The equations of the lines passing through (1 | vedclass

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(D) The given line is $x+y=0$, which can be written as $y = -x$. The slope of this line is $m_1 = -1$.Let the slope of the required lines be $m$. The

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$x-1=0, y-1=0$

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(D) The given line is $x+y=0$, which can be written as $y = -x$. The slope of this line is $m_1 = -1$.Let the slope of the required lines be $m$. The angle between the lines is $45^{\circ}$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m m_1} ight|$, we have:$	an 45^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} ight|$$1 = \left| \frac{m+1}{1-m} ight|$This gives two cases:Case $1$: $\frac{m+1}{1-m} = 1 \implies m+1 = 1-m \implies 2m = 0 \implies m = 0$.The equation of the line passing through $(1,1)$ with slope $0$ is $y-1 = 0(x-1) \implies y-1 = 0$.Case $2$: $\frac{m+1}{1-m} = -1 \implies m+1 = -1+m \implies 1 = -1$, which is impossible (this implies the line is vertical).$A$ vertical line passing through $(1,1)$ is $x=1$, or $x-1=0$.Thus, the equations of the lines are $x-1=0$ and $y-1=0$.

The equations of the lines passing through $(1,1)$ and making an angle of $45^{\circ}$ with the line $x+y=0$ are

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