Equation of line passing through point of intersection of the two lines and having slope calculator

SolutionMethods

1. Find the equation of a line passing through the point of intersection of lines

x-4y+18=0 $x-4y+18=0$ and

x+y-12=0 $x+y-12=0$ and having slope

2 $2$

2. Find the equation of a line passing through the point of intersection of lines

2x+3y+4=0 $2x+3y+4=0$ and

3x+6y-8=0 $3x+6y-8=0$ and having slope

2 $2$

3. Find the equation of a line passing through the point of intersection of lines

x=3y $x=3y$ and

3x=2y+7 $3x=2y+7$ and having slope

-12 $-1/2$

Example

1. Find the equation of a line passing through the point of intersection of lines x-4y+18=0 $x-4y+18=0$ and x+y-12=0 $x+y-12=0$ and having slope 2 $2$

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

x-4y+18=0 $x-4y+18=0$

∴x-4y=-18 $:.x-4y=-18$

and

x+y-12=0 $x+y-12=0$

∴x+y=12 $:.x+y=12$

x-4y=-18→(1) $x-4y=-18 ->(1)$

x+y=12→(2) $x+y=12 ->(2)$

Substracting

⇒-5y=-30 $=>-5y=-30$

⇒5y=30 $=>5y=30$

⇒y=305 $=>y=30/5$

⇒y=6 $=>y=6$

Putting

y=6 $y=6$ in equation

(2) $(2)$ , we have

x+6=12 $x+6=12$

⇒x=12-6 $=>x=12-6$

⇒x=6 $=>x=6$

∴x=6 and y=6 $:.x=6" and "y=6$

∴(6,6) $:. (6,6)$ is the intersection point of the given two lines.

The equation of a line with slope m and passing through

(x1,y1) $(x_1,y_1)$ is

y-y1=m(x-x1) $y-y_1=m(x-x_1)$

Here Point

(x1,y1)=(6,6) $(x_1,y_1)=(6,6)$ and Slope

m=2 $m=2$ (given)

∴y-6=2(x-6) $:. y-6=2(x-6)$

∴y-6=2x-12 $:. y -6=2x -12$

∴2x-y-6=0 $:. 2x-y-6=0$

Hence, The equation of line is

2x-y-6=0 $2x-y-6=0$

2. Find the equation of a line passing through the point of intersection of lines 2x+3y+4=0 $2x+3y+4=0$ and 3x+6y-8=0 $3x+6y-8=0$ and having slope 2 $2$

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

2x+3y+4=0 $2x+3y+4=0$

∴2x+3y=-4 $:.2x+3y=-4$

and

3x+6y-8=0 $3x+6y-8=0$

∴3x+6y=8 $:.3x+6y=8$

2x+3y=-4→(1) $2x+3y=-4 ->(1)$

3x+6y=8→(2) $3x+6y=8 ->(2)$

equation

(1)×3⇒6x+9y=-12 $(1) xx 3 =>6x+9y=-12$

equation

(2)×2⇒6x+12y=16 $(2) xx 2 =>6x+12y=16$

Substracting

⇒-3y=-28 $=>-3y=-28$

⇒3y=28 $=>3y=28$

⇒y=283 $=>y=28/3$

Putting

y=283 $y=28/3$ in equation

(1) $(1)$ , we have

2x+3(283)=-4 $2x+3(28/3)=-4$

⇒2x=-4-28 $=>2x=-4-28$

⇒2x=-32 $=>2x=-32$

⇒x=-16 $=>x=-16$

∴x=-16 and y=283 $:.x=-16" and "y=28/3$

∴(-16,283) $:. (-16,28/3)$ is the intersection point of the given two lines.

The equation of a line with slope m and passing through

(x1,y1) $(x_1,y_1)$ is

y-y1=m(x-x1) $y-y_1=m(x-x_1)$

Here Point

(x1,y1)=(-16,283) $(x_1,y_1)=(-16,28/3)$ and Slope

m=2 $m=2$ (given)

∴y-283=2(x+16) $:. y-28/3=2(x+16)$

∴y-283=2x+32 $:. y -28/3=2x +32$

∴2x-y+1243=0 $:. 2x-y+124/3=0$

∴6x-3y+124=0 $:. 6x-3y+124=0$

Hence, The equation of line is

6x-3y+124=0 $6x-3y+124=0$

3. Find the equation of a line passing through the point of intersection of lines x=3y $x=3y$ and 3x=2y+7 $3x=2y+7$ and having slope -12 $-1/2$

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

x=3y $x=3y$

∴x-3y=0 $:.x-3y=0$

and

3x=2y+7 $3x=2y+7$

∴3x-2y=7 $:.3x-2y=7$

x-3y=0→(1) $x-3y=0 ->(1)$

3x-2y=7→(2) $3x-2y=7 ->(2)$

equation

(1)×3⇒3x-9y=0 $(1) xx 3 =>3x-9y=0$

equation

(2)×1⇒3x-2y=7 $(2) xx 1 =>3x-2y=7$

Substracting

⇒-7y=-7 $=>-7y=-7$

⇒7y=7 $=>7y=7$

⇒y=77 $=>y=7/7$

⇒y=1 $=>y=1$

Putting

y=1 $y=1$ in equation

(2) $(2)$ , we have

3x-2(1)=7 $3x-2(1)=7$

⇒3x=7+2 $=>3x=7+2$

⇒3x=9 $=>3x=9$

⇒x=3 $=>x=3$

∴x=3 and y=1 $:.x=3" and "y=1$

∴(3,1) $:. (3,1)$ is the intersection point of the given two lines.

The equation of a line with slope m and passing through

(x1,y1) $(x_1,y_1)$ is

y-y1=m(x-x1) $y-y_1=m(x-x_1)$

Here Point

(x1,y1)=(3,1) $(x_1,y_1)=(3,1)$ and Slope

m=-12 $m=-1/2$ (given)

∴y-1=-12(x-3) $:. y-1=-1/2(x-3)$

∴2(y-1)=-1(x-3) $:. 2(y-1)=-1(x-3)$

∴2y-2=-x+3 $:. 2y -2=-x +3$

∴x+2y-5=0 $:. x+2y-5=0$

Hence, The equation of line is

x+2y-5=0 $x+2y-5=0$