1. Find the equation of a line passing through the point of intersection of lines `x-4y+18=0` and `x+y-12=0` and having slope `2`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`x-4y+18=0`
`:.x-4y=-18`
and `x+y-12=0`
`:.x+y=12`
`x-4y=-18 ->(1)`
`x+y=12 ->(2)`
Substracting `=>-5y=-30`
`=>5y=30`
`=>y=30/5`
`=>y=6`
Putting `y=6` in equation `(2)`, we have
`x+6=12`
`=>x=12-6`
`=>x=6`
`:.x=6" and "y=6`
`:. (6,6)` is the intersection point of the given two lines.
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)=(6,6)` and Slope `m=2` (given)
`:. y-6=2(x-6)`
`:. y -6=2x -12`
`:. 2x-y-6=0`
Hence, The equation of line is `2x-y-6=0`
2. Find the equation of a line passing through the point of intersection of lines `2x+3y+4=0` and `3x+6y-8=0` and having slope `2`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`2x+3y+4=0`
`:.2x+3y=-4`
and `3x+6y-8=0`
`:.3x+6y=8`
`2x+3y=-4 ->(1)`
`3x+6y=8 ->(2)`
equation`(1) xx 3 =>6x+9y=-12`
equation`(2) xx 2 =>6x+12y=16`
Substracting `=>-3y=-28`
`=>3y=28`
`=>y=28/3`
Putting `y=28/3 ` in equation `(1)`, we have
`2x+3(28/3)=-4`
`=>2x=-4-28`
`=>2x=-32`
`=>x=-16`
`:.x=-16" and "y=28/3`
`:. (-16,28/3)` is the intersection point of the given two lines.
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)=(-16,28/3)` and Slope `m=2` (given)
`:. y-28/3=2(x+16)`
`:. y -28/3=2x +32`
`:. 2x-y+124/3=0`
`:. 6x-3y+124=0`
Hence, The equation of line is `6x-3y+124=0`
3. Find the equation of a line passing through the point of intersection of lines `x=3y` and `3x=2y+7` and having slope `-1/2`
Solution:
The point of intersection of the lines can be obtainted by solving the given equations
`x=3y`
`:.x-3y=0`
and `3x=2y+7`
`:.3x-2y=7`
`x-3y=0 ->(1)`
`3x-2y=7 ->(2)`
equation`(1) xx 3 =>3x-9y=0`
equation`(2) xx 1 =>3x-2y=7`
Substracting `=>-7y=-7`
`=>7y=7`
`=>y=7/7`
`=>y=1`
Putting `y=1` in equation `(2)`, we have
`3x-2(1)=7`
`=>3x=7+2`
`=>3x=9`
`=>x=3`
`:.x=3" and "y=1`
`:. (3,1)` is the intersection point of the given two lines.
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)=(3,1)` and Slope `m=-1/2` (given)
`:. y-1=-1/2(x-3)`
`:. 2(y-1)=-1(x-3)`
`:. 2y -2=-x +3`
`:. x+2y-5=0`
Hence, The equation of line is `x+2y-5=0`
This material is intended as a summary. Use your textbook for detail explanation.
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