1.

Find the equation of the line, which makes intercepts `3` and 2 on the x and y axes respectively.

Answer» `xcostheta +4sintheta = p`
let p=normal distance from origin
`theta`=angle normal makes with +ve x direction
given `P= 4`
`theta=15^@`
since we know that
`cos2theta=2costheta^2-1`
`take theta=15^@`
`cos30^@=2cos^2 15-1`
`sqrt 3/2+1=2cos^2 15`
`cos15=sqrt((sqrt 3/4+2)`
now `(sqrt 3+1)^2 =3+2sqrt 3+1`
`=sqrt((sqrt 3 +1)^2/2 -2+2)/4 `
`cos15^@=sqrt((sqrt3+1)^2)/8`
`=(sqrt3 +1)/(2sqrt2)`
`sin15=sqrt1-cos^2 15`
`=sqrt(1-(sqrt3+2)/4)`
`=sqrt(4-(sqrt3+))/4`
`sin15=sqrt(2-15)/4`
`(sqrt3-1)^2=1+3-2sqrt3`
`-sqrt3=((sqrt3-1)^2-4)/2`
`sin15=sqrt(2+((sqrt3-1)^2/2)-2)/4`
`sin15=sqrt(sqrt3-1)^2)/8`
`=(sqrt3-1)/(2sqrt2)`
`xcostheta+4sintheta=P`
`x((sqrt3+1)/(2sqrt2))+4((sqrt3-1)/(2sqrt2)) =4`
`x((sqrt3+1))+4((sqrt3-1))=8sqrt2`


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