Given pair of lines are x² + 4xy + y² = 0
⇒ (y/x) ² + 4 y/x + 1 = 0
⇒ y/x = -4±2√3/2 = -2±√3,
∴ The lines y = (-2 + √3) x and y = (-2 - √3) x and x - y = 4 forms an equilateral triangle
Clearly the pair of lines x² + 4xy +y² = 0 intersect at origin,
The perpendicular distance form origin to x - y = 4 is the height of the
h = 2 √ 2
∵ Area of triangle = h²/√3 = 8/√3
Answer:
Correct option: 2 -> sqrt(37)
Step-by-step explanation:
To solve this problem, we just need to use the law of cosines. This law is used to find the third side of a triangle, when we have the two other sides and the angle between them.
The equation of the law of cosines is:
c^2 = a^2 + b^2 - 2 * a * b * cos(C)
So, we have that:
c^2 = 7^2 + 3^2 - 2 * 7 * 3 * (1/2)
c^2 = 49 + 9 - 21
c^2 = 37
c = sqrt(37)
Correct option: 2
Answer:
c) -4/5
d) 4/5
Step-by-step explanation:
The given equation is 25x^2 -16 = 0
Adding 16 on both sides, we get
25x^2 = 16
Dividing both sides by 25, we get
x^2 = 16/25
Taking square root on both sides, we get
x = √(16/25)
x = ±4/5
Therefore, x = 4/5 and x = -4/5
Answers: C) and D)
Hope this will helpful to understand the concept.
Thank you.
