Question

If one diagonal of a square is along the line x = 2y and one of its vertex is (3, 0), then its sides through this vertex are given by the equations-

• A. y – 3x + 9 = 0, x – 3y – 3 = 0
• B. y – 3x + 9 = 0, x – 3y – 3 = 0
• C. y + 3x – 9 = 0, x + 3y – 3 = 0
• D. y – 3x + 9 = 0, x + 3y – 3 = 0

Answer 1

Answer: (D)

y – 3x + 9 = 0, x + 3y – 3 = 0

Answer 2

The required equations are

y = m(x – 3)

where m = $\frac{\frac{1}{2} + \tan 45{^\circ}}{1 - \frac{\tan 45{^\circ}}{2}}$, $\frac{\frac{1}{2}–\tan 45{^\circ}}{1 + \frac{\tan 45{^\circ}}{2}}$

m = 3 , $\frac{- 1}{3}$

Hence, the equations are

Y = 3(x – 3) and y = $\frac{- 1}{3}$ (x – 3)

i.e. 3x – y – 9 = 0 and x + 3y – 3 = 0.