1. Find the equation of the line passing through the point `A(1,1)` and perpendicular to the line `2x-3y+2=0`

Solution:
Here Point `(x_1,y_1)=(1,1)` and line `2x-3y+2=0` (given)

When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.

We shall first find the slope of line `2x-3y+2=0`

`2x-3y+2=0`

`:. 3y=2x+2`

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

`:.` Slope `=2/3`

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



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

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

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

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

`:. 3x+2y-5=0`

Hence, The equation of line is `3x+2y-5=0`

---

2. Find the equation of the line passing through the point `A(0,0)` and perpendicular to the line `7x-3y+4=0`

Solution:
Here Point `(x_1,y_1)=(0,0)` and line `7x-3y+4=0` (given)

When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.

We shall first find the slope of line `7x-3y+4=0`

`7x-3y+4=0`

`:. 3y=7x+4`

`:. y=(7x)/(3)+4/3`

`:.` Slope `=7/3`

`:.` Slope of perpendicular line`=(-1)/(7/3)=-3/7` (`:' m_1*m_2=-1`)



Here Point `(x_1,y_1)=(0,0)` and Slope `m=-3/7` (given)

`:. y-0=-3/7(x-0)`

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

`:. 7y +0=-3x +0`

`:. 3x+7y=0`

Hence, The equation of line is `3x+7y=0`

---

3. Find the equation of the line passing through the point `A(3,2)` and perpendicular to the line `y=5x+2`

Solution:
Here Point `(x_1,y_1)=(3,2)` and line `y=5x+2` (given)

When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.

We shall first find the slope of line `y=5x+2`

`y=5x+2`

`:. y=5x+2`

`:.` Slope `=5`

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



Here Point `(x_1,y_1)=(3,2)` and Slope `m=-1/5` (given)

`:. y-2=-1/5(x-3)`

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

`:. 5y -10=-x +3`

`:. x+5y-13=0`

Hence, The equation of line is `x+5y-13=0`

---

4. Find the equation of the line passing through the point `A(-1,1)` and perpendicular to the line `2x+3y+4=0`

Solution:
Here Point `(x_1,y_1)=(-1,1)` and line `2x+3y+4=0` (given)

When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.

We shall first find the slope of line `2x+3y+4=0`

`2x+3y+4=0`

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

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

`:.` Slope `=-2/3`

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



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

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

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

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

`:. 3x-2y+5=0`

Hence, The equation of line is `3x-2y+5=0`