Question

The point of intersection of the straight lines
represented by 6x² + xy - 40y²- 35x -83y +11=0​

Answer 1

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  [  $\frac{(hf-bg) }{ab-hx^{2} } ,\frac{gh-a}{ab-hx^{2} }$  ]

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,(  $\frac{-2883 X 4}{4 X -961} , \frac{961 X 4}{4 X -961}$   )

          Point of Intersection =(3, -1)

Answer 2

(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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