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Coordinate Geometry Exercises

🎓 Class 10 Mathematics CBSE Theory Ch 7 — Coordinate Geometry ⏱ ~35 min
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This MCQ module is based on: Coordinate Geometry Exercises

This mathematics assessment will be based on: Coordinate Geometry Exercises
Targeting Class 10 level in Coordinate Geometry, with Intermediate difficulty.

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Exercise 7.1 — Distance Formula

Q1. Find the distance between the following pairs of points: (i) (2, 3) and (4, 1); (ii) (−5, 7) and (−1, 3); (iii) (a, b) and (−a, −b).
(i) √(4+4) = 2√2 units. (ii) √(16+16) = 4√2 units. (iii) √{(2a)² + (2b)²} = 2√(a²+b²) units.
Q2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
Distance = √(1296 + 225) = √1521 = 39 units. Yes — with A at origin and B at (36, 15), AB = 39 km.
Q3. Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
AB = √(1+4)=√5, BC = √(16+196)=√212, AC = √(9+256)=√265. Since AB + BC ≠ AC and no pair sums to the third, the points are not collinear.
Q4. Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
AB = √(1+36)=√37, BC = √(1+36)=√37, AC = √(4+0)=2. Two sides equal ⇒ isosceles.
Q5. In a classroom, 4 friends are seated at A(3, 4), B(6, 7), C(9, 4) and D(6, 1). Champa says ABCD is a square. Do you agree?
AB = BC = CD = DA = 3√2. Diagonals: AC = 6, BD = 6. All sides equal, diagonals equal ⇒ square. Yes, Champa is right.
Q6. Name the quadrilateral formed by the points (−1, −2), (1, 0), (−1, 2) and (−3, 0).
All four sides = 2√2 (equal). Diagonals = 4 and 4. Equal sides with equal diagonals ⇒ square.
Q7. Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).
P(x, 0). (x−2)²+25 = (x+2)²+81 ⇒ −4x+25 = 4x+81 ⇒ 8x = −56 ⇒ x = −7. Point: (−7, 0).
Q8. Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.
64 + (y+3)² = 100 ⇒ (y+3)² = 36 ⇒ y+3 = ±6 ⇒ y = 3 or −9.
Q9. If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find the values of x. Also find the distances QR and PR.
QP² = 25+16 = 41, QR² = x²+25. Equate: x² = 16 ⇒ x = ±4. QR = √41. For x = 4: PR = √(1+81)=√82. For x = −4: PR = √(81+81)=9√2.

Exercise 7.2 — Section Formula

Q1. Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
x = (2·4 + 3·(−1))/5 = 5/5 = 1, y = (2·(−3) + 3·7)/5 = 15/5 = 3. Point: (1, 3).
Q2. Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Trisectors divide in 1:2 and 2:1. P = ((−2+8)/3, (−3−2)/3) = (2, −5/3). Q = ((−4+4)/3, (−6−1)/3) = (0, −7/3).
Q3. To conduct Sports Day activities, in your rectangular school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD. Niharika runs (1/4) the distance AD on the 2nd line and posts a green flag. Preet runs (1/5) the distance on the 8th line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the two flags, where should she post it?
Green flag at G(2, 25), Red flag at R(8, 20). GR = √(36+25) = √61 m. Midpoint = (5, 22.5) on the 5th line, 22.5 m from AD.
Q4. Find the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by (−1, 6).
Let ratio k : 1. −1 = (6k−3)/(k+1) ⇒ −k−1 = 6k−3 ⇒ 7k = 2 ⇒ k = 2/7. Ratio = 2 : 7.
Q5. Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
On x-axis, y = 0. Let ratio k : 1. 0 = (5k − 5)/(k+1) ⇒ k = 1. Ratio 1 : 1. x = (−4+1)/2 = −3/2. Point: (−3/2, 0).
Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Diagonals bisect: midpoint of (1,2)–(x,6) = midpoint of (4,y)–(3,5) ⇒ (1+x)/2 = 7/2 and (2+6)/2 = (y+5)/2 ⇒ x = 6, y = 3.
Q7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
Centre is midpoint of AB: 2 = (1+x)/2 ⇒ x = 3; −3 = (4+y)/2 ⇒ y = −10. A = (3, −10).
Q8. If A and B are (−2, −2) and (2, −4) respectively, find the coordinates of P such that AP = 3/7 AB and P lies on segment AB.
AP : PB = 3 : 4. x = (3·2+4·(−2))/7 = −2/7, y = (3·(−4)+4·(−2))/7 = −20/7. P = (−2/7, −20/7).
Q9. Find the coordinates of the points which divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.
Ratios 1:3, 2:2, 3:1. P₁ = (−1, 7/2), P₂ = (0, 5), P₃ = (1, 13/2).
Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order. [Hint: Area = ½ · d₁ · d₂.]
Diagonals AC and BD: AC = √(16+16) = 4√2, BD = √(36+36) = 6√2. Area = ½ · 4√2 · 6√2 = 24 sq units.
Activity — Plot and Verify

