Coordinates of R will be (−5+r1‌cos‌θ,−4+r1sin‌θ) ‌ Sub in ‌x+3y+2=0 −5+r1‌cos‌θ+3(−4+r1sin‌θ)+2=0 r1(cos‌θ+3sin‌θ)=15 r1=‌
15
cos‌θ+3sin‌θ
Coordinates of S will be (−5+r2‌cos‌θ,−4+r2sin‌θ) Sub in 2x+3y+4=0 2(−5+r2‌cos‌θ)+3(−4+r2sin‌θ)+4=0 r2(2‌cos‌θ+3sin‌θ)=18 r2=‌
18
2‌cos‌θ+3sin‌θ
Coordinates of T will be (−5+r3‌cos‌θ,−4+r3sin‌θ) ⇒r3=‌
6
cos‌θ+sin‌θ
Substituting in (‌
15
r1
)2+(‌
10
r2
)2=(‌
6
r3
)2 we get tan‌θ=18 now slope of mx−y+5m−4=0 is m ∴m=tan‌θ=18