Angle between two straight lines Questions in English

(A) The line $L$ passes through $(-2, -3)$ and has an undefined slope,so its equation is $x = -2$ (a vertical line).
The angle between $x = -2$ and the line $ax - 2y + 3 = 0$ is $45^{\circ}$.
The slope of the line $ax - 2y + 3 = 0$ is $m_1 = \frac{a}{2}$.
The angle $\theta$ between a vertical line and a line with slope $m_1$ is given by $|\tan(90^{\circ} - \theta)| = |\frac{1}{m_1}|$.
Since the angle is $45^{\circ}$,we have $|\frac{1}{a/2}| = \tan(45^{\circ}) = 1$.
Thus,$\frac{2}{a} = 1$,which gives $a = 2$.
Now,substitute $a = 2$ into the equation $x + ay - 4 = 0$ to get $x + 2y - 4 = 0$.
The slope of this line is $m = -\frac{1}{2}$.
The angle $\theta$ made with the positive $X$-axis is given by $\tan \theta = -\frac{1}{2}$.
Since the slope is negative,the angle is in the second quadrant,so $\theta = \pi - \tan^{-1}\left(\frac{1}{2}\right)$.

(B) The slope of the line is $m$. The slope of the line $x + y - 3 = 0$ is $m_2 = -1$.
Given $\tan \theta = \left|\frac{m - m_2}{1 + m m_2}\right| = \frac{5}{7}$.
Substituting $m_2 = -1$,we get $\left|\frac{m + 1}{1 - m}\right| = \frac{5}{7}$.
This implies $7|m + 1| = 5|1 - m|$.
Case $1$: $7(m + 1) = 5(1 - m)$ $\Rightarrow 7m + 7 = 5 - 5m$ $\Rightarrow 12m = -2$ $\Rightarrow m = -1/6$.
Case $2$: $7(m + 1) = -5(1 - m)$ $\Rightarrow 7m + 7 = -5 + 5m$ $\Rightarrow 2m = -12$ $\Rightarrow m = -6$.
Since $m \in \mathbb{Z}$,we must have $m = -6$.
The equation of the line passing through $(1, 1)$ with slope $m = -6$ is $y - 1 = -6(x - 1)$.
$y - 1 = -6x + 6 \Rightarrow 6x + y - 7 = 0$.
Comparing with $ax + y + c = 0$,we get $a = 6$ and $c = -7$.
Therefore,$ac = 6 \times (-7) = -42$.

(D) The angle $\theta$ between two lines with slopes $m$ and $m'$ is given by $\tan \theta = \left| \frac{m - m'}{1 + m m'} \right|$.
Given $\theta = \frac{\pi}{6}$ and $m' = 2$,we have $\tan \frac{\pi}{6} = \left| \frac{m - 2}{1 + 2m} \right|$.
Since $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \left| \frac{m - 2}{1 + 2m} \right|$.
Squaring both sides,we get $\frac{1}{3} = \frac{(m - 2)^2}{(1 + 2m)^2}$.
$(1 + 2m)^2 = 3(m - 2)^2$.
$1 + 4m + 4m^2 = 3(m^2 - 4m + 4)$.
$1 + 4m + 4m^2 = 3m^2 - 12m + 12$.
$m^2 + 16m - 11 = 0$.
This is a quadratic equation in $m$ whose roots are $m_1$ and $m_2$.
Using the sum of roots formula,$m_1 + m_2 = -\frac{b}{a} = -\frac{16}{1} = -16$.

(A) The slope of the given line $x+y-3=0$ is $m_1 = -1$. Let the slope of the required line be $m$. The angle between the two lines is $60^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m m_1} \right|$,we have $\tan 60^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} \right|$.
$\sqrt{3} = \left| \frac{m+1}{1-m} \right|$.
This gives two cases: $\frac{m+1}{1-m} = \sqrt{3}$ or $\frac{m+1}{1-m} = -\sqrt{3}$.
Case $1$: $m+1 = \sqrt{3} - \sqrt{3}m$ $\Rightarrow m(1+\sqrt{3}) = \sqrt{3}-1$ $\Rightarrow m = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$.
The equation of the line is $y-1 = (2-\sqrt{3})(x-1)$ $\Rightarrow y-1 = (2-\sqrt{3})x - 2 + \sqrt{3}$ $\Rightarrow (2-\sqrt{3})x - y + (1-\sqrt{3}) = 0$.
Case $2$: $m+1 = -\sqrt{3} + \sqrt{3}m$ $\Rightarrow m(1-\sqrt{3}) = -\sqrt{3}-1$ $\Rightarrow m = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}$.
The equation of the line is $y-1 = (2+\sqrt{3})(x-1)$ $\Rightarrow y-1 = (2+\sqrt{3})x - 2 - \sqrt{3}$ $\Rightarrow (2+\sqrt{3})x - y - (1+\sqrt{3}) = 0$.
Comparing with the options,option $A$ is $(1+\sqrt{3})x+(1-\sqrt{3})y-2=0$ which simplifies to $y-1 = \frac{1+\sqrt{3}}{\sqrt{3}-1}(x-1) = (2+\sqrt{3})(x-1)$,matching Case $2$.

