Problem: line through (5,5) and perpendicular to 2x+3y+4=0 [ Calculator, Method and examples ]
Solution:
Your problem `->` line through (5,5) and perpendicular to 2x+3y+4=0
Here Point `(x_1,y_1)=(5,5)` and line `2x+3y+4=0` (given)
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=-0.6667x-1.3333`
`:.` Slope `=-0.6667`
`:.` Slope of perpendicular line`=(-1)/(-0.6667)=1.5` (`:' m_1*m_2=-1`)
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)=(5,5)` and Slope `m=1.5` (given)
`:. y-5=1.5(x-5)`
`:. 2(y-5)=3(x-5)`
`:. 2y -10=3x -15`
`:. 3x-2y-5=0`
Hence, The equation of line is `3x-2y-5=0`
