IMO Maths Olympiad Class 10 Chapter 07 Coordinate Geometry MCQs with Answers

Answer: (a) 2ac

Explanation:
1. Area \( = \dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| \).
2. \( = \dfrac{1}{2}\left|a(b-c-c)+a(c-b-c)+(-a)(b+c-b+c)\right| = \dfrac{1}{2}\left|a(b-2c)-ab-2ac\right| \).
3. \( = \dfrac{1}{2}\left|-4ac\right| \) = 2ac.

Teacher's Note:
1. Write the three terms carefully; most errors come from signs.
2. Two vertices share x = a, so the base is the vertical side of length 2c and the height is 2a: ½ × 2c × 2a = 2ac.

Answer: (a) 20.5 sq. units

Explanation:
1. Let A(1, 1), B(3, 4), C(5, −2), D(4, −7). Area ABCD = area △ABC + area △ACD.
2. Area △ABC \( = \dfrac{1}{2}\left|1(4+2)+3(-2-1)+5(1-4)\right| = \dfrac{1}{2}\left|6-9-15\right| = 9 \).
3. Area △ACD \( = \dfrac{1}{2}\left|1(-2+7)+5(-7-1)+4(1+2)\right| = \dfrac{1}{2}\left|5-40+12\right| = 11.5 \).
4. Area of quadrilateral = 9 + 11.5 = 20.5 sq. units.

Teacher's Note:
1. Split the quadrilateral along a diagonal into two triangles.
2. Take the absolute value of each triangle's area before adding.

Answer: (a) Rectangle

Explanation:
1. Let A(1, 1), B(−1, 5), C(7, 9), D(9, 5). AB = CD = √(4 + 16) = √20 and BC = DA = √(64 + 16) = √80.
2. Diagonals: AC = √(36 + 64) = 10 and BD = √(100 + 0) = 10 are equal.
3. Opposite sides equal, adjacent sides unequal and equal diagonals, so ABCD is a rectangle.

Teacher's Note:
1. A parallelogram with equal diagonals is a rectangle.
2. If all four sides were also equal it would be a square.

Answer: (d) \( \dfrac{1}{a}+\dfrac{1}{b}=1 \)

Explanation:
1. Collinear points form a triangle of zero area: \( \dfrac{1}{2}\left|a(b-1)+0(1-0)+1(0-b)\right| = 0 \).
2. So ab − a − b = 0, i.e. a + b = ab.
3. Dividing by ab: \( \dfrac{1}{b}+\dfrac{1}{a}=1 \).

Teacher's Note:
1. Three points are collinear exactly when the triangle they form has zero area.
2. This is the intercept form x/a + y/b = 1 with the point (1, 1) on the line.

Answer: (a) \( \left(3, \dfrac{7}{3}\right) \)

Explanation:
1. A triangle and the triangle formed by the mid-points of its sides have the same centroid.
2. Centroid \( = \left(\dfrac{4+3+2}{3}, \dfrac{2+3+2}{3}\right) = \left(3, \dfrac{7}{3}\right) \).

Teacher's Note:
1. Adding the mid-points gives the same total as adding the vertices, so the averages match.
2. You do not need to find the vertices.

Answer: (d) 4 sq. units

Explanation:
1. Area \( = \dfrac{1}{2}\left|t\{(t+2)-t\}+(t+2)\{t-(t-2)\}+(t+3)\{(t-2)-(t+2)\}\right| \).
2. \( = \dfrac{1}{2}\left|2t+2t+4-4t-12\right| = \dfrac{1}{2}\left|-8\right| \) = 4 sq. units.

Teacher's Note:
1. The answer does not depend on t: shifting a triangle does not change its area.
2. Subtracting t from every coordinate gives (0, −2), (2, 2), (3, 0), which is quicker.

Answer: (a) \( \left(\dfrac{20}{3}, \dfrac{5}{3}\right) \)

Explanation:
1. AB = √(16 + 9) = 5 and AC = √(36 + 64) = 10.
2. By the angle bisector theorem, BD : DC = AB : AC = 5 : 10 = 1 : 2.
3. D \( = \left(\dfrac{1 \times 8 + 2 \times 6}{3}, \dfrac{1 \times 9 + 2 \times (-2)}{3}\right) = \left(\dfrac{20}{3}, \dfrac{5}{3}\right) \).

Teacher's Note:
1. The bisector of an angle divides the opposite side in the ratio of the adjacent sides.
2. In the section formula, the ratio m : n pairs m with the far end point.

