Math Olympiad Test: Coordinate Geometry- 4 - Class 10 MCQ

Since, ΔABC is equilateral
∴ AB = AC ⇒ AB² = AC²

⇒ (x – 3)² + (y – 4)²
= (x + 2)² + (y – 3)²
⇒ 5x + y – 6 = 0 ... (i)
Now, area of equilateral triangle = 
∴ Area of ΔABC = 

⇒ x – 5y = 13√3 − 17 ...(ii)
or x – 5y = −(13√3 + 17) ...(iii)
Solving (i) and (ii), we get,

Also, on solving (i) and (iii), we get

AD is the angle bisector of ∠BAC.
So, by the angle bisector theorem in ΔABC, we have
 ...(i) 

Now, AB = 5 and AC = 10
∴ 1/2 = BD/DC [using (i)]
Thus, D divides BC in the ratio 1 : 2
∴ 

Let P(x, 0) be the point on x-axis which is equidistant from points A(–3, 4) and B(2, 5)
Distance AP = Distance BP
⇒
⇒ (x + 3)² + 16 = (x – 2)² + 25
⇒ x² + 9 + 6x + 16 = x² + 4 – 4x + 25
⇒ 10x = 4 ⇒ x = 2/5
Co-ordinate of P (x, 0) is (2/5, 0).

We know, co-ordinates on Y-axis is (0, y). Now, let P(0, y) divides the line segment joining the points A(–3, 2) and B(6, 1) in λ : 1.

Then, 
∴ Required ratio = 1 : 2.

According to question,
Distance covered from House to Bank

Distance covered from bank to school

Distance covered from school to office

∴ Total distance covered by Ayush = 5 + 10 + 12 = 27 km
Also, Distance covered directly from House to office 
∴ Required distance = 27 – 11√5 = 2.40 km

Since, centroid of triangle formed by (a, b), (b, c) and (c, a) is at the origin,
 ⇒ a + b + c = 0 ...(i)
Also, a³ + b³ + c³ – 3abc = 1/2 [a + b + c] × [(a – b)² + (b – c)² + (c – a)²]
⇒ a³ + b³ + c³ – 3abc = 0 (from (i))
⇒ a³ + b³ + c³ = 3abc

Mohit’s position will be (7, 5).

From the given figure, Coordinates of A, B, C and D are respectively (3, 5), (7, 1), (11, 5) and (7, 9)
Distance from A to C 
Distance from A to D

∴ Required difference = 8 – 4√2 = 2.34 units

Area of banner
1/2= | [7(1− 1) + 1(1− 3) + 6(3 − 1)] = 5 cm²
Cost of 1 cm² of banner = ₹2
∴ Cost of 5 cm² of banner = 2 × 5 = ₹10

According to question,
Coordinates of Chaitanya’s house is (6, 5)
Coordinates of Hamida’s house is (2, 2)
∴ Shortest distance between their houses