + (-2-2)^2 = (-4)^2 + (-4)^2 = 16 + 16 = 32 \)
\( BD = \sqrt{32} \)
Since the diagonals AC and BD are also equal, the figure ABCD is a square. A square is a special type of rhombus where the diagonals are equal.
In simple words: To prove that the points make a square, we first check if all four sides are the same length. If they are, it could be a square or a rhombus. Then, we check if the two diagonals are also the same length. If both these conditions are true, then it is definitely a square.

🎯 Exam Tip: For a square, all four sides must be equal, AND both diagonals must be equal. For a rhombus, only all four sides need to be equal.

Question 3. Show that the point A (3, 7) B (6, 5) and C (15, -1) are collinear.
Answer: To show that points A, B, and C are collinear, we need to prove that the sum of the lengths of any two smaller segments equals the length of the longest segment (e.g., AB + BC = AC).
First, we calculate the distance between each pair of points using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Distance AB: \( AB = \sqrt{(6-3)^2 + (5-7)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)
Distance BC: \( BC = \sqrt{(15-6)^2 + (-1-5)^2} = \sqrt{9^2 + (-6)^2} = \sqrt{81 + 36} = \sqrt{117} \)
We can simplify \( \sqrt{117} \) as \( \sqrt{9 \times 13} = 3\sqrt{13} \)
Distance AC: \( AC = \sqrt{(15-3)^2 + (-1-7)^2} = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} \)
We can simplify \( \sqrt{208} \) as \( \sqrt{16 \times 13} = 4\sqrt{13} \)
Now, we check if AB + BC = AC:
\( \sqrt{13} + 3\sqrt{13} = 4\sqrt{13} \)
Since \( 4\sqrt{13} = 4\sqrt{13} \), the points A, B, and C are collinear. When points are collinear, they all lie on the same straight line.
In simple words: To prove that three points are on the same straight line, we find the distance between each pair of points. If the two shorter distances add up to the longest distance, then the points are collinear.

🎯 Exam Tip: Another way to check for collinearity is to calculate the slope between AB and BC; if they are the same, the points are collinear.

Question 4. Find the type of triangle formed by (-1, -1), (1, 1) and (-\(\sqrt{3}\), \(\sqrt{3}\)).
Answer: Let the points be A(-1, -1), B(1, 1), and C(-\(\sqrt{3}\), \(\sqrt{3}\)).
We use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the length of each side.
Distance AB: \( AB = \sqrt{(1-(-1))^2 + (1-(-1))^2} = \sqrt{(1+1)^2 + (1+1)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \)
Distance BC: \( BC = \sqrt{(-\sqrt{3}-1)^2 + (\sqrt{3}-1)^2} \)
\( = \sqrt{(\sqrt{3}+1)^2 + (\sqrt{3}-1)^2} \)
\( = \sqrt{(3+1+2\sqrt{3}) + (3+1-2\sqrt{3})} \)
\( = \sqrt{4+2\sqrt{3} + 4-2\sqrt{3}} \)
\( = \sqrt{8} \)
Distance AC: \( AC = \sqrt{(-\sqrt{3}-(-1))^2 + (\sqrt{3}-(-1))^2} \)
\( = \sqrt{(-\sqrt{3}+1)^2 + (\sqrt{3}+1)^2} \)
\( = \sqrt{(3+1-2\sqrt{3}) + (3+1+2\sqrt{3})} \)
\( = \sqrt{4-2\sqrt{3} + 4+2\sqrt{3}} \)
\( = \sqrt{8} \)
Since AB = BC = AC = \( \sqrt{8} \), all three sides of the triangle are equal in length. Therefore, triangle ABC is an equilateral triangle. An equilateral triangle has all sides equal and all angles equal to 60 degrees.
In simple words: We calculate the length of all three sides of the triangle. If all three sides turn out to be the same length, then the triangle is called an equilateral triangle.

🎯 Exam Tip: An equilateral triangle has all three sides equal. An isosceles triangle has two sides equal. A scalene triangle has no sides equal.

Question 5. Find x such that PQ = QR where P(6, -1) Q(1, 3) and R(x, 8) respectively.
Answer: We are given that PQ = QR. We will use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
First, calculate the distance PQ:
\( PQ = \sqrt{(1-6)^2 + (3-(-1))^2} \)
\( PQ = \sqrt{(-5)^2 + (4)^2} \)
\( PQ = \sqrt{25 + 16} \)
\( PQ = \sqrt{41} \)
Next, calculate the distance QR:
\( QR = \sqrt{(x-1)^2 + (8-3)^2} \)
\( QR = \sqrt{(x-1)^2 + 5^2} \)
\( QR = \sqrt{(x-1)^2 + 25} \)
Since PQ = QR, we can set their squared distances equal to avoid the square root:
\( PQ^2 = QR^2 \)
\( 41 = (x-1)^2 + 25 \)
Now, solve for x:
\( (x-1)^2 = 41 - 25 \)
\( (x-1)^2 = 16 \)
Take the square root of both sides:
\( x-1 = \pm\sqrt{16} \)
\( x-1 = \pm 4 \)
This gives two possible values for x:
Case 1: \( x-1 = 4 \implies x = 4+1 \implies x = 5 \)
Case 2: \( x-1 = -4 \implies x = -4+1 \implies x = -3 \)
So, the possible values for x are 5 or -3. This problem shows that there can be multiple points satisfying a distance condition.
In simple words: We are told that the distance from P to Q is the same as the distance from Q to R. We use the distance formula to write down equations for PQ and QR. We set these two equations equal to each other and then solve for 'x', which gives us two possible answers.

