1. Find the point of intersection of the lines `x+y=1` and `x-y=1`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`x+y=1`
and `x-y=1`
`x+y=1 ->(1)`
`x-y=1 ->(2)`
Substracting `=>2y=0`
`=>y=0/2`
`=>y=0`
Putting `y=0` in equation `(1)`, we have
`x+0=1`
`=>x=1`
`:.x=1" and "y=0`
`:. (1,0)` is the intersection point of the given two lines.
2. Find the point of intersection of the lines `3y+1=0` and `x+y-2=0`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`3y+1=0`
`:.3y=-1`
and `x+y-2=0`
`:.x+y=2`
`3y=-1 ->(1)`
`x+y=2 ->(2)`
Taking equation `(1)`, we have
`=>3y=-1`
`=>y=(-1)/3 ->(3)`
Putting `y=(-1)/3` in equation `(2)`, we get
`=>x+y=2`
`=>x+((-1)/3)=2`
`=>3x-1=6`
`=>3x=6+1`
`=>3x=7`
`=>x=7/3`
`:.y=-1/3" and "x=7/3`
`:. (7/3,-1/3)` is the intersection point of the given two lines.
3. Find the point of intersection of the lines `x-y+1=0` and `2x-3y+5=0`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`x-y+1=0`
`:.x-y=-1`
and `2x-3y+5=0`
`:.2x-3y=-5`
`x-y=-1 ->(1)`
`2x-3y=-5 ->(2)`
equation`(1) xx 2 =>2x-2y=-2`
equation`(2) xx 1 =>2x-3y=-5`
Substracting `=>y=3`
Putting `y=3` in equation`(1)`, we have
`x-(3)=-1`
`=>x=-1+3`
`=>x=2`
`:.x=2" and "y=3`
`:. (2,3)` is the intersection point of the given two lines.
4. Find the point of intersection of the lines `2x+y-5=0` and `x+y-3=0`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`2x+y-5=0`
`:.2x+y=5`
and `x+y-3=0`
`:.x+y=3`
`2x+y=5 ->(1)`
`x+y=3 ->(2)`
Substracting `=>x=2`
Putting `x=2` in equation`(2)`, we have
`2+y=3`
`=>y=3-2`
`=>y=1`
`:.x=2" and "y=1`
`:. (2,1)` is the intersection point of the given two lines.
5. Find the point of intersection of the lines `x-2y+15=0` and `3x+y-4=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.
2. Find the point of intersection of the lines `5x+2y-11=0` and `3x-y+11=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.
This material is intended as a summary. Use your textbook for detail explanation.
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