One diagonal of a square is along the line `8x-15 y=0` and one of its vertex is (1, 2). Then the equations of the sides of the square passing through this vertex are `23 x+7y=9,7x+23 y=53` `23 x-7y+9=0,7x+23 y+53=0` `23 x-7y-9=0,7x+23 y-53=0` none of these

Options

A. 7x-8y+9=0, 8x+7y-22=0
B. 9x-8y+7=0.8x+9y-26=0
C. 23x-7y-9=0,7x+23y-53=0
D. none of these

Answer

C

Explanation

The slope of BD is 8/15 and the angle made by BD with DC and BC is `45^(@)`. So, let the slope of DC be m. Then,
`"tan"45^(@) =+-(m-(8)/(15))/(1+(8)/(15)m)`
`"or " (15+8m)=+-(15m-8)`
`"or "m=(23)/(7) " and " -(7)/(23)`
Hence, the equations of DC and BC are
`y-2=(23)/(7)(x-1)`
or 23x-7y-9=0
`"and " y-2=-(7)/(23)(x-1)`
or 7x+23y-53=0