Answer: (c) \( \left(\dfrac{1+\sqrt{3}}{2}, \dfrac{7-5\sqrt{3}}{2}\right) \) or \( \left(\dfrac{1-\sqrt{3}}{2}, \dfrac{7+5\sqrt{3}}{2}\right) \)

Explanation:
1. Let the third vertex be (x, y). Equal distances from (3, 4) and (−2, 3) give (x − 3)² + (y − 4)² = (x + 2)² + (y − 3)², i.e. 5x + y = 6.
2. The vector (1, −5) is perpendicular to the side (−5, −1) and has length √26, equal to the side, while the height is \( \dfrac{\sqrt{3}}{2} \) × side. So the third vertex is \( \left(\dfrac{1}{2}, \dfrac{7}{2}\right) \pm \dfrac{\sqrt{3}}{2}(1, -5) \) (it also satisfies 5x + y = 6).
3. Third vertex \( = \left(\dfrac{1 \pm \sqrt{3}}{2}, \dfrac{7 \mp 5\sqrt{3}}{2}\right) \), which is option (c).

Teacher's Note:
1. The third vertex lies on the perpendicular bisector of the given side.
2. There are always two such vertices, one on each side of the given segment.

Answer: (d) 3 abc

Explanation:
1. Centroid at the origin gives \( \dfrac{a+b+c}{3} = 0 \), so a + b + c = 0.
2. a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 0.
3. So a³ + b³ + c³ = 3abc.

Teacher's Note:
1. Remember: if a + b + c = 0 then a³ + b³ + c³ = 3abc.
2. Check with a = 1, b = 1, c = −2: 1 + 1 − 8 = −6 = 3 × (−2).

Answer: (d) 2.3 units

Explanation:
1. From the figure, A(3, 5), B(7, 1), C(11, 5) and D(7, 9).
2. AC = 11 − 3 = 8 units and AD = √(4² + 4²) = 4√2 ≈ 5.66 units.
3. Difference = 8 − 4√2 ≈ 2.34, i.e. about 2.3 units.

Teacher's Note:
1. Points on the same horizontal line: just subtract the x-coordinates.
2. Use √2 ≈ 1.414 for the final estimate.

Answer: (d) 5 units

Explanation:
1. Taking Road - I as the x-axis and Road - II as the y-axis, Chaitanya’s house is (6, 5) and Hamida’s house is (2, 2).
2. Distance = √((6 − 2)² + (5 − 2)²) = √(16 + 9) = 5 units.

Teacher's Note:
1. House number gives the x-coordinate and lane number gives the y-coordinate.
2. A 3-4-5 triangle appears again.

Answer: (d) ₹10

Explanation:
1. Area \( = \dfrac{1}{2}\left|7(1-1)+1(1-3)+6(3-1)\right| = \dfrac{1}{2}\left|0-2+12\right| = 5 \) cm².
2. Cost = 5 × ₹2 = ₹10.

Teacher's Note:
1. The base from (1, 1) to (6, 1) is horizontal: base 5, height 3 − 1 = 2, area ½ × 5 × 2 = 5.
2. Do not stop at the area (₹5 is a trap).

Answer: (c) (i) 4 sq. units, (ii) 1 sq. units, (iii) 1 : 4

Explanation:
1. (i) Area △ABC \( = \dfrac{1}{2}\left|2(4-6)+4(6-2)+2(2-4)\right| = \dfrac{1}{2}\left|-4+16-4\right| = 4 \) sq. units.
2. (ii) D(3, 3), E(3, 5), F(2, 4): area △DEF = ½|3(5 − 4) + 3(4 − 3) + 2(3 − 5)| = ½ × 2 = 1 sq. unit.
3. (iii) Ratio = 1 : 4. So the answer is 4, 1, 1 : 4.

Teacher's Note:
1. The triangle joining the mid-points always has one-fourth of the area.
2. So once (i) is known, (ii) and (iii) follow at once.

Answer: (a) −1, 17

Explanation:
1. Split the pentagon into △ABC, △ACD and △ADE.
2. Area △ABC \( = \dfrac{1}{2}\left|1(5+1)-2(-1-3)-3(3-5)\right| = 10 \) and area △ACD \( = \dfrac{1}{2}\left|1(-1+2)-3(-2-3)+0\right| = 8 \).
3. Area △ADE \( = \dfrac{1}{2}\left|1(-2-t)+0+2(3+2)\right| = \dfrac{1}{2}\left|t-8\right| \).
4. \( 10 + 8 + \dfrac{1}{2}\left|t-8\right| = \dfrac{45}{2} \Rightarrow \left|t-8\right| = 9 \), so t = 17 or t = −1: option (a).

Teacher's Note:
1. Split a polygon into triangles from one vertex and add their areas.
2. Strictly, only t = −1 keeps ABCDE a simple (non-crossing) pentagon; for t = 17 the side DE crosses AB, so always sketch the figure.

Answer: (d) \( \sqrt{58} \) units

Explanation:
1. R is the mid-point of QS, so Q = (2 × 3 − 6, 2 × 5 − 5) = (0, 5).
2. Q is the mid-point of PS, so P = (2 × 0 − 6, 2 × 5 − 5) = (−6, 5).
3. TU is the median of △TPQ, so U is the mid-point of PQ = (−3, 5).
4. TU = √((4 + 3)² + (8 − 5)²) = √(49 + 9) = √58 units.

Teacher's Note:
1. If M is the mid-point of XY, then X = 2M − Y.
2. A median joins a vertex to the mid-point of the opposite side.