1.

Find the equation of the line through the intersection of lines `x+ 2y 3 = 0`and `4x y+ 7 =0` and which is parallel to `5x+ 4y 20 = 0`

Answer» `5x+4y-20=0 Rightarrow y=(-5)/(4)x+5`
`therefore "slope of the given lne"=(-5)/(4)`
and slope of the required line =`(-5)/(4)`
Now, the equation of any line through the intersection of hte given line of the form
`(x+2y-3)+k(4x-y+7)=0`
`Rightarrow (1+4k)x+(2-k)y+(7k-3)=0`
`Rightarrow (2-k)y=-(1+4k)x+(3-7k)`
`Rightarrow y=(-(1+4k))/((2-k))x+((3-7k))/((2-k))`
`Rightarrow y=((1+4k))/((k-2))x+((3-7k))/((2-k))`
Slope of this line =`((1+4k))/((k-2))`
`therefore ((1+4k))/((k-2))=(-5)/(4)Rightarrow 4+16k=-5k+10`
`Rightarrow 21k=6Rightarrow =(6)/(21)=(2)/(7)`
Substituting `k=(2)/(7)` in (i), we get
`(x+2y-3)+(2)/(7)(4x-y+7)=0`
`Rightarrow (7x+14y-21)+(8x-2y+14)=0`
`Rightarrow 15x+12y-7=0,` which is the required equation.


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