(C) The slope of the line joining the points $(1, 2)$ and $(3, 4)$ is $m = \frac{4-2}{3-1} = \frac{2}{2} = 1$.
Let the slopes of the required lines be $m_1$ and $m_2$. The angle between these lines and the given line is $30^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have $\tan 30^{\circ} = \left| \frac{m - 1}{1 + m(1)} \right| = \left| \frac{m - 1}{1 + m} \right|$.
$\frac{1}{\sqrt{3}} = \left| \frac{m - 1}{m + 1} \right| \Rightarrow \pm \frac{1}{\sqrt{3}} = \frac{m - 1}{m + 1}$.
Case $1$: $\frac{1}{\sqrt{3}} = \frac{m - 1}{m + 1}$ $\Rightarrow m + 1 = \sqrt{3}m - \sqrt{3}$ $\Rightarrow m(\sqrt{3} - 1) = \sqrt{3} + 1$ $\Rightarrow m = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.
Case $2$: $-\frac{1}{\sqrt{3}} = \frac{m - 1}{m + 1}$ $\Rightarrow -m - 1 = \sqrt{3}m - \sqrt{3}$ $\Rightarrow m(\sqrt{3} + 1) = \sqrt{3} - 1$ $\Rightarrow m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 + 1 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.
Since $m_1 > m_2$,we have $m_1 = 2 + \sqrt{3}$ and $m_2 = 2 - \sqrt{3}$.
Therefore,$\frac{m_1}{m_2} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} = \frac{(2 + \sqrt{3})^2}{4 - 3} = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$.

(B) The given line is $\sqrt{3} x + y = 1$,which can be written as $y = -\sqrt{3} x + 1$. The slope $m_1 = -\sqrt{3}$.
Let the slope of the required line be $m_2$. The angle between the two lines is $60^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\tan 60^{\circ} = \left| \frac{-\sqrt{3} - m_2}{1 + (-\sqrt{3})m_2} \right|$
$\sqrt{3} = \left| \frac{-\sqrt{3} - m_2}{1 - \sqrt{3} m_2} \right|$
Squaring both sides:
$3 = \frac{(-\sqrt{3} - m_2)^2}{(1 - \sqrt{3} m_2)^2}$
$3(1 - \sqrt{3} m_2)^2 = (\sqrt{3} + m_2)^2$
$3(1 + 3m_2^2 - 2\sqrt{3} m_2) = 3 + m_2^2 + 2\sqrt{3} m_2$
$3 + 9m_2^2 - 6\sqrt{3} m_2 = 3 + m_2^2 + 2\sqrt{3} m_2$
$8m_2^2 - 8\sqrt{3} m_2 = 0$
$8m_2(m_2 - \sqrt{3}) = 0$
So,$m_2 = 0$ or $m_2 = \sqrt{3}$.
For $m_2 = \sqrt{3}$,the equation of the line passing through $(3, 2)$ is:
$y - 2 = \sqrt{3}(x - 3)$
$y - 2 = \sqrt{3} x - 3\sqrt{3}$
$\sqrt{3} x - y + (2 - 3\sqrt{3}) = 0$.

(C) The slopes of the lines $x-2y+3=0$ and $kx-y+2=0$ are $m_1 = \frac{1}{2}$ and $m_2 = k$ respectively.
Given the angle $\theta = 45^{\circ}$,we use the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan 45^{\circ} = \left| \frac{\frac{1}{2} - k}{1 + \frac{k}{2}} \right| = 1$.
$|1 - 2k| = |2 + k|$.
Case $1$: $1 - 2k = 2 + k$ $\Rightarrow 3k = -1$ $\Rightarrow k = -\frac{1}{3}$.
Case $2$: $1 - 2k = -(2 + k)$ $\Rightarrow 1 - 2k = -2 - k$ $\Rightarrow k = 3$.
Since $k_1 > k_2$,we have $k_1 = 3$ and $k_2 = -\frac{1}{3}$.
Then $k_1 - 2 = 3 - 2 = 1$.
Checking the options: $-3k_2 = -3(-\frac{1}{3}) = 1$.
Thus,$k_1 - 2 = -3k_2$.

