1. 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`
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 `3x+4y=12`
`3x+4y=12`
`:. 4y=-3x+12`
`:. y=-(3x)/(4)+3`
`:.` Slope `=-3/4`
`:.` Slope of parallel line `=-3/4` `(:' m_1=m_2)`
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=-3/4` (given)
`:. y-3=-3/4(x-4)`
`:. 4(y-3)=-3(x-4)`
`:. 4y -12=-3x +12`
`:. 3x+4y-24=0`
Hence, The equation of line is `3x+4y-24=0`
2. 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 `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 parallel line `=-5/6` `(:' m_1=m_2)`
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=-5/6` (given)
`:. y-3=-5/6(x-4)`
`:. 6(y-3)=-5(x-4)`
`:. 6y -18=-5x +20`
`:. 5x+6y-38=0`
Hence, The equation of line is `5x+6y-38=0`
3. 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 parallel 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 parallel line `=2/3` `(:' m_1=m_2)`
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=2/3` (given)
`:. y-7=2/3(x+1)`
`:. 3(y-7)=2(x+1)`
`:. 3y -21=2x +2`
`:. 2x-3y+23=0`
Hence, The equation of line is `2x-3y+23=0`
4. Find the equation of the line passing through the point of intersection of the lines `5x+2y-11=0` and `3x-y+11=0` and parallel to the line `4x-3y+2=0`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`5x+2y-11=0`
`:.5x+2y=11`
and `3x-y+11=0`
`:.3x-y=-11`
`5x+2y=11 ->(1)`
`3x-y=-11 ->(2)`
equation`(1) xx 1 =>5x+2y=11`
equation`(2) xx 2 =>6x-2y=-22`
Adding `=>11x=-11`
`=>x=-11/11`
`=>x=-1`
Putting `x=-1` in equation `(2)`, we have
`3(-1)-y=-11`
`=>-y=-11+3`
`=>-y=-8`
`=>y=8`
`:.x=-1" and "y=8`
`:. (-1,8)` is the intersection point of the given two lines.
Now, the slope of the line `4x-3y+2=0`
`4x-3y+2=0`
`:. 3y=4x+2`
`:. y=(4x)/(3)+2/3`
`:.` Slope `=4/3`
`:.` Slope of parallel line `=4/3` `(:' m_1=m_2)`
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)=(-1,8)` and Slope `m=4/3` (given)
`:. y-8=4/3(x+1)`
`:. 3(y-8)=4(x+1)`
`:. 3y -24=4x +4`
`:. 4x-3y+28=0`
Hence, The equation of line is `4x-3y+28=0`
This material is intended as a summary. Use your textbook for detail explanation.
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