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Find line through intersection point of 5x+2y-11=0, 3x-y+11=0 and perpendicular to 4x-3y+2=0

Solution:
Your problem `->` line through intersection point of 5x+2y-11=0, 3x-y+11=0 and perpendicular to 4x-3y+2=0


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.


`4x-3y+2=0`

`:. 3y=4x+2`

`:. y=(4x)/(3)+2/3`

`:.` Slope`=4/3`

`:.` Slope of perpendicular line`=-3/4`

The equation of a line with slope m and passing through `(x_1,y_1)` is `y-y_1=m(x-x_1)`

Putting `(-1,8)` for `(x_1,y_1)` and `m=-3/4,` we get

`:. y-8=-3/4(x+1)`

`:. 4(y-8)=-3(x+1)`

`:. 4y -32=-3x -3`

`:. 3x+4y-29=0`








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