Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive xaxis.(i) `x-sqrt(3)y+8=0`, (ii) `y -2= 0`, (iii) `x -y = 4`.

Explanation

`(i) x-sqrt(3)y+8=0`
`implies -x+sqrt(3)y=8`
Divide both sides by `sqrt((-1)^(2)+(sqrt(3))^(2))=2`
`(-x)/(2)+(sqrt(3))/(2)y=4`
Comparing with `x cos alpha+ y sin alpha=p`
`cos alpha=-(1)/(2)` and `sin alpha =(sqrt(3))/(2)`
`implies alpha=(2pi)/(3)` and `p=4`
`:. ` Perpendicular form of line is, `x cos"(2pi)/(3)+y sin "(2pi)/(3)=4`
Therefore, perpendicular distance of line from origin`=4` unit
and angle between perpendicular and `x`-axis `=(2pi)/(3)`
`(ii) y-2=0`
`implies 0*x+1*y=2`
Comparing with `x cos alpha+y sin alpha=p`
`cos alpha=0`, `sin alpha=1`, `p=2`
`implies alpha=(pi)/(2)`
`:. ` Perpendicular form of line is,
`x cos"(pi)/(2)+y sin "(pi)/(2)=2`
Therefore, perpendicular distance of line from origin`=2` unit
and angle between perpendicular and `x`-axis `=(pi)/(2)`
`(iii) x-y=4`
Dividing both sides by `sqrt(1^(2)+(-1)^(2))=sqrt(2)`
`(1)/(sqrt(2))x-(1)/(sqrt(2))y=(4)/(sqrt(2))`
`implies(1)/(srt(2))x-(1)/(sqrt(2))y=2sqrt(2)`
Comparing with `x cos alpha+ y sin alpha=p`
`cos alpha=(1)/(sqrt(2))`, `sin alpha=-(1)/(sqrt(2))`, `p=2sqrt(2)`
`implies alpha=315^(@)`
`:. ` Perpendicular form of line is,
`x cos315^(@)+y sin 315^(@)=2sqrt(2)`
Therefore, perpendicular distance of line from origin`=2sqrt(2)` unit
and angle between perpendicular and `x`-axis `=315^(@)`