If the image of point P(lambda, 2lambda) in t | vedclass

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(B) Let the point $P = (\lambda, 2\lambda)$ and its image $Q = (\mu, 3\lambda)$ with respect to the line $x - y + 2 = 0$.The midpoint $F$ of $PQ$ is $

Vedclass · Accepted Answer

$\lambda + \mu = 1$

Vedclass · Answer

(B) Let the point $P = (\lambda, 2\lambda)$ and its image $Q = (\mu, 3\lambda)$ with respect to the line $x - y + 2 = 0$.The midpoint $F$ of $PQ$ is $\left( \frac{\lambda + \mu}{2}, \frac{2\lambda + 3\lambda}{2} ight) = \left( \frac{\lambda + \mu}{2}, \frac{5\lambda}{2} ight)$.Since $F$ lies on the line $x - y + 2 = 0$,we have $\frac{\lambda + \mu}{2} - \frac{5\lambda}{2} + 2 = 0$,which simplifies to $\frac{\mu - 4\lambda}{2} + 2 = 0$,or $\mu - 4\lambda + 4 = 0$  $(1)$.The slope of $PQ$ is $m_{PQ} = \frac{3\lambda - 2\lambda}{\mu - \lambda} = \frac{\lambda}{\mu - \lambda}$.Since $PQ$ is perpendicular to the line $x - y + 2 = 0$ (which has slope $1$),we have $m_{PQ} 	imes 1 = -1$,so $\frac{\lambda}{\mu - \lambda} = -1$,which implies $\lambda = -\mu + \lambda$,so $\mu = 0$.Substituting $\mu = 0$ into $(1)$,we get $0 - 4\lambda + 4 = 0$,so $\lambda = 1$.Thus,$\lambda = 1$ and $\mu = 0$. Checking the options:$A) \mu - \lambda = 0 - 1 = -1 
eq 0$$B) \lambda + \mu = 1 + 0 = 1$$C) \lambda - \mu = 1 - 0 = 1 
eq -1$$D) \lambda + \mu = 1 + 0 = 1 
eq 0$The correct option is $B$.

If the image of point $P(\lambda, 2\lambda)$ in the line mirror $x - y + 2 = 0$ is $Q(\mu, 3\lambda)$,then choose the correct option.

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