Question

prove that the three lines whose equations are 3x+4y+6=0, 6x + 5y + 9 = 0,3 x + 3y + 5 =0 all meet in a
point.
​

Answer 1

The point of intersection of equations (1) ,(2) and (3) is ($\frac{-2}{3}$,-1)

Therefore the three lines meets at ($\frac{-2}{3}$,-1)

Therefore the given three lines are concurrent

Step-by-step explanation:

Given equations are 3x+4y+6=0 can be written as $3x+4y=-6\hfill (1)$

6x+5y+9=0 can be written as $6x+5y=-9\hfill (2)$

3x+3y+5=0 can be written as $3x+3y=-5\hfill (3)$

To find the point in which all lines meet :

By solving the given three equations  

solving (1) and (2)

Mutliply the equation (1) into 2 we get $6x+8y=-12\hfill (4)$

Now subtracting the equations (2) and (4) we get

$6x+8y=-12$

$6x+5y=-9$

______________

3y=-3

$=\frac{-3}{3}$

Therefore y=-1

Substitute the value of y=-1 in equation (1)  we get

3x+4(-1)=-6

3x=-6+4

$x=\frac{-2}{3}$

The point of intersection of equations (1) and (2) is ($\frac{-2}{3}$,-1)

Now solving (2) and (3)

Mutliply the equation (3) into 2 we get $6x+6y=-10\hfill (5)$

Now subtracting the equations (2) and (5) we get

$6x+6y=-10$

$6x+5y=-9$

______________

y=-1

Therefore y=-1

Substitute the value of y=-1 in equation (2)  we get

6x+5(-1)=-9

6x=-9+5

$=\frac{-4}{6}$

$x=\frac{-2}{3}$

The point of intersection of equations (2) and (3) is ($\frac{-2}{3}$,-1)

solving (1) and (3) we get

Subtracting equations (1) and (3)

3x+4y=-6

3x+3y=-5

_________

y=-1

Therefore y=-1

Substitute the value of y=-1 in equation (3)  we get

3x+3(-1)=-5

3x=-5+3

$x=\frac{-2}{3}$

The point of intersection of equations (1) and (3) is ($\frac{-2}{3}$,-1)

Therefore the three lines meets at ($\frac{-2}{3}$,-1)

Therefore the given three lines are concurrent