(D) For two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be perpendicular,the product of their slopes must be $-1$,or $a_1a_2 + b_1b_2 = 0$.
Here,$a_1 = 2, b_1 = (3 - 2k)$ and $a_2 = k, b_2 = (k - 1)$.
Applying the condition $a_1a_2 + b_1b_2 = 0$:
$2(k) + (3 - 2k)(k - 1) = 0$
$2k + (3k - 3 - 2k^2 + 2k) = 0$
$-2k^2 + 7k - 3 = 0$
$2k^2 - 7k + 3 = 0$
$(2k - 1)(k - 3) = 0$
Thus,$k = 3$ or $k = \frac{1}{2}$.
Given $\alpha > \beta$,we have $\alpha = 3$ and $\beta = \frac{1}{2}$.
Therefore,$\alpha^2 + 2\beta = (3)^2 + 2(\frac{1}{2}) = 9 + 1 = 10$.

(D) Given lines are $L_1: \frac{x}{a} + \frac{y}{b} = 1$ and $L_2: \frac{x}{b} + \frac{y}{a} = 1$.
Slope of $L_1$ is $m_1 = -\frac{b}{a}$.
Slope of $L_2$ is $m_2 = -\frac{a}{b}$.
The angle $\theta$ between two lines is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the slopes: $\tan \theta = \left| \frac{-\frac{b}{a} - (-\frac{a}{b})}{1 + (-\frac{b}{a})(-\frac{a}{b})} \right| = \left| \frac{\frac{a}{b} - \frac{b}{a}}{1 + 1} \right| = \left| \frac{a^2 - b^2}{2ab} \right|$.
Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{|a^2 - b^2|}{|2ab|}$,the hypotenuse is $\sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4 + b^4 - 2a^2b^2 + 4a^2b^2} = \sqrt{(a^2 + b^2)^2} = |a^2 + b^2|$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \left| \frac{a^2 - b^2}{a^2 + b^2} \right|$.

(B) Let the given line be $L_1: \sqrt{3}x + y - 9 = 0$. The slope of $L_1$ is $m_1 = -\sqrt{3}$.
Let the slope of the required line $L$ be $m$. The angle between $L$ and $L_1$ is $60^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$,we have:
$\tan 60^{\circ} = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} \right|$ $\Rightarrow \sqrt{3} = \left| \frac{m + \sqrt{3}}{1 - \sqrt{3}m} \right|$.
Case $1$: $\sqrt{3} = \frac{m + \sqrt{3}}{1 - \sqrt{3}m}$ $\Rightarrow \sqrt{3} - 3m = m + \sqrt{3}$ $\Rightarrow 4m = 0$ $\Rightarrow m = 0$.
The equation is $y - (-3) = 0(x - 5) \Rightarrow y + 3 = 0$. This line is parallel to the $X$-axis and does not intersect it at a unique point (it is the $X$-axis itself if $y=0$,but here $y=-3$).
Case $2$: $-\sqrt{3} = \frac{m + \sqrt{3}}{1 - \sqrt{3}m}$ $\Rightarrow -\sqrt{3} + 3m = m + \sqrt{3}$ $\Rightarrow 2m = 2\sqrt{3}$ $\Rightarrow m = \sqrt{3}$.
The equation of line $L$ is $y - (-3) = \sqrt{3}(x - 5)$ $\Rightarrow y + 3 = \sqrt{3}x - 5\sqrt{3}$ $\Rightarrow \sqrt{3}x - y - 3 - 5\sqrt{3} = 0$.

(C) The slope $m_1$ of the line passing through $(1, -2)$ and $(3, 2)$ is given by $m_1 = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2$.
The slope $m_2$ of the line $x + 2y - 7 = 0$ is given by $m_2 = -\frac{1}{2}$.
Since $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$,the two lines are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(C) The slopes of the lines $4x - y + 7 = 0$ and $kx - 5y - 9 = 0$ are $m_1 = 4$ and $m_2 = \frac{k}{5}$,respectively.
Given the angle $\theta = 45^{\circ}$ between the lines,we use the formula:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
$\tan 45^{\circ} = \left| \frac{4 - \frac{k}{5}}{1 + 4 \cdot \frac{k}{5}} \right|$
$1 = \left| \frac{20 - k}{5 + 4k} \right|$
This implies $\frac{20 - k}{5 + 4k} = 1$ or $\frac{20 - k}{5 + 4k} = -1$.
Case $1$: $20 - k = 5 + 4k$ $\Rightarrow 5k = 15$ $\Rightarrow k = 3$.
Case $2$: $20 - k = -5 - 4k$ $\Rightarrow 3k = -25$ $\Rightarrow k = -\frac{25}{3}$.
Since $k > 0$,the only valid solution is $k = 3$.

