The image of the family of lines (lambda + 2) | vedclass

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(A) The given equation is $(\lambda + 2)x + (\lambda - 1)y - (8\lambda + 1) = 0$.Rearranging the terms to isolate $\lambda$,we get $(2x - y - 1) + \la

Vedclass · Accepted Answer

$(\mu + 1)x - (2\mu + 1)y + \mu - 2 = 0$

Vedclass · Answer

(A) The given equation is $(\lambda + 2)x + (\lambda - 1)y - (8\lambda + 1) = 0$.Rearranging the terms to isolate $\lambda$,we get $(2x - y - 1) + \lambda(x + y - 8) = 0$.This represents a family of lines passing through the intersection of $2x - y - 1 = 0$ and $x + y - 8 = 0$.Solving these equations: Adding them gives $3x - 9 = 0$,so $x = 3$. Substituting $x = 3$ into $x + y - 8 = 0$ gives $y = 5$.Thus,the family of lines passes through the fixed point $P(3, 5)$.The image of point $P(3, 5)$ in the line mirror $y = x$ is $P'(5, 3)$.Any line passing through $P'(5, 3)$ can be represented as $a(x - 5) + b(y - 3) = 0$,which simplifies to $ax + by - (5a + 3b) = 0$.Comparing this with the options,we check for the form $(\mu + 1)x - (2\mu + 1)y + \mu - 2 = 0$.For this line to pass through $(5, 3)$,we substitute $x = 5$ and $y = 3$: $(\mu + 1)(5) - (2\mu + 1)(3) + \mu - 2 = 5\mu + 5 - 6\mu - 3 + \mu - 2 = 0$.Since the result is $0$,the line passes through $(5, 3)$ for all $\mu$.

The image of the family of lines $(\lambda + 2)x + (\lambda - 1)y - (8\lambda + 1) = 0$ in the line mirror $y = x$ is (where $\lambda$ and $\mu$ are parameters):

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