Exams › JEE Advanced › Maths
If a² + 9b² - 4c² = 6ab, find the point(s) at which the family of lines a*x + b*y + c = 0 is concurrent.
- (1/2, 3/2)
- (-1/2, -3/2)
- (-1/2, 3/2)
- (1/2, -3/2)
Correct answer: (-1/2, 3/2)
Solution
a² + 9b² - 4c² = 6ab => (a-3b)² = 4c² => a-3b = +/-2c. Case 1: a = 3b+2c: line becomes (3b+2c)x+by+c=0 => b(3x+y)+c(2x+1)=0, concurrent at 3x+y=0 and 2x+1=0 => x=-1/2, y=3/2. Case 2: a=3b-2c: concurrent at x=1/2, y=-3/2. Both are valid.
Related JEE Advanced Maths questions
- Determine the range of β for which the point (0, β) is located on or within the triangle formed by the equations y + 3x + 2 = 0, 3y − 2x − 5 = 0, and 4y − x − 14 = 0.
- For the equation a² + b² − c² − 2ab = 0, the common intersection point of the set of lines represented by ax + by + c = 0 is located on which line?
- If the points (a³/(a − 3), (a² − 3)/(a − 1)), (b³/(b − 3), (b² − 3)/(b − 1)), and (c³/(c − 3), (c² − 3)/(c − 1)), where a, b, and c are not equal to 1, are situated on the line l₁x + m₁y + n₁ = 0, then which of the following is true?
- In triangle ABC, the median AD is divided into two equal parts at point E. Line BE intersects AC at point F. What is the ratio of AF to AC?
- Let m1 and m2 be the slopes of two adjacent sides of a square of side length a such that a² + 11a + 3(m1² + m2²) = 220. One vertex of the square is at the point (10(cos(alpha) - sin(alpha)), 10(sin(alpha) + cos(alpha))), where alpha is in (0, pi/2), and one diagonal of the square has equation (cos(alpha) - sin(alpha))x + (sin(alpha) + cos(alpha))y = 10. Find the value of 72(sin⁴(alpha) + cos⁴(alpha)) + a² - 3a + 13.
- Let P be any point on the line x - y + 3 = 0 and let A = (3, 4) be a fixed point. A family of lines given by (3 sec(t) + 5 cosec(t)) x + (7 sec(t) - 3 cosec(t)) y + 11(sec(t) - cosec(t)) = 0 passes through a fixed point B for all permissible values of t. If the maximum value of |PA - PB| is expressed as 2 * sqrt(2n) where n is a natural number, find n.
⚔️ Practice JEE Advanced Maths free + battle 1v1 →