A point equidistant from the line `4x + 3y + 10 = 0, 5x-12y + 26 = 0` and `7x + 24y-50 = 0 `is

Options

A. `(1,-1)`
B. `(1,1)`
C. `(0,0)`
D. `(0,1)`

Answer

C

Explanation

The given equation of lines are `4x+3y+10=0` ......(i)
`rArr4x-12y+26=0` ......(ii)
`rArr7x+24y-50=0` .......(iii)
Let the point (h,k) which is equidistant from these lines.
Distance from line (i) `=(|4h+3k+10|)/(sqrt(16+9))`
Distance from line (ii) `=(|5h-12k+26|)/(sqrt(25+144))`
Distance from line (iii) `=(|7h+24k-50|)/(sqrt(7^2+24^2))`
So, the point (h,k) is equidistant form lines (i), (ii) and (iii).
`therefore (4h+3k+10)/(sqrt(16+9))= (5h-12k+26)/(sqrt(25+144))=(7h+24k+50)/(sqrt(49+576))`
`rArr(|4h+3k+10|)/(5)= (|5h-12k+26|)/(13)=(7h+24k+50)/(25)`
Clearly, if h=0=0, then `(10)/(5)=(26)/(13)=(50)/(25)=2`
Hence, the required point is (0,0).