The point of intersection of the straight lines
represented by 6x² + xy - 40y²- 35x -83y +11=0
Answers
Given: 6x² + xy - 40y²- 35x -83y +11=0
To Find: point of intersection of the straight line
Solution:
Standard form of line =ax² + by² + 2hxy + 2gx +2fy + c= 0
On compairing => a =6 , b=49 , h=1/2, g= -35/2 ,f=-83/2 ,c=11
Point of intersection [ ]
Therefore, ab-h = bx - 40 - 1/4
=-240 - 1/4 = -967 /4
Similarly hf - bg = 1/2x - 83/2 -(40 x -35/2)
= -83/4 -700 = -2883/4
gh -af =(-35/2 X 1/2 ) - (6 X -83/2)
Now, to find (x-coordinate ,y - coordinate)
Therefore,( )
Point of Intersection =(3, -1)
(3 , -1) is the required intersection point of lines represented by 6x² + xy - 40y²- 35x -83y +11=0
Given :
6x² + xy - 40y²- 35x -83y +11=0 representing pair of straight lines
To Find :
The point of intersection of the straight lines
Solution:
6x² + xy - 40y²- 35x -83y +11=0
=> (3x + 8y - 1)(2x - 5y - 11) = 0
3x + 8y - 1 = 0 Eq1
2x - 5y - 11 = 0 Eq2
5 * Eq1 + 8 * Eq2
=> 15x - 5 + 16x - 88 = 0
=> 31x = 93
=> x = 3
Substitute in Eq1
=> 3(3) + 8y - 1 = 0
=> 9 + 8y - 1 = 0
=> 8y = -8
=> y = -1
(3 , -1) is the required intersection point
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