Find the equations of the lines through the point of intersection of the lines `x-y+1=0a n d2x-3y+5=0` whose distance from the point `(3,2)` is 7/5.

Answer

N/a

Explanation

Given equation of lines `x-y+1=0`
and `2x-3y+5=0`
From Eq. (i) `x=y-1`
Now, put the value of x in Eq.(ii) we get
`2(y-1)-3y+5=0`
`rArr 2y-2-3y+5=0`
`rArr 3-y=0rArry=3`
y=3puit in Eq.(i) , we get
`x=2`
Since, the point of intersection is `(2,3)`.
Let slope of the required line be m.
`therefore` Equation of line is `y-3=m(x-2)`
`rArr mx-y+3-2m=0`
Since, the distance from (3,2) to line (iii) is `(7)/(5)`.
`therefore (7)/(5)=|(3m-2+3-2m)/(sqrt(1+m^(2)))|`
`rArr(49)/(25)=(m+1)^2/(1+m^2)`
`rArr49+49m^2=25(m^2+2m+1)`
`rArr49+49m^2=25m^2+50m+25`
`rArr24m^2-50m+24=0`
`rArr12m^2-25m+12=0`
`therefore m=(25pmsqrt(625-4.12.12))/(24)`
`=(25pmsqrt(49))/(24)=(25pm7)/(24)=(32)/(24)or (18)/(24)=(4)/(3)or (3)/(4)`
`therefore` First equation of a line is `y-3=(4)/(3)(x-2)`
`rArr3y-9=4x-8`
`rArr4x-3y+1=0`
and second equation of line is `y-3=(3)/(4)(x-2)`
`rArr4y-12=3x-6`
`rArr3x-4y+6=0`