The equations of the diagonals of the square formed by the pairs of straight lines 3x2+8xy−3y2=0 and 3x2+8xy−3y2+2x−4y−1=0 are?
A
x=2y,4x+2y+1=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2x+y=0,2x=4y+1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x=2y,2x=4y+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2x+y=0,4x+2y+1=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B2x+y=0,2x=4y+1 Given pair of straight lines are 3x2+8xy−3y2=03x2+8xy−3y2+2x−4y−1=0−(2) Partial differentiate with respect to x⇒6x+8y+2=0−(3) Partial differentiate with respect to y⇒8x−6y−4=0− (4)
To find point of intersection of (3) and (4) x=525=125;y=−25 To find equation of diagonals slope of AC=−2/5−01/5−0=−2 Slope of DB=y=−2x=12
Equation of BD: y+15=12(x−110)⇒10x−20y=5⇒2x−4y=1 Equation of AC which is 1 to BC⇒−4x+2y=k(−15,110)⇒45+15=k⇒k=1AC:2x+y=0 then, (B) is the correct option.