(B) Let the slope of the required line be $m$. The line passes through $(1, 2)$,so its equation is $(y - 2) = m(x - 1)$.
Given that the angle between this line and $y = 2x + 1$ is $45^{\circ}$.
The slope of the given line is $m_1 = 2$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$,we have:
$\tan 45^{\circ} = \left| \frac{m - 2}{1 + 2m} \right|$
$1 = \left| \frac{m - 2}{1 + 2m} \right|$
This gives two cases:
Case $1$: $\frac{m - 2}{1 + 2m} = 1$ $\Rightarrow m - 2 = 1 + 2m$ $\Rightarrow m = -3$.
The equation is $(y - 2) = -3(x - 1)$ $\Rightarrow y - 2 = -3x + 3$ $\Rightarrow 3x + y = 5$.
Case $2$: $\frac{m - 2}{1 + 2m} = -1$ $\Rightarrow m - 2 = -1 - 2m$ $\Rightarrow 3m = 1$ $\Rightarrow m = \frac{1}{3}$.
The equation is $(y - 2) = \frac{1}{3}(x - 1)$ $\Rightarrow 3y - 6 = x - 1$ $\Rightarrow x - 3y + 5 = 0$.
Comparing with the options,$3x + y = 5$ is correct.

(B) Let the line joining the points $(1, -2)$ and $(3, 2)$ be $L_1$. The slope $m_1$ of $L_1$ is given by $m_1 = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2$.
The equation of the second line $L_2$ is $x + 2y - 7 = 0$,which can be written as $2y = -x + 7$ or $y = -\frac{1}{2}x + \frac{7}{2}$.
The slope $m_2$ of $L_2$ is $-\frac{1}{2}$.
Now,calculate the product of the slopes: $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$.
Since the product of the slopes is $-1$,the two lines are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(B) Given that the two lines $L_1: \frac{3}{2}x + (2a - 1)y = 3$ and $L_2: 2x + a^2y = -3$ are perpendicular.
Since $L_1 \perp L_2$,the product of their slopes $m_1 \times m_2 = -1$.
The slope of $L_1$ is $m_1 = -\frac{3/2}{2a - 1} = -\frac{3}{4a - 2}$.
The slope of $L_2$ is $m_2 = -\frac{2}{a^2}$.
Thus,$(-\frac{3}{4a - 2}) \times (-\frac{2}{a^2}) = -1 \Rightarrow \frac{6}{a^2(4a - 2)} = -1$.
$6 = -4a^3 + 2a^2$ $\Rightarrow 4a^3 - 2a^2 + 6 = 0$ $\Rightarrow 2a^3 - a^2 + 3 = 0$.
Testing $a = -1$: $2(-1)^3 - (-1)^2 + 3 = -2 - 1 + 3 = 0$. So,$(a + 1)$ is a factor.
Dividing $2a^3 - a^2 + 3$ by $(a + 1)$ gives $2a^2 - 3a + 3 = 0$,which has no real roots $(D = 9 - 24 < 0)$.
So,$a = -1$.
Substituting $a = -1$ into the lines:
$L_1: \frac{3}{2}x + (2(-1) - 1)y = 3$ $\Rightarrow \frac{3}{2}x - 3y = 3$ $\Rightarrow x - 2y = 2$.
$L_2: 2x + (-1)^2y = -3 \Rightarrow 2x + y = -3$.
Solving the system:
$x - 2y = 2$ $(i)$
$2x + y = -3$ (ii)
From (ii),$y = -3 - 2x$. Substituting into $(i)$: $x - 2(-3 - 2x) = 2$ $\Rightarrow x + 6 + 4x = 2$ $\Rightarrow 5x = -4$ $\Rightarrow x = -\frac{4}{5}$.
Then $y = -3 - 2(-\frac{4}{5}) = -3 + \frac{8}{5} = -\frac{7}{5}$.
The intersection point is $(-\frac{4}{5}, -\frac{7}{5})$.
The distance from $(1, 1)$ is $\sqrt{(1 - (-\frac{4}{5}))^2 + (1 - (-\frac{7}{5}))^2} = \sqrt{(\frac{9}{5})^2 + (\frac{12}{5})^2} = \sqrt{\frac{81 + 144}{25}} = \sqrt{\frac{225}{25}} = \sqrt{9} = 3$.

