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 the vertex are given by the equations?

Ans: `y-3x+9=0, x+3y-3=0`

Answer 1

see explanation

Explanation:

As `A(3,0)` does not satisfy the diagonal `x=2y`,
`=> A(3,0)` does not lie on the diagonal `x=2y`.
Let `m` be the slope of a side passing through `A(3,0)`,
`=>` equation of the side is : `y-0=m(x-3)`,
`=> y=m(x-3) ------color(red)(EQ1)`,
formula to find the angle between two lines :
`tantheta=|(m_1-m_2)/(1+m_1m_2)|`,
where `theta` is the angle between the two lines and `m_1 and m_2` are the slope of the two lines.
slope of the diagonal `x=2y` is `1/2`
`=> tan45=|(m-1/2)/(1+m*1/2)|`
`=> 1=|(m-1/2)/(1+m/2)|`
`=> m=3 or -1/3`
Substituting `m=3, and -1/3` in `color(red)(EQ1)`, we get :
`y=3(x-3), => color(red)(y-3x+9=0)`,
and,
`y=-1/3(x-3), => color(red)(3y+x-3=0)`