Let $l_{1}$ be the line $4x + 3y = 3$ and $l_{2}$ be the line $y = 8x$. $L_{1}$ is the line formed by reflecting $l_{1}$ across the line $y = x$ and $L_{2}$ is the line formed by reflecting $l_{2}$ across the x-axis. If $ heta$ is the acute angle between $L_{1}$ and $L_{2}$ such that $ an heta…

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Let $l_{1}$ be the line $4x + 3y = 3$ and $l_{2}$ be the line $y = 8x$. $L_{1}$ is the line formed by reflecting $l_{1}$ across the line $y = x$ and $L_{2}$ is the line formed by reflecting $l_{2}$ across the x-axis. If $	heta$ is the acute angle between $L_{1}$ and $L_{2}$ such that $	an 	heta = \frac{a}{b}$, where $a$ and $b$ are coprime then find $(a + b)$.

Aicharya · Accepted Answer

$L_{1}: 4x + 3y = 3$ To take reflection w.r.t $y = x$, exchange $x$ and $y$ we get $L_{1}: 4y + 3x = 3$ $L_{2}: y = 8x$ To take reflection w.r.t x-axis, replace $y$ with $-y$ $L_{2}: -y = 8x$ To find angle between $L_{1}$ and $L_{2}$ $	an{	heta} = \left| \frac{3/2 - 3}{2/4 + 4} ight| = \left| \frac{2/9}{2/8} ight| = \frac{a}{b}$ $	herefore a + b = 57$

Let $l_{1}$ be the line $4x + 3y = 3$ and $l_{2}$ be the line $y = 8x$. $L_{1}$ is the line formed by…

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