(D) Let the given lines be $L_1: 3x + 4y - 5 = 0$ and $L_2: 4x - 3y - 15 = 0$. The slopes are $m_1 = -\frac{3}{4}$ and $m_2 = \frac{4}{3}$.
Since $m_1 \times m_2 = -1$,the lines are perpendicular.
Let the line passing through $(1, 2)$ have slope $m$. Its equation is $y - 2 = m(x - 1)$,or $mx - y + (2 - m) = 0$.
Since $AB = AC$,the line $BC$ must make equal angles with $L_1$ and $L_2$. Since $L_1 \perp L_2$,the line $BC$ must make an angle of $45^\circ$ with both lines.
The angle between the line with slope $m$ and $L_1$ is given by $\tan 45^\circ = |\frac{m - (-3/4)}{1 + m(-3/4)}| = 1$.
$|\frac{4m + 3}{4 - 3m}| = 1$.
Case $1$: $\frac{4m + 3}{4 - 3m} = 1$ $\Rightarrow 4m + 3 = 4 - 3m$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$.
The equation is $y - 2 = \frac{1}{7}(x - 1)$ $\Rightarrow 7y - 14 = x - 1$ $\Rightarrow x - 7y + 13 = 0$.
Case $2$: $\frac{4m + 3}{4 - 3m} = -1$ $\Rightarrow 4m + 3 = -4 + 3m$ $\Rightarrow m = -7$.
The equation is $y - 2 = -7(x - 1)$ $\Rightarrow y - 2 = -7x + 7$ $\Rightarrow 7x + y = 9$.
Thus,the lines are $7x + y = 9$ and $x - 7y + 13 = 0$.

(A) Given lines are $y-3kx+4=0$ $(i)$ and $(2k-1)x-(8k-1)y-6=0$ (ii).
The slope of line $(i)$ is $m_1 = \frac{-(-3k)}{1} = 3k$.
The slope of line (ii) is $m_2 = \frac{-(2k-1)}{-(8k-1)} = \frac{2k-1}{8k-1}$.
Since the lines are perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 m_2 = -1$.
$3k \times \frac{2k-1}{8k-1} = -1$
$3k(2k-1) = -(8k-1)$
$6k^2 - 3k = -8k + 1$
$6k^2 + 5k - 1 = 0$
$6k^2 + 6k - k - 1 = 0$
$6k(k+1) - 1(k+1) = 0$
$(6k-1)(k+1) = 0$
Therefore,$k = 1/6$ or $k = -1$.

(B) Given line: $\sqrt{3}x + y = 1$,which can be written as $y = -\sqrt{3}x + 1$. The slope $m_2 = -\sqrt{3}$.
Let the slope of the required line be $m_1$. The angle between the lines is $\theta = 60^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\tan 60^{\circ} = \left| \frac{m_1 - (-\sqrt{3})}{1 + m_1(-\sqrt{3})} \right|$
$\sqrt{3} = \left| \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} \right|$
This gives two cases:
Case $1$: $\sqrt{3} = \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} \implies \sqrt{3} - 3m_1 = m_1 + \sqrt{3} \implies 4m_1 = 0 \implies m_1 = 0$.
The equation of the line passing through $(3, -2)$ with slope $0$ is $y - (-2) = 0(x - 3) \implies y + 2 = 0$.
Case $2$: $-\sqrt{3} = \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} \implies -\sqrt{3} + 3m_1 = m_1 + \sqrt{3} \implies 2m_1 = 2\sqrt{3} \implies m_1 = \sqrt{3}$.
The equation of the line passing through $(3, -2)$ with slope $\sqrt{3}$ is $y - (-2) = \sqrt{3}(x - 3) \implies y + 2 = x\sqrt{3} - 3\sqrt{3} \implies y - x\sqrt{3} + 2 + 3\sqrt{3} = 0$.
Comparing with the given options,the correct equation is $y - x\sqrt{3} + 2 + 3\sqrt{3} = 0$.

(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 $\tan \theta = \left| \frac{m - m_1}{1 + m m_1} \right|$,we have:
$\tan 45^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} \right|$
$1 = \left| \frac{m+1}{1-m} \right|$
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 $y-1=0$.