This MCQ module is based on: Coordinates: Exercises & Summary
TOPIC 3 OF 25
Coordinates: Exercises & Summary
🎓 Class 9
Mathematics
CBSE
Theory
Ch 1 — Orienting Yourself: The Use of Coordinates
⏱ ~30 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: Coordinates: Exercises & Summary
Targeting Class 9 level in Coordinate Geometry, with Intermediate difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
End-of-Chapter Exercises
The following questions consolidate plotting, quadrant identification, axis location, and use of the distance formula.
Q1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
The intersection of the axes is the origin O, with coordinates \((0, 0)\).
Q2. Point W has x-coordinate \(-5\). Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?
All points on a line through W parallel to the y-axis have x-coordinate \(-5\). So \(H = (-5, y)\) for some \(y\). If \(y>0\) → Quadrant II; if \(y<0\) → Quadrant III; if \(y=0\) → on x-axis.
Q3. Consider the points \(B(3, 0)\), \(A(0, -1)\), \(K(-3, -2)\), \(M(-2, 4)\) and \(S(3, 2)\). If they are joined in this order, what figure is formed?
Plotting and connecting B→A→K→M→S→B yields a convex pentagon spread across all four quadrants and the axes. Side lengths: BA = \(\sqrt{10}\), AK = \(\sqrt{10}\), KM = \(\sqrt{37}\), MS = \(\sqrt{29}\), SB = 2.
Q4. (i) Two sides of RAMP are along the line through the y-axis parallel to it. (ii) One side of RAMP is parallel to one of the axes. Verify your predictions for given coordinates of three points.
For two sides parallel to the y-axis, two pairs of vertices must share x-coordinates. Two sides parallel to the same axis only requires one pair sharing that coordinate. Without specific points, conceptually the latter is easier to satisfy.
Q5. Plot point \(Z(5, -6)\) on the Cartesian plane. Construct a right-angled triangle \(IZN\) and find the lengths of the three sides.
Choose, e.g., \(I(5, 0)\) and \(N(0, -6)\). Then \(IZ = 6\), \(ZN = 5\), \(IN = \sqrt{25+36} = \sqrt{61}\). Right angle is at Z.
Q6. Would a square with vertices M(−3, −4), A(−3, 2), R(2, 2), Y(2, −4) have all sides equal? Without plotting, determine the side lengths.
MA = \(|2-(-4)| = 6\); AR = \(|2-(-3)| = 5\); RY = 6; YM = 5. The figure is a rectangle (6 by 5), not a square.
Q7. Are the points M(−3, 4), A(0, 0), G(6, −8) collinear? Use distances.
MA = \(\sqrt{9+16} = 5\); AG = \(\sqrt{36+64} = 10\); MG = \(\sqrt{81+144}=15\). Since \(5+10 = 15\), the three points are collinear.
Q8. (i) Plot points L(2, 0), R(−5, 0), S(−4, 0), T(7, 0). What is common? (ii) Plot points C(0, 3), A(0, −2), R(0, −5), D(0, 4). What is common?
(i) All four lie on the x-axis (y-coordinate = 0). (ii) All four lie on the y-axis (x-coordinate = 0).
Q9. Using the origin as one vertex, plot the vertices of: (i) A right-angled isosceles triangle. (ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
(i) \(O(0,0), A(4,0), B(0,4)\) — right angle at O, legs equal. (ii) \(O(0,0), P(-3,-2), Q(3,-2)\) — base PQ horizontal, OP = OQ = \(\sqrt{13}\).
Q10. Find the coordinates of points S, M, T such that S is the midpoint between two given points etc. (NCERT Q10 from page 13).
Reading the NCERT table: for instance, midpoint of S(−3, 0) and T(3, 0) is M(0, 0). Verify each row of the table by averaging the x-coordinates and y-coordinates.
Activity: Build a Coordinate Treasure Hunt
L4 AnalyseMaterials: Graph paper, ruler, friend
Predict: How precise can your friend's "treasure" guess become with only 3 coordinate clues?
- Hide an imaginary treasure at a secret point \(T(x_T, y_T)\) on a 10×10 grid.
- Give your friend three clue points \(C_1, C_2, C_3\) along with the distance from each to T (e.g. "T is 5 units from \(C_1(2,3)\)").
- Friend must use the distance formula to set up three equations and find T.
- Verify by plotting the three circles centred at \(C_1, C_2, C_3\) — the unique intersection is T.
Three circles in the plane intersect (generically) in at most 2 points; only when all three constraints are consistent does a unique point T exist. This is the geometric basis of GPS trilateration — the same maths used by your phone to find your location from satellites.
Competency-Based Questions
Scenario: An urban planner overlays a city map on a coordinate grid (1 unit = 1 km). Three civic buildings stand at \(L(0, 0)\) — Library, \(H(8, 0)\) — Hospital, and \(S(4, 6)\) — School. The planner wants to install a single Wi-Fi tower equidistant from all three.
