IMO Practice Test — Straight Lines
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
What type of triangle is formed by the straight lines x + y = 0, 3x + y − 4 = 0, and x + 3y − 4 = 0?
Equilateral
Isosceles
Right-angled
Scalene
Explanation: Slopes are −1, −3, −1/3. Angle between L1 & L2: tan A = |(−1 − (−3))/(1+3)| = 1/2. Between L1 & L3: tan B = |(−1 − (−1/3))/(1+1/3)| = 1/2. Since two angles are equal, it's isosceles.
Question 2 of 6
hard
The normal form x cos α + y sin α = p represents a line whose perpendicular distance from origin is:
α
p
p²
1/p
Explanation: In normal form, p is distance from origin.
Question 3 of 6
hard
The diagonals of a square are along the coordinate axes. If the length of each side is √2, find the equations of its sides.
x ± y = ±1
x ± y = ±√2
x ± y = ±2
2x ± y = ±1
Explanation: Since diagonals are on axes, vertices are on axes. Side length √2 means hypotenuse of right triangle with legs a,a is √2. a²+a² = 2 → 2a² = 2 → a=1. Vertices are (±1, 0) and (0, ±1). Sides are x ± y = ±1.
Question 4 of 6
hard
Find one of the equations of the angle bisectors between the lines 3x − 4y + 7 = 0 and 12x − 5y − 8 = 0.
21x + 27y − 131 = 0
9x − 7y + 15 = 0
11x − 3y + 9 = 0
21x − 27y + 131 = 0
Explanation: Bisector formula: (3x−4y+7)/5 = ±(12x−5y−8)/13. Taking positive sign: 39x − 52y + 91 = 60x − 25y − 40 → 21x + 27y − 131 = 0.
Question 5 of 6
hard
A light ray passing through (1, 2) reflects on the x-axis at point A. The reflected ray passes through (5, 3). Find the x-coordinate of A.
13/5
11/5
3
5/2
Explanation: Image of (1,2) across x-axis is (1,−2). The reflected ray lies on the line joining (1,−2) and (5,3). Equation: y+2 = (5/4)(x−1) → 4y+8 = 5x−5. At x-axis, y=0 → 8 = 5x−5 → 5x=13 → x=13/5.
Question 6 of 6
hard
If a, b, c are in Arithmetic Progression (AP), then the family of lines ax + by + c = 0 always passes through a fixed point. Find this point.
(1, −2)
(1, 2)
(−1, 2)
(2, −1)
Explanation: Since a, b, c are in AP, 2b = a + c → a − 2b + c = 0. Comparing this with ax + by + c = 0 gives x = 1 and y = −2. The fixed point is (1, −2).