Question

Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y – 4 = 0 to the point of intersection of 7x – 3y = – 12 and 2y = x + 3

Answer 1

The given lines are

                                    3x + y + 2 = 0
                                          3x + y = − 2     ...(1)
                                    x − 2y − 4 = 0
                                         x − 2y  = 4         ...(2)
(1) × 2 ⇒                         6x + 2y = − 4      ...(3)
(2) × 1 ⇒                           x − 2y = 4         ...(4)
By adding (3) and (4) ⇒  7x         = 0 

x = `0/7` = 0

Substitute the value of x = 0 in (1)

3(0) + y = −2

y = − 2

The point of intersection is (0, −2).

The given equation is

                                         7x − 3y  = − 12     ...(5)
                                                 2y  = x + 3     
                                        − x + 2y  = 3          ...(6)
(5) × 1 ⇒                           7x − 3y  = − 12    ...(7)
(6) × 7 ⇒                       − 7x + 14y = 21       ...(8)     
By adding (7) and (8) ⇒            11y = 9

y = `9/11`

Substitute the value of y = `9/11` in (6)

`- x + 2(9/11)` = 3

⇒ `-x + 18/11` = 3

− x = `3 - 18/11`

= `(33 - 18)/11`

= `15/11`

x = `- 15/11`

The point of intersection is `(- 15/11, 9/11)`

Equation of the line joining the points (0, −2) and `(- 15/11, 9/11)` is

`(y - y_1)/(y_2 - y_1) = (x - x_1)/(x_2 - x_1)`

`(y + 2)/(9/11 + 2) = (x - 0)/(- 15/11 - 0)`

`(y+ 2)/(9 / 11 + 2) = x/(- 15/11)`

`(y + 2)/(31/11) = - (11x)/15`

⇒ `(y + 2) 11/31 = - (11x)/15`

31 × (– 11x) = 11 × 15 (y + 2)

= 165 (y + 2)

– 341x = 165y + 330

– 341x – 165y – 330 = 0

341x + 165y + 330 = 0

⇒ 31x + 15y + 30 = 0    ...(÷ by 11)

The required equation is 31x + 15y + 30 = 0