🎯 Exam Tip: When you have \( (x-a)^2 = k \), remember to consider both the positive and negative square roots of k, leading to two possible solutions for x.

Question 6. Find the coordinate of the point of trisection of the line segment joining (4, -1) and (-2, -3).
Answer: Let the line segment be AB, with A(4, -1) and B(-2, -3). Trisection means dividing the segment into three equal parts.
Let P(a, b) and Q(c, d) be the points of trisection such that AP = PQ = QB.
Point P divides AB in the ratio 1:2.
Using the section formula \( \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \), where m=1 and n=2:
For point P:
\( a = \frac{1(-2) + 2(4)}{1+2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2 \)
\( b = \frac{1(-3) + 2(-1)}{1+2} = \frac{-3 - 2}{3} = \frac{-5}{3} \)
So, the coordinates of P are \( \left(2, -\frac{5}{3}\right) \).
Point Q divides AB in the ratio 2:1.
Using the section formula again, where m=2 and n=1:
For point Q:
\( c = \frac{2(-2) + 1(4)}{2+1} = \frac{-4 + 4}{3} = \frac{0}{3} = 0 \)
\( d = \frac{2(-3) + 1(-1)}{2+1} = \frac{-6 - 1}{3} = \frac{-7}{3} \)
So, the coordinates of Q are \( \left(0, -\frac{7}{3}\right) \).
The two points of trisection are \( \left(2, -\frac{5}{3}\right) \) and \( \left(0, -\frac{7}{3}\right) \). These points divide the line segment into three equal lengths.
In simple words: Trisection means cutting a line into three equal pieces. We find the coordinates of the two points that do this by using a special formula called the section formula. For the first point, the ratio is 1:2, and for the second point, the ratio is 2:1.

🎯 Exam Tip: Remember that for trisection, there are two points, dividing the segment in ratios 1:2 and 2:1 respectively.

Question 7. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Answer: Let the points be A(-3, 10) and B(6, -8). Let the point P(-1, 6) divide the line segment AB in the ratio m:n.
Using the section formula for internal division: \( (x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \)
Substitute the given coordinates and the division point:
\( (-1, 6) = \left( \frac{m(6) + n(-3)}{m+n}, \frac{m(-8) + n(10)}{m+n} \right) \)
We can equate either the x-coordinates or the y-coordinates to find the ratio m:n.
Using the x-coordinates:
\( -1 = \frac{6m - 3n}{m+n} \)
\( -1(m+n) = 6m - 3n \)
\( -m - n = 6m - 3n \)
\( -n + 3n = 6m + m \)
\( 2n = 7m \)
\( \frac{m}{n} = \frac{2}{7} \)
So, the ratio m:n is 2:7.
We can also verify this using the y-coordinates:
\( 6 = \frac{-8m + 10n}{m+n} \)
\( 6(m+n) = -8m + 10n \)
\( 6m + 6n = -8m + 10n \)
\( 6m + 8m = 10n - 6n \)
\( 14m = 4n \)
\( \frac{m}{n} = \frac{4}{14} = \frac{2}{7} \)
Both calculations give the same ratio. Thus, the point P(-1, 6) divides the line segment joining (-3, 10) and (6, -8) in the ratio 2:7. The positive ratio indicates internal division.
In simple words: We want to find how a point splits a line segment. We use a formula that connects the coordinates of the points with a ratio (m:n). By setting the x-coordinates (or y-coordinates) equal in the formula, we can solve for this ratio.

🎯 Exam Tip: You only need to use one set of coordinates (either x or y) to find the ratio. Using both can serve as a useful check for your answer.

Question 8. If (1, 2) (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find 'x' and 'y'.
Answer: Let the vertices of the parallelogram be A(1, 2), B(4, y), C(x, 6), and D(3, 5).
In a parallelogram, the diagonals bisect each other, which means their midpoints are the same.
Midpoint of diagonal AC \( = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) \)
\( = \left( \frac{1+x}{2}, \frac{2+6}{2} \right) \)
\( = \left( \frac{1+x}{2}, \frac{8}{2} \right) \)
\( = \left( \frac{1+x}{2}, 4 \right) \)
Midpoint of diagonal BD \( = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2} \right) \)
\( = \left( \frac{4+3}{2}, \frac{y+5}{2} \right) \)
\( = \left( \frac{7}{2}, \frac{y+5}{2} \right) \)
Since the midpoints are the same, we equate the corresponding coordinates:
Equating x-coordinates:
\( \frac{1+x}{2} = \frac{7}{2} \)
\( 1+x = 7 \)
\( x = 7-1 \)
\( x = 6 \)
Equating y-coordinates:
\( \frac{y+5}{2} = 4 \)
\( y+5 = 4 \times 2 \)
\( y+5 = 8 \)
\( y = 8-5 \)
\( y = 3 \)
Thus, the values of x and y are 6 and 3 respectively. This property of diagonals is fundamental to parallelograms.
In simple words: For a shape called a parallelogram, the middle point of one diagonal is the exact same as the middle point of the other diagonal. We use the midpoint formula for both diagonals and then set their x-parts equal and their y-parts equal to find the missing 'x' and 'y' values.

🎯 Exam Tip: The key property of a parallelogram for coordinate geometry problems is that its diagonals bisect each other, meaning they share a common midpoint.