1. Find the equation of the line passing through the point `A(-3,5)` and perpendicular to line passing through the points `B(2,5)` and `C(-3,6)`

Solution:
Given points are `A(-3,5)`, `B(2,5)` and `C(-3,6)`

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 `B(2,5)` and `C(-3,6)`

Points are `B(2,5),C(-3,6)`

`:. x_1=2,y_1=5,x_2=-3,y_2=6`

Slope of the line, `m=(y_2-y_1)/(x_2-x_1)`

`:. m=(6-5)/(-3-2)`

`:. m=(1)/(-5)`

`:. m=-1/5`

`:.` Slope `=-1/5`

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



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

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

`:. y -5=5x +15`

`:. 5x-y+20=0`

Hence, The equation of line is `5x-y+20=0`

---

2. Find the equation of the line passing through the point `A(5,2)` and perpendicular to line passing through the points `B(2,3)` and `C(3,-1)`

Solution:
Given points are `A(5,2)`, `B(2,3)` and `C(3,-1)`

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 `B(2,3)` and `C(3,-1)`

Points are `B(2,3),C(3,-1)`

`:. x_1=2,y_1=3,x_2=3,y_2=-1`

Slope of the line, `m=(y_2-y_1)/(x_2-x_1)`

`:. m=(-1-3)/(3-2)`

`:. m=(-4)/(1)`

`:. m=-4`

`:.` Slope `=-4`

`:.` Slope of perpendicular line`=(-1)/(-4)=1/4` (`:' m_1*m_2=-1`)



Here Point `(x_1,y_1)=(5,2)` and Slope `m=1/4` (given)

`:. y-2=1/4(x-5)`

`:. 4(y-2)=(x-5)`

`:. 4y -8=x -5`

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

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

---

3. Find the equation of the line passing through the point `A(0,3)` and perpendicular to line passing through the points `B(-3,2)` and `C(9,1)`

Solution:
Given points are `A(0,3)`, `B(-3,2)` and `C(9,1)`

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 `B(-3,2)` and `C(9,1)`

Points are `B(-3,2),C(9,1)`

`:. x_1=-3,y_1=2,x_2=9,y_2=1`

Slope of the line, `m=(y_2-y_1)/(x_2-x_1)`

`:. m=(1-2)/(9+3)`

`:. m=(-1)/(12)`

`:.` Slope `=-1/12`

`:.` Slope of perpendicular line`=(-1)/(-1/12)=12` (`:' m_1*m_2=-1`)



Here Point `(x_1,y_1)=(0,3)` and Slope `m=12` (given)

`:. y-3=12(x-0)`

`:. y -3=12x +0`

`:. 12x-y+3=0`

Hence, The equation of line is `12x-y+3=0`

---

4. Find the equation of the line passing through the point `A(2,-3)` and perpendicular to line passing through the points `B(5,7)` and `C(-6,3)`

Solution:
Given points are `A(2,-3)`, `B(5,7)` and `C(-6,3)`

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 `B(5,7)` and `C(-6,3)`

Points are `B(5,7),C(-6,3)`

`:. x_1=5,y_1=7,x_2=-6,y_2=3`

Slope of the line, `m=(y_2-y_1)/(x_2-x_1)`

`:. m=(3-7)/(-6-5)`

`:. m=(-4)/(-11)`

`:. m=4/11`

`:.` Slope `=4/11`

`:.` Slope of perpendicular line`=(-1)/(4/11)=-11/4` (`:' m_1*m_2=-1`)



Here Point `(x_1,y_1)=(2,-3)` and Slope `m=-11/4` (given)

`:. y+3=-11/4(x-2)`

`:. 4(y+3)=-11(x-2)`

`:. 4y +12=-11x +22`

`:. 11x+4y-10=0`

Hence, The equation of line is `11x+4y-10=0`

---

5. Find the equation of the line passing through the point `A(5,7)` and perpendicular to line passing through the points `B(-1,-1)` and `C(-4,1)`

Solution:
Given points are `A(5,7)`, `B(-1,-1)` and `C(-4,1)`

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 `B(-1,-1)` and `C(-4,1)`

Points are `B(-1,-1),C(-4,1)`

`:. x_1=-1,y_1=-1,x_2=-4,y_2=1`

Slope of the line, `m=(y_2-y_1)/(x_2-x_1)`

`:. m=(1+1)/(-4+1)`

`:. m=(2)/(-3)`

`:. m=-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)=(5,7)` and Slope `m=3/2` (given)

`:. y-7=3/2(x-5)`

`:. 2(y-7)=3(x-5)`

`:. 2y -14=3x -15`

`:. 3x-2y-1=0`

Hence, The equation of line is `3x-2y-1=0`

---

6. Find the equation of the line passing through the point `A(4,2)` and perpendicular to line passing through the points `B(1,-1)` and `C(3,2)`

Solution:
Given points are `A(4,2)`, `B(1,-1)` and `C(3,2)`

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 `B(1,-1)` and `C(3,2)`

Points are `B(1,-1),C(3,2)`

`:. x_1=1,y_1=-1,x_2=3,y_2=2`

Slope of the line, `m=(y_2-y_1)/(x_2-x_1)`

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

`:. m=(3)/(2)`

`:.` Slope `=3/2`

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



Here Point `(x_1,y_1)=(4,2)` and Slope `m=-2/3` (given)

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

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

`:. 3y -6=-2x +8`

`:. 2x+3y-14=0`

Hence, The equation of line is `2x+3y-14=0`

---

7. Find the equation of the line passing through the point `A(5,5)` and perpendicular to line passing through the points `B(1,-2)` and `C(-5,2)`

Solution:
Given points are `A(5,5)`, `B(1,-2)` and `C(-5,2)`

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 `B(1,-2)` and `C(-5,2)`

Points are `B(1,-2),C(-5,2)`

`:. x_1=1,y_1=-2,x_2=-5,y_2=2`

Slope of the line, `m=(y_2-y_1)/(x_2-x_1)`

`:. m=(2+2)/(-5-1)`

`:. m=(4)/(-6)`

`:. m=-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)=(5,5)` and Slope `m=3/2` (given)

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

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

`:. 2y -10=3x -15`

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

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

---

---