CBSE Class 9 Maths Coordinate Geometry Assignment Set 03

Section - A

Q.1 On which axes do the given points lie?
(i) (7, 0) (ii) (0, -3) (iii) (0, 6) (iv) (-5, 0)
Answer:
(i) (7, 0) lies on the X-axis because its y-coordinate is 0.
(ii) (0, -3) lies on the Y-axis because its x-coordinate is 0.
(iii) (0, 6) lies on the Y-axis because its x-coordinate is 0.
(iv) (-5, 0) lies on the X-axis because its y-coordinate is 0.
In simple words: If the y-coordinate is zero, the point lies on the X-axis. If the x-coordinate is zero, it lies on the Y-axis.

Exam Tip: Remember that the coordinate \((x, 0)\) is always on the X-axis and \((0, y)\) is always on the Y-axis.

Q.2 In which quadrants do the given points lie?
(i) (4, -2) (ii) (-3, 7) (iii) (-1, -2) (iv) (3, 6)
Answer:
(i) (4, -2): \((+, -)\) signs indicate Quadrant IV.
(ii) (-3, 7): \((-, +)\) signs indicate Quadrant II.
(iii) (-1, -2): \((-, -)\) signs indicate Quadrant III.
(iv) (3, 6): \((+, +)\) signs indicate Quadrant I.
In simple words: Check the positive or negative signs of the x and y numbers to find the correct quadrant.

Exam Tip: Quadrants go counter-clockwise starting from the top-right: I (+,+), II (-,+), III (-,-), IV (+,-).

Q.3 Is P (3, 2) & Q(2, 3) represent the same point?
Answer:
No, P(3, 2) and Q(2, 3) do not represent the same point because ordered pairs depend strictly on the position of numbers. Here, the x-coordinate and y-coordinate are interchanged, meaning they represent two different locations on the Cartesian plane.
In simple words: Order matters in coordinate geometry, so \((3, 2)\) is not the same as \((2, 3)\).

Exam Tip: Always note that \((a, b) \neq (b, a)\) unless \(a = b\).

Q.4 In which quadrant points P(3,0), Q(6,0) , R (-7,0), S (0,-6), lie?
Answer:
These points do not lie in any quadrant. Points P(3,0), Q(6,0), and R(-7,0) lie on the X-axis, while point S(0,-6) lies on the Y-axis.
In simple words: Points with a zero coordinate lie directly on an axis, not inside any quadrant.

Exam Tip: Axes are boundaries between quadrants and do not belong to any single quadrant.

Q.5 If a<0 and b<0, then the point P(a,b) lies in
(a) quadrant IV
(b) quadrant II
(c) quadrant III
(d) quadrant I
Answer: (c) quadrant III
Since both \(a < 0\) and \(b < 0\), both the x-coordinate and y-coordinate are negative. Points with negative x and negative y coordinates lie in Quadrant III.
In simple words: Both coordinates are negative, which places the point in the bottom-left region (Quadrant III).

Exam Tip: Memorize the signs for each quadrant: I(+ , +), II(- , +), III(- , -), IV(+ , -).

Q.6 The points (other than the origin) for which the abscissa is equal to the ordinate lie in
(a) Quadrant I only
(b) Quadrant I and II
(c) Quadrant I & III
(d) Quadrant II only.
Answer: (c) Quadrant I & III
In Quadrant I, both abscissa (x) and ordinate (y) are positive and equal (e.g., (2, 2)). In Quadrant III, both are negative and equal (e.g., (-2, -2)).
In simple words: When x and y have the same numerical value and the same sign, they fall in Quadrant I or Quadrant III.

Exam Tip: The line \(y = x\) passes through Quadrant I and Quadrant III.

Q.7 The perpendicular distance of the point P(4,3) from the y axis is
(a) 3 Units
(b) 4 Units
(c) 5 Units
(d) 7 Units
Answer: (b) 4 Units
The perpendicular distance of any point \(P(x, y)\) from the Y-axis is given by the absolute value of its x-coordinate, which is \(|4| = 4\) units.
In simple words: Distance from the Y-axis is simply the x-coordinate value.

Q.8 The area of triangle OAB with O(0,0), A(4,0) & B(0,6) is
(a) 8 sq. unit
(b) 12 sq. units
(c) 16 sq. units
(d) 24 sq. units
Answer: (b) 12 sq. units
Here, O is the origin \((0,0)\), A lies on the X-axis at distance 4, and B lies on the Y-axis at distance 6. This forms a right-angled triangle with base \(OA = 4\) units and height \(OB = 6\) units.
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area = \(\frac{1}{2} \times 4 \times 6 = 12\) sq. units
In simple words: Use the right-angled triangle area formula with the lengths along the axes as base and height.

Exam Tip: When vertices are on the axes with the origin, the legs of the right triangle are simply the non-zero coordinates.

Section - B

Q.9 Write down the coordinates of each of the points P,Q, R, S and T as shown in the following figure?
Answer:
Based on standard Cartesian grid reading from the worksheet figure:
P = (-2, 3)
Q = (3, 1)
R = (-3, -2)
S = (4, -3)
T = (0, -4)
In simple words: Read the horizontal x-value first, followed by the vertical y-value for each marked point.

Exam Tip: Always trace perpendicular lines from the point to the X-axis and Y-axis to read coordinates accurately.

