The equation of the line joining the point `(3,5)` to the point of intersection of the lines `4x+y-1=0` and `7x-3y-35=0` is equidistant from the points `(0,0)` and `(8,34)`.

Answer

True

Explanation

Given equation of lines are
`4x+y-1=0" .....(i)"`
and `7x-3y-35=0" .....(ii)"`
From Eq. (i), on putting `y=1-4x` in Eq. (ii), we get
`7x-3+12x-35=0`
`rArr 19x-38=0rArrx=2`
On putting x=2 in Eq. (i), we get
`8+y-1=0rArry=-7`
Now, the equation of a line passing through (3,5) and (2,-7) is
`y-5=(-7-5)/(2-3)(x-3)`
`rArr y-5=12(x-3)`
`rArr 12x-y-31=0`
Distance from (0,0) to the line (iii),
`d_(1)=(|-31|)/(sqrt(144+1))=(31)/(sqrt(145))`
`:.` Distance from (8,34) to the line (iii),
`d_(2)=(|96-34-31|)/(sqrt(145))=(31)/(sqrt(145))`
`:. d_(1)=d_(2)`
Hence, the statement is true