Equation of line passing through a given point and perpendicular to the line passing given points calculator

SolutionMethods

1. Find the equation of the line passing through the point

A(-3,5) $A(-3,5)$ and perpendicular to line passing through the points

B(2,5) $B(2,5)$ and

C(-3,6) $C(-3,6)$

2. Find the equation of the line passing through the point

A(5,2) $A(5,2)$ and perpendicular to line passing through the points

B(2,3) $B(2,3)$ and

C(3,-1) $C(3,-1)$

3. Find the equation of the line passing through the point

A(0,3) $A(0,3)$ and perpendicular to line passing through the points

B(-3,2) $B(-3,2)$ and

C(9,1) $C(9,1)$

4. Find the equation of the line passing through the point

A(2,-3) $A(2,-3)$ and perpendicular to line passing through the points

B(5,7) $B(5,7)$ and

C(-6,3) $C(-6,3)$

5. Find the equation of the line passing through the point

A(5,7) $A(5,7)$ and perpendicular to line passing through the points

B(-1,-1) $B(-1,-1)$ and

C(-4,1) $C(-4,1)$

6. Find the equation of the line passing through the point

A(4,2) $A(4,2)$ and perpendicular to line passing through the points

B(1,-1) $B(1,-1)$ and

C(3,2) $C(3,2)$

7. Find the equation of the line passing through the point

A(5,5) $A(5,5)$ and perpendicular to line passing through the points

B(1,-2) $B(1,-2)$ and

C(-5,2) $C(-5,2)$

Example

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

Solution:
Given points are

A(-3,5) $A(-3,5)$ ,

B(2,5) $B(2,5)$ and

C(-3,6) $C(-3,6)$

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) $B(2,5)$ and

C(-3,6) $C(-3,6)$

Points are

B(2,5),C(-3,6) $B(2,5),C(-3,6)$

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

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

$=(-1)/(-1/5)=5$ (

$:' m_1*m_2=-1$ )

The equation of a line with slope m and passing through

$(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

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

$=(-1)/(-4)=1/4$ (

$:' m_1*m_2=-1$ )

The equation of a line with slope m and passing through

$(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

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

$=(-1)/(-1/12)=12$ (

$:' m_1*m_2=-1$ )

The equation of a line with slope m and passing through

$(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

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

$=(-1)/(4/11)=-11/4$ (

$:' m_1*m_2=-1$ )

The equation of a line with slope m and passing through

$(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

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

$=(-1)/(-2/3)=3/2$ (

$:' m_1*m_2=-1$ )

The equation of a line with slope m and passing through

$(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

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

$=(-1)/(3/2)=-2/3$ (

$:' m_1*m_2=-1$ )

The equation of a line with slope m and passing through

$(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

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

$=(-1)/(-2/3)=3/2$ (

$:' m_1*m_2=-1$ )

The equation of a line with slope m and passing through

$(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

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

$3x-2y-5=0$