On graph paper, plot the vertices of any quadrilateral using the formulas of this chapter. Compute the four side-lengths and the two diagonal-lengths. Classify the quadrilateral (square / rectangle / rhombus / parallelogram / kite / general). Verify your classification visually.

Remember: all sides equal + equal diagonals ⇒ square; all sides equal + unequal diagonals ⇒ rhombus; only opposite sides equal + equal diagonals ⇒ rectangle.
Competency-Based Questions
Q1. Points A(1, 2), B(4, y), C(x, 6) and D(3, 5) are vertices of a parallelogram. Find x + y.
L3 Apply
From midpoints equality: x = 6, y = 3, so x + y = 9.
Q2. A drone starts at (0, 0) and flies in a straight line to a target at (12, 5). A charging stop is needed at exactly 1/3 of the way. Where should the stop be placed, and what is the remaining distance?
L4 Analyse
Ratio 1:2 ⇒ (4, 5/3). Remaining distance = 2/3 × 13 = 26/3 ≈ 8.67 units.
Q3. Evaluate whether the points (2, 1), (5, 3) and (8, 5) could all lie on a straight road.
L5 Evaluate
AB = √(9+4) = √13, BC = √(9+4) = √13, AC = √(36+16) = √52 = 2√13. Since AB + BC = AC, the points are collinear — yes, they lie on a straight road.
Q4. Design a city layout where four parks P, Q, R, S are placed at the vertices of a square with diagonal 10 units centred at the origin, with sides parallel to x and y axes.
L6 Create
If sides are parallel to axes and diagonal is 10, side = 10/√2 = 5√2, half-side = 5√2/2. Vertices: (5√2/2, 5√2/2), (−5√2/2, 5√2/2), (−5√2/2, −5√2/2), (5√2/2, −5√2/2). (Alternative with diagonal on axes: (5,0), (0,5), (−5,0), (0,−5).)
Assertion & Reason
A: The distance between (0, 0) and (a, b) equals √(a² + b²).
R: The distance from the origin is a special case of the distance formula.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. Substituting (x₁, y₁) = (0, 0) in the distance formula gives √(a² + b²).
A: A triangle with vertices (1, 1), (4, 1), (1, 5) is right-angled.
R: If the square of the longest side equals the sum of squares of the other two, the triangle is right-angled.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. Sides: 3, 4, 5. 3² + 4² = 25 = 5². Pythagoras' converse applies.

7.4 Summary

Key Results
  1. Distance between \(P(x_1,y_1)\) and \(Q(x_2,y_2)\): \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
  2. Distance of \(P(x,y)\) from origin: \(\sqrt{x^2+y^2}\).
  3. Section formula (internal, ratio \(m_1:m_2\)): \(\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)\).
  4. Midpoint (ratio 1 : 1): \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
A Note to the Reader
In Section 7.3 we derived the section formula for internal division. An analogous formula handles external division: just replace \(+m_2\) by \(-m_2\). For the area of a triangle from coordinates, a determinant formula will be introduced in higher classes.

Frequently Asked Questions

What is the summary of Chapter 7 Coordinate Geometry?
Chapter 7 introduces the distance formula, the section formula for internal division, the midpoint as a special case and applications including collinearity and quadrilateral type identification using coordinates.
How do you check if four points form a parallelogram?
Compute the midpoints of the two diagonals. If both diagonals have the same midpoint, the four points form a parallelogram because the diagonals bisect each other.
How do you verify three points form an isosceles triangle?
Compute the three side lengths using the distance formula. If any two sides are equal, the triangle is isosceles; if all three are equal, it is equilateral.
What typical problem uses the section formula?
A question like 'Find the coordinates of the point dividing the segment from A(2,3) to B(10,7) in the ratio 3:1 internally' applies the section formula directly to get (8, 6).
How does Chapter 7 connect to later topics?
It links to trigonometry (Chapter 8), applications of trigonometry (Chapter 9), straight lines and conics in Class 11 and analytical geometry in advanced mathematics.
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