Q.10 Draw the lines X'OX and YOY' as the axes on the plane of a paper and plot the given points.
(i) A(5,3) (ii) B (-3, 2) (iii) C(-5, -4) (iv) D(2,-6)
Answer:
Steps to plot the points:
1. Draw horizontal X-axis and vertical Y-axis intersecting at origin \(O(0,0)\).
2. For A(5, 3): Move 5 units to the right along the X-axis and 3 units upward.
3. For B(-3, 2): Move 3 units to the left along the X-axis and 2 units upward.
4. For C(-5, -4): Move 5 units to the left along the X-axis and 4 units downward.
5. For D(2, -6): Move 2 units to the right along the X-axis and 6 units downward.
In simple words: Locate each point by walking horizontally first and then vertically according to the given numbers.

Exam Tip: Always mark equal scale divisions on both axes before plotting.

Section - C

Q.11 Find the mirror images of the following point using x-axis & y-axis as mirror.
(i) A(2,3)
(ii) B(2,-3)
(iii) C(-2,3)
(iv) D(-2,-3)
Answer:
When reflecting a point \((x, y)\):
- Reflection across X-axis changes the sign of the y-coordinate: \((x, -y)\).
- Reflection across Y-axis changes the sign of the x-coordinate: \((-x, y)\).
(i) For A(2,3): Image across X-axis is \((2, -3)\), across Y-axis is \((-2, 3)\).
(ii) For B(2,-3): Image across X-axis is \((2, 3)\), across Y-axis is \((-2, -3)\).
(iii) For C(-2,3): Image across X-axis is \((-2, -3)\), across Y-axis is \((2, 3)\).
(iv) For D(-2,-3): Image across X-axis is \((-2, 3)\), across Y-axis is \((2, -3)\).
In simple words: To reflect over the X-axis, flip the y sign. To reflect over the Y-axis, flip the x sign.

Exam Tip: Mirror image across X-axis keeps x same and negates y. Mirror image across Y-axis keeps y same and negates x.

Q.12 Draw the graph of the following equations
(i) \(y = 3x + 2\)
(ii) \(y = x\)
Answer:
(i) For \(y = 3x + 2\):
- If \(x = 0\), \(y = 3(0) + 2 = 2\) (Point: \((0, 2)\))
- If \(x = -1\), \(y = 3(-1) + 2 = -1\) (Point: \((-1, -1)\))
- If \(x = 1\), \(y = 3(1) + 2 = 5\) (Point: \((1, 5)\))
(ii) For \(y = x\):
- If \(x = 0\), \(y = 0\) (Point: \((0, 0)\))
- If \(x = 2\), \(y = 2\) (Point: \((2, 2)\))
- If \(x = -2\), \(y = -2\) (Point: \((-2, -2)\))
In simple words: Find at least two solutions for each equation, plot them as points on graph paper, and draw a straight line through them.

Exam Tip: Always plot at least three points to verify that your straight line graph is correct.

Q.13 Draw a triangle with vertices O(0,0) A(3,0) B(3,4). Classify the triangle and also find its area.
Answer:
1. Vertices are \(O(0,0)\), \(A(3,0)\), and \(B(3,4)\).
2. Length of base \(OA = 3 - 0 = 3\) units (along X-axis).
3. Length of height \(AB = 4 - 0 = 4\) units (along Y-axis parallel).
4. Since angle at A is \(90^\circ\), triangle OAB is a right-angled triangle.
5. Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6\) square units.
In simple words: The triangle has a 90-degree corner at point A, making it a right-angled triangle with area 6 square units.

Exam Tip: Use coordinate differences to find lengths of sides parallel to the coordinate axes.

Q.14 Draw a quadrilateral with vertices A(2,2) B(2,-2) C(-2,-2), D(-2,2). Classify the quadrilateral and also find its area.
Answer:
1. Given vertices: \(A(2,2)\), \(B(2,-2)\), \(C(-2,-2)\), \(D(-2,2)\).
2. Length of side \(AB = 2 - (-2) = 4\) units.
3. Length of side \(BC = 2 - (-2) = 4\) units.
4. All four sides are equal (\(AB = BC = CD = DA = 4\)) and interior angles are \(90^\circ\). Therefore, the quadrilateral is a square.
5. Area of square = \(\text{side}^2 = 4 \times 4 = 16\) square units.
In simple words: All four sides are 4 units long and meet at right angles, so it is a square with an area of 16 square units.

Q.15 Find the coordinates of point which are equidistant from these two points P(3,0) and Q(-3,0). How many points are possible satisfying this condition?
Answer:
1. The given points are \(P(3,0)\) and \(Q(-3,0)\), which lie on the X-axis.
2. Any point lying on the perpendicular bisector of segment PQ is equidistant from P and Q.
3. The perpendicular bisector of segment PQ is the Y-axis.
4. Therefore, every point on the Y-axis of the form \((0, y)\) satisfies this condition.
5. Since there are infinitely many points on a line, there are infinitely many points possible satisfying this condition.
In simple words: Any point on the Y-axis is equally distant from \((3,0)\) and \((-3,0)\), so there are infinite possible points.

Exam Tip: The locus of points equidistant from two fixed points is always the perpendicular bisector line joining them.