1. Find the equation of the line passing through the point of intersection of the lines `x-y=1` and `2x-3y+1=0` and parallel to the line `3x+4y=12`

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

`x-y=1`

and `2x-3y+1=0`

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

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

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

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

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

Substracting `=>y=3`

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

`x-(3)=1`

`=>x=1+3`

`=>x=4`

`:.x=4" and "y=3`

`:. (4,3)` is the intersection point of the given two lines.


Now, the slope of the line `3x+4y=12`

`3x+4y=12`

`:. 4y=-3x+12`

`:. y=-(3x)/(4)+3`

`:.` Slope `=-3/4`

`:.` Slope of parallel line `=-3/4` `(:' m_1=m_2)`



Here Point `(x_1,y_1)=(4,3)` and Slope `m=-3/4` (given)

`:. y-3=-3/4(x-4)`

`:. 4(y-3)=-3(x-4)`

`:. 4y -12=-3x +12`

`:. 3x+4y-24=0`

Hence, The equation of line is `3x+4y-24=0`

---

2. Find the equation of the line passing through the point of intersection of the lines `x-y=1` and `2x-3y+1=0` and parallel to the line `5x+6y=7`

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

`x-y=1`

and `2x-3y+1=0`

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

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

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

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

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

Substracting `=>y=3`

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

`x-(3)=1`

`=>x=1+3`

`=>x=4`

`:.x=4" and "y=3`

`:. (4,3)` is the intersection point of the given two lines.


Now, the slope of the line `5x+6y=7`

`5x+6y=7`

`:. 6y=-5x+7`

`:. y=-(5x)/(6)+7/6`

`:.` Slope `=-5/6`

`:.` Slope of parallel line `=-5/6` `(:' m_1=m_2)`



Here Point `(x_1,y_1)=(4,3)` and Slope `m=-5/6` (given)

`:. y-3=-5/6(x-4)`

`:. 6(y-3)=-5(x-4)`

`:. 6y -18=-5x +20`

`:. 5x+6y-38=0`

Hence, The equation of line is `5x+6y-38=0`

---

3. Find the equation of the line passing through the point of intersection of the lines `x-2y+15=0` and `3x+y-4=0` and parallel to the line `2x-3y+7=0`

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

`x-2y+15=0`

`:.x-2y=-15`

and `3x+y-4=0`

`:.3x+y=4`

`x-2y=-15 ->(1)`

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

equation`(1) xx 1 =>x-2y=-15`

equation`(2) xx 2 =>6x+2y=8`

Adding `=>7x=-7`

`=>x=-7/7`

`=>x=-1`

Putting `x=-1` in equation `(1)`, we have

`-1-2y=-15`

`=>-2y=-15+1`

`=>-2y=-14`

`=>y=7`

`:.x=-1" and "y=7`

`:. (-1,7)` is the intersection point of the given two lines.


Now, the slope of the line `2x-3y+7=0`

`2x-3y+7=0`

`:. 3y=2x+7`

`:. y=(2x)/(3)+7/3`

`:.` Slope `=2/3`

`:.` Slope of parallel line `=2/3` `(:' m_1=m_2)`



Here Point `(x_1,y_1)=(-1,7)` and Slope `m=2/3` (given)

`:. y-7=2/3(x+1)`

`:. 3(y-7)=2(x+1)`

`:. 3y -21=2x +2`

`:. 2x-3y+23=0`

Hence, The equation of line is `2x-3y+23=0`

---

4. Find the equation of the line passing through the point of intersection of the lines `5x+2y-11=0` and `3x-y+11=0` and parallel to the line `4x-3y+2=0`

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

`5x+2y-11=0`

`:.5x+2y=11`

and `3x-y+11=0`

`:.3x-y=-11`

`5x+2y=11 ->(1)`

`3x-y=-11 ->(2)`

equation`(1) xx 1 =>5x+2y=11`

equation`(2) xx 2 =>6x-2y=-22`

Adding `=>11x=-11`

`=>x=-11/11`

`=>x=-1`

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

`3(-1)-y=-11`

`=>-y=-11+3`

`=>-y=-8`

`=>y=8`

`:.x=-1" and "y=8`

`:. (-1,8)` is the intersection point of the given two lines.


Now, the slope of the line `4x-3y+2=0`

`4x-3y+2=0`

`:. 3y=4x+2`

`:. y=(4x)/(3)+2/3`

`:.` Slope `=4/3`

`:.` Slope of parallel line `=4/3` `(:' m_1=m_2)`



Here Point `(x_1,y_1)=(-1,8)` and Slope `m=4/3` (given)

`:. y-8=4/3(x+1)`

`:. 3(y-8)=4(x+1)`

`:. 3y -24=4x +4`

`:. 4x-3y+28=0`

Hence, The equation of line is `4x-3y+28=0`

---

---