Straight Lines Test

Result

Straight Lines Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

The pair of straight lines joining the origin to the points of intersection of the circles x² + y² = a² and x² + y² + 2(gx + fy) = 0 is

A
a² (x² + y²) – 2(gx + fy)² = 12

B
a²(x² + y²) – 4(gx + fy)² = 0

C
a² (x² + y²) – (gx + fy)² = 2a²

D
a² (x² + y²) + 4(gx + fy)² = 8

Solution

The equation of lines joining the origin to the points of intersection will be obtained by making one equation homogeneous with the help of the other.

x² + y² = a² ...(1)
and, x₂ + y² + 2(gx + fy) = 0 ....(2)

From (1) and (2),
2(gx + fy) = -a²

squaring both the sides, we get
4(gx + fy)² = a⁴ = a²(x² + y²)
⇒ a²(x² + y²) -4(gx + fy)² = 0,
Which is the required equation
Question 2
1 / -0

Statement -1: The combined equation is l₁, l₂ is 2x² + 6xy + y² = 0 and that of m₁, m₂ is 4x² + 18xy + y² = 0. If the angle between l₁, m₂ is α then angle between the lines l₂, m₁ is α.
Statement -2: If the pair of lines l₁ l₂ = 0 and m₁ m₂ = 0, are equally inclined then angle between l₁ and m₂ = angle between l₂ and m₁.

A
Both Statements–1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1

B
Both Statements-1 and Statement-2 are true and Statement-2 is not the correct explanation of Statement-1

C
Statement-1 is true and Statement-2 is false

D
Statement-1 is false and Statement-2 is true

Solution

The pair of bisectors of
2x² + 6xy + y² = 0 and 4x² + 18xy + y² = 0 coincides
Therfore, angle between l₁, m₂ is same as angle between l₂, m₁.
Both are true and it is correct reason.
Question 3
1 / -0

If one of the lines ax² + 2hxy + by² = 0 bisects the angle between the coordinate axes, then which of the following can be true

A
a + b = 2h

B
(a + b)² = 4h²

C
(a – b)² = 4h²

D
a + b = –2h

Solution

one of the lines must be y = x or y = -x
So, a + b = 2h
or, a + b = -2h
⇒ (a + b)² = 4h²
Question 4
1 / -0

If 4x² + ay² + 2by = c² represents a pair of perpendicular straight lines then

A
a = –2, b = –c

B
a = –4, b = 2c

C
a = –8, b = c

D
a = –4, b = –2c

Solution

We have, 4 + a = 0
⇒ a = -4
also, 4 × a × -c² + 0 - 0 - 4b² + 0 = 0
⇒ 16c² - 4b² = 0
⇒ a = -4 and b = ±2c