1. Find the equation of a line passing through the point of intersection of lines `x-4y+18=0` and `x+y-12=0` and having slope `2`

Solution:
The point of intersection of the lines can be obtainted by solving the given equations

`x-4y+18=0`

`:.x-4y=-18`

and `x+y-12=0`

`:.x+y=12`

`x-4y=-18 ->(1)`

`x+y=12 ->(2)`

Substracting `=>-5y=-30`

`=>5y=30`

`=>y=30/5`

`=>y=6`

Putting `y=6` in equation `(2)`, we have

`x+6=12`

`=>x=12-6`

`=>x=6`

`:.x=6" and "y=6`

`:. (6,6)` is the intersection point of the given two lines.




Here Point `(x_1,y_1)=(6,6)` and Slope `m=2` (given)

`:. y-6=2(x-6)`

`:. y -6=2x -12`

`:. 2x-y-6=0`

Hence, The equation of line is `2x-y-6=0`

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2. Find the equation of a line passing through the point of intersection of lines `2x+3y+4=0` and `3x+6y-8=0` and having slope `2`

Solution:
The point of intersection of the lines can be obtainted by solving the given equations

`2x+3y+4=0`

`:.2x+3y=-4`

and `3x+6y-8=0`

`:.3x+6y=8`

`2x+3y=-4 ->(1)`

`3x+6y=8 ->(2)`

equation`(1) xx 3 =>6x+9y=-12`

equation`(2) xx 2 =>6x+12y=16`

Substracting `=>-3y=-28`

`=>3y=28`

`=>y=28/3`

Putting `y=28/3 ` in equation `(1)`, we have

`2x+3(28/3)=-4`

`=>2x=-4-28`

`=>2x=-32`

`=>x=-16`

`:.x=-16" and "y=28/3`

`:. (-16,28/3)` is the intersection point of the given two lines.




Here Point `(x_1,y_1)=(-16,28/3)` and Slope `m=2` (given)

`:. y-28/3=2(x+16)`

`:. y -28/3=2x +32`

`:. 2x-y+124/3=0`

`:. 6x-3y+124=0`

Hence, The equation of line is `6x-3y+124=0`

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3. Find the equation of a line passing through the point of intersection of lines `x=3y` and `3x=2y+7` and having slope `-1/2`

Solution:
The point of intersection of the lines can be obtainted by solving the given equations

`x=3y`

`:.x-3y=0`

and `3x=2y+7`

`:.3x-2y=7`

`x-3y=0 ->(1)`

`3x-2y=7 ->(2)`

equation`(1) xx 3 =>3x-9y=0`

equation`(2) xx 1 =>3x-2y=7`

Substracting `=>-7y=-7`

`=>7y=7`

`=>y=7/7`

`=>y=1`

Putting `y=1` in equation `(2)`, we have

`3x-2(1)=7`

`=>3x=7+2`

`=>3x=9`

`=>x=3`

`:.x=3" and "y=1`

`:. (3,1)` is the intersection point of the given two lines.




Here Point `(x_1,y_1)=(3,1)` and Slope `m=-1/2` (given)

`:. y-1=-1/2(x-3)`

`:. 2(y-1)=-1(x-3)`

`:. 2y -2=-x +3`

`:. x+2y-5=0`

Hence, The equation of line is `x+2y-5=0`

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