If the area of the triangle formed by the lines 3x^2-2xy-8y^2=0 and the line 2x+y-k=0 is 5 sq. Units then k =
Answers
Given info : If the area of the triangle formed by the lines 3x^2-2xy-8y^2=0 and the line 2x+y-k=0 is 5 sq unit.
To find : the value of k
Solution : a pair of linear equations is 3x² - 2xy - 8y² = 0
⇒3x² - 6xy + 4xy - 8y² = 0
⇒3x(x - 2y) + 4y(x - 2y) = 0
⇒(3x + 4y)(x - 2y) = 0
⇒3x + 4y = 0 and x - 2y = 0, these both lines intersect at (0,0)
now, 3x + 4y = 0 and 2x + y - k = 0
2(3x + 4y) - 3(2x + y -k) = 0
⇒6x + 8y - 6x - 3y + 3k = 0
⇒5y = - 3k
⇒y = -3k/5 so x = -4(-3k/5)/3 = 4k/5
so, (4k/5, -3k/5)
Solving x - 2y = 0 and 2x + y - k = 0
x - 2y - 2(2x + y - k) = 0
⇒x - 2y - 4x - 2y + 2k = 0
⇒-3x + 2k = 0
⇒x = 2k/3 , so y = x/2 = k/3
so, (2k/3, k/3)
Therefore the area of triangle formed by points (0,0) , (2k/3, k/3) and (4k/5, -3k/5) is given by,
∆ = 1/2|0 + 2k/3(-3k/5 - 0) + 4k/5(0 - k/3)|
⇒5 = 1/2|-6k²/15 - 4k²/15|
⇒10 = |-10k²/15|
⇒10 = 10k²/15
⇒k = ±√15
therefore the values of k are √15 and -√15.
Answer:
Step-by-step explanation:
