Find the equations the sides of an isosceles right angled triangle the equation of whose hypotenuse is `3x+4y=4` and the opposite vertex is the point (2, 2).

Answer

N/a

Explanation

Let slope of line AC be m and slope of line BC is `(-3)/(4)` and let angle between line AC and BC be `theta`
`therefore tan theta=|(m+3/(4))/(1-(3m)/(4))|rArrtan 45^@=pm[(m+3/(4))/(1-(3m)/(4))]`
Taking positive sign, `1=(m+3/4)/(1-(3m)/4)`
`rArr m+3/4=1-(3m)/(4)`
`rArr m+(3m)/4=1-(3)/(4)`
`rArr (7m)/4=(1)/(4)rArrm=(1)/(7)`
Taking negative sign, `1=(m+(3)/(4))/(1-(3m)/(4))`
`rArrm+(3)/(4)=1-(3m)/(4)`
`rArrm+(3m)/(4)=1-(3)/(4)`
`rArr(7m)/(4)=1/(4)rArrm=(1)/(7)`
Taking negative sign,
`1=((m+(3)/(4))/(1-(3m)/(4)))rArr1-(3m)/(4)=-m-(3)/(4)`
`rArrm-(3m)/(4)=-1-(3)/(4)`
`rArr(3m)/(4)=(-7)/(4)rArrm=-7`
`therefore` Equation of side AC having slope `((1)/(7))` is
`y-2=(1)/(7)(x-2)`
`rArr7y-14=x-2`
`rArrx-7y+12=0`
and equation of side AB having slope `(-7)` is
`y-2=-7(x-2)`
`rArry-2=-7x+14`
`rArr7x+y-16=0`