The line MN whose equation is x-y-2=0 cuts th | vedclass

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(D) $1$. Find the coordinates of $M$: The line $x-y-2=0$ cuts the $X$-axis where $y=0$. Substituting $y=0$ into the equation,we get $x-0-2=0$,so $x=2$

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$x=2$

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(D) $1$. Find the coordinates of $M$: The line $x-y-2=0$ cuts the $X$-axis where $y=0$. Substituting $y=0$ into the equation,we get $x-0-2=0$,so $x=2$. Thus,$M = (2,0)$.$2$. Find the slope of the original line $MN$: The equation $x-y-2=0$ can be written as $y=x-2$. Comparing this with $y=mx+c$,the slope $m_1 = 1$. This corresponds to an angle of $	heta_1 = 45^{\circ}$.$3$. Find the new slope: The line is rotated by $45^{\circ}$ anticlockwise. The new angle $	heta_2 = 45^{\circ} + 45^{\circ} = 90^{\circ}$.$4$. Find the equation of the new line: $A$ line passing through $M(2,0)$ with an angle of $90^{\circ}$ is a vertical line. The equation of a vertical line passing through $(2,0)$ is $x=2$.

The line $MN$ whose equation is $x-y-2=0$ cuts the $X$-axis at $M$ and the coordinates of $N$ are $(4,2)$. The line $MN$ is rotated about $M$ through $45^{\circ}$ in the anticlockwise direction. What is the equation of the line $MN$ in the new position?

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