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10. Find the equation of a line passing through point of intersection of Line-1, Line-2 and parallel or perpendicular to Line-3 example
( Enter your problem )
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- Find the equation of the line passing through the point of intersection of the lines x-y=1 and 2x-3y+1=0 and parallel to the line 3x+4y=12
- 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
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2. 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
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`)
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)=(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`
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`)
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)=(-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`
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`)
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)=(3,2)` 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`
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`)
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)=(4,3)` 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`
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`)
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)=(-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`
This material is intended as a summary. Use your textbook for detail explanation.
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