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Find line through (5,5) and perpendicular to 2x+3y+4=0

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Your problem `->` line through (5,5) and perpendicular to 2x+3y+4=0


When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.
We shall first find the slope of line `2x+3y+4=0`

`2x+3y+4=0`

`:. 3y=-2x-4`

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

`:.` Slope`=-2/3`

`:.` Slope of perpendicular line`=3/2`

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

Putting `(5,5)` for `(x_1,y_1)` and `m=3/2,` we get

`:. y-5=3/2(x-5)`

`:. 2(y-5)=3(x-5)`

`:. 2y -10=3x -15`

`:. 3x-2y-5=0`








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