1. Find the equation of the line passing through the point of intersection of the lines `2x+3y=1` and `3x+4y=6` and perpendicular to the line `5x-2y=7`

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

`2x+3y=1`

and `3x+4y=6`

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

`3x+4y=6 ->(2)`

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

equation`(2) xx 2 =>6x+8y=12`

Substracting `=>y=-9`

Putting `y=-9` in equation`(1)`, we have

`2x+3(-9)=1`

`=>2x=1+27`

`=>2x=28`

`=>x=14`

`:.x=14" and "y=-9`

`:. (14,-9)` is the intersection point of the given two lines.


Now, the slope of the line `5x-2y=7`

`5x-2y=7`

`:. 2y=5x-7`

`:. y=(5x)/(2)-7/2`

`:.` Slope `=5/2`

`:.` Slope of perpendicular line`=(-1)/(5/2)=-2/5` (`:' m_1*m_2=-1`)



Here Point `(x_1,y_1)=(14,-9)` and Slope `m=-2/5` (given)

`:. y+9=-2/5(x-14)`

`:. 5(y+9)=-2(x-14)`

`:. 5y +45=-2x +28`

`:. 2x+5y+17=0`

Hence, The equation of line is `2x+5y+17=0`

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2. Find the equation of the line passing through the point of intersection of the lines `5x-6y=1` and `3x+2y+5=0` and perpendicular to the line `3x-5y+11=0`

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

`5x-6y=1`

and `3x+2y+5=0`

`:.3x+2y=-5`

`5x-6y=1 ->(1)`

`3x+2y=-5 ->(2)`

equation`(1) xx 1 =>5x-6y=1`

equation`(2) xx 3 =>9x+6y=-15`

Adding `=>14x=-14`

`=>x=-14/14`

`=>x=-1`

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

`3(-1)+2y=-5`

`=>2y=-5+3`

`=>2y=-2`

`=>y=-1`

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

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


Now, the slope of the line `3x-5y+11=0`

`3x-5y+11=0`

`:. 5y=3x+11`

`:. y=(3x)/(5)+11/5`

`:.` Slope `=3/5`

`:.` Slope of perpendicular line`=(-1)/(3/5)=-5/3` (`:' m_1*m_2=-1`)



Here Point `(x_1,y_1)=(-1,-1)` and Slope `m=-5/3` (given)

`:. y+1=-5/3(x+1)`

`:. 3(y+1)=-5(x+1)`

`:. 3y +3=-5x -5`

`:. 5x+3y+8=0`

Hence, The equation of line is `5x+3y+8=0`

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3. Find the equation of the line passing through the point of intersection of the lines `2x-3y=0` and `4x-5y=2` and perpendicular to the line `x+2y+1=0`

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

`2x-3y=0`

and `4x-5y=2`

`2x-3y=0 ->(1)`

`4x-5y=2 ->(2)`

equation`(1) xx 2 =>4x-6y=0`

equation`(2) xx 1 =>4x-5y=2`

Substracting `=>-y=-2`

`=>y=2`

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

`2x-3(2)=0`

`=>2x=0+6`

`=>2x=6`

`=>x=3`

`:.x=3" and "y=2`

`:. (3,2)` is the intersection point of the given two lines.


Now, the slope of the line `x+2y+1=0`

`x+2y+1=0`

`:. 2y=-x-1`

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

`:.` Slope `=-1/2`

`:.` Slope of perpendicular line`=(-1)/(-1/2)=2` (`:' m_1*m_2=-1`)



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

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

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

`:. 2x-y-4=0`

Hence, The equation of line is `2x-y-4=0`

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4. Find the equation of the line passing through the point of intersection of the lines `x-y=1` and `2x-3y+1=0` and perpendicular 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 perpendicular line`=(-1)/(-5/6)=6/5` (`:' m_1*m_2=-1`)



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

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

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

`:. 5y -15=6x -24`

`:. 6x-5y-9=0`

Hence, The equation of line is `6x-5y-9=0`

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5. 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 perpendicular 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 perpendicular line`=(-1)/(2/3)=-3/2` (`:' m_1*m_2=-1`)



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

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

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

`:. 2y -14=-3x -3`

`:. 3x+2y-11=0`

Hence, The equation of line is `3x+2y-11=0`