Q1. Find the type of triangle LHS based on side lengths.
L3 ApplyLH = 8, HS = \(\sqrt{16+36} = \sqrt{52}\approx 7.21\), SL = \(\sqrt{16+36} \approx 7.21\). Triangle is isosceles (HS = SL).
Q2. Determine the coordinates of the Wi-Fi tower (the circumcentre — equidistant from all three vertices).
L4 AnalyseLet tower = \((x, y)\). Equating squares of distances from L and H: \(x^2 + y^2 = (x-8)^2 + y^2 \Rightarrow x = 4\). From L and S: \(x^2 + y^2 = (x-4)^2 + (y-6)^2\). Substituting \(x=4\): \(16 + y^2 = 0 + (y-6)^2\) → \(16 + y^2 = y^2 - 12y + 36\) → \(12y = 20\) → \(y = 5/3 \approx 1.67\). Tower at \((4, 5/3)\).
Q3. The mayor proposes placing the tower at the centroid \((4, 2)\) instead. Evaluate whether this still gives equal coverage.
L5 EvaluateFrom centroid \((4,2)\): to L = \(\sqrt{16+4} = \sqrt{20}\); to H = \(\sqrt{16+4} = \sqrt{20}\); to S = \(\sqrt{0+16} = 4\). Distances are not equal — School would have weaker signal. The mayor's plan is mathematically incorrect for equal coverage.
Q4. Design coordinates for a fourth building (Park) such that LHSP forms a parallelogram. Provide reasoning.
L6 CreateFor LHSP a parallelogram, vector \(\vec{LH} = \vec{PS}\). \(\vec{LH} = (8,0)\), so \(P = S - (8,0) = (-4, 6)\). Park at \((-4, 6)\). (Other valid orderings give different points.)
Assertion–Reason Questions
A: The midpoint of segment from \((-2, 4)\) to \((6, -2)\) is \((2, 1)\).
R: The midpoint formula is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\).
R: The midpoint formula is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\).
Answer: (a) — \(\left(\tfrac{-2+6}{2}, \tfrac{4-2}{2}\right) = (2, 1)\). R explains A.
A: A circle of radius 5 around the origin passes through \((3, 4)\) but not through \((4, 4)\).
R: A point lies on a circle of radius \(r\) around \(O\) iff its distance from \(O\) equals \(r\).
R: A point lies on a circle of radius \(r\) around \(O\) iff its distance from \(O\) equals \(r\).
Answer: (a) — Distance to (3,4) = 5, but to (4,4) = \(\sqrt{32}\approx 5.66 \neq 5\). R is precisely the criterion.
Chapter Summary
Key Take-aways
- To locate a point in a plane we need two perpendicular reference lines: the x-axis (horizontal) and y-axis (vertical).
- The plane formed is the Cartesian plane; the axes meet at the origin \(O(0, 0)\).
- The plane is split into four quadrants with sign patterns (+,+), (−,+), (−,−), (+,−).
- Coordinates of any point on the x-axis are \((x, 0)\); on the y-axis are \((0, y)\).
- If \(x = y\), then \((x, y) = (y, x)\); otherwise the order matters.
- Distance between points \((x_1, y_1)\) and \((x_2, y_2)\) is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) — derived from the Baudhāyana–Pythagoras Theorem.
- Midpoint formula: \(\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)\).
Frequently Asked Questions
What are the coordinates of the point where the two axes intersect?
The two axes intersect at the origin O. Its coordinates are (0, 0): zero distance along the x-axis and zero distance along the y-axis.
How do you tell which quadrant a point lies in?
Look at the signs of its coordinates. (+,+) is Quadrant I (top-right), (-,+) is Quadrant II (top-left), (-,-) is Quadrant III (bottom-left) and (+,-) is Quadrant IV (bottom-right). Points on an axis belong to no quadrant.
What is the midpoint formula in Class 9 Maths?
The midpoint of the segment joining (x1, y1) and (x2, y2) is ((x1+x2)/2, (y1+y2)/2). For example, the midpoint of (-2, 4) and (6, -2) is (2, 1).
How can the distance formula prove three points are collinear?
Compute all three pairwise distances. If the largest equals the sum of the other two (for example MA + AG = MG), the points lie on a single straight line.
Is a quadrilateral with vertices M(-3,-4), A(-3,2), R(2,2), Y(2,-4) a square?
No. The sides MA and RY have length 6 while AR and YM have length 5. With unequal adjacent sides it is a 6-by-5 rectangle, not a square.
What is the GPS connection mentioned in the chapter activity?
Locating a point from its known distances to three reference points is the geometric idea of trilateration. Your phone uses the same maths to combine signals from three or more satellites and calculate your position on Earth.
AI Tutor
Mathematics Class 9 — Ganita Manjari
Ready