Problems related to triangle and quadrilateral Questions in English

(B) The given line is $2x - 3y + 7 = 0$.
Any line perpendicular to the given line is of the form $3x + 2y + k = 0$.
To find the intercepts,set $y = 0$: $3x + k = 0 \Rightarrow x = -\frac{k}{3}$.
Set $x = 0$: $2y + k = 0 \Rightarrow y = -\frac{k}{2}$.
The area of the triangle formed with the coordinate axes is given by $\frac{1}{2} |x_{intercept} \cdot y_{intercept}| = 3$.
$\frac{1}{2} |(-\frac{k}{3}) \cdot (-\frac{k}{2})| = 3$.
$\frac{1}{2} |\frac{k^2}{6}| = 3$.
$|k^2| = 36 \Rightarrow k = \pm 6$.
Substituting $k$ into the equation $3x + 2y + k = 0$,we get $3x + 2y = \pm 6$.

(A) Let the vertices of the square be $A(-4, 5)$ and $B(3, 4)$. The diagonal $L$ is $7x - y + 8 = 0$.
Since $A$ and $B$ are vertices of the square,the midpoint of the diagonal $AB$ is $M = (\frac{-4+3}{2}, \frac{5+4}{2}) = (-0.5, 4.5)$.
Check if $M$ lies on $L$: $7(-0.5) - 4.5 + 8 = -3.5 - 4.5 + 8 = 0$. Thus,$AB$ is the other diagonal.
The slope of $AB$ is $m_1 = \frac{4-5}{3-(-4)} = \frac{-1}{7}$.
The slope of diagonal $L$ is $m_2 = 7$. Since $m_1 \times m_2 = -1$,the diagonals are perpendicular,which is consistent with a square.
The vertices on $L$ are equidistant from $M$ and at a distance equal to half the length of diagonal $AB$.
Length $AB = \sqrt{(3 - (-4))^2 + (4 - 5)^2} = \sqrt{7^2 + (-1)^2} = \sqrt{50} = 5\sqrt{2}$.
Half-length $d = \frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}$.
The line $L$ has direction vector $(1, 7)$. The unit vector is $u = \pm(\frac{1}{\sqrt{50}}, \frac{7}{\sqrt{50}}) = \pm(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}})$.
The vertices on $L$ are $M \pm d \cdot u = (-0.5, 4.5) \pm \frac{5}{\sqrt{2}} \cdot (\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}) = (-0.5, 4.5) \pm (0.5, 3.5)$.
Vertex $1: (-0.5 + 0.5, 4.5 + 3.5) = (0, 8)$.
Vertex $2: (-0.5 - 0.5, 4.5 - 3.5) = (-1, 1)$.
Thus,the vertices are $(0, 8)$ and $(-1, 1)$.

(B) The given lines are $L_1: x+2y+3=0$,$L_2: 2x+4y+9=0$,$L_3: x-2y+3=0$,and $L_4: 3x-6y+11=0$.
Notice that $L_1$ and $L_2$ are parallel,and $L_3$ and $L_4$ are parallel.
Let $L_1: x+2y+3=0$ and $L_2: x+2y+4.5=0$. The distance between these parallel lines is $d_1 = \frac{|4.5-3|}{\sqrt{1^2+2^2}} = \frac{1.5}{\sqrt{5}} = \frac{3}{2\sqrt{5}}$.
Let $L_3: x-2y+3=0$ and $L_4: x-2y+\frac{11}{3}=0$. The distance between these parallel lines is $d_2 = \frac{|11/3-3|}{\sqrt{1^2+(-2)^2}} = \frac{2/3}{\sqrt{5}} = \frac{2}{3\sqrt{5}}$.
The area of a parallelogram formed by two pairs of parallel lines $a_1x+b_1y+c_1=0, a_1x+b_1y+c_2=0$ and $a_2x+b_2y+d_1=0, a_2x+b_2y+d_2=0$ is given by $\frac{|c_1-c_2||d_1-d_2|}{|a_1b_2-a_2b_1|}$.
Here,$a_1=1, b_1=2, c_1=3, c_2=4.5$ and $a_2=1, b_2=-2, d_1=3, d_2=11/3$.
The denominator is $|(1)(-2) - (1)(2)| = |-2-2| = 4$.
The area is $\frac{|3-4.5| \times |3-11/3|}{4} = \frac{1.5 \times 2/3}{4} = \frac{1}{4}$.

(D) Let the vertices of the square be $A(4,3)$ and $C(1,-2)$ as the endpoints of a diagonal. The angle between the diagonal and any side of a square is $45^{\circ}$.
Slope of diagonal $AC = \frac{3 - (-2)}{4 - 1} = \frac{5}{3}$.
Let $m$ be the slope of a side. Then,$\tan 45^{\circ} = \left| \frac{m - 5/3}{1 + m(5/3)} \right|$.
$1 = \left| \frac{3m - 5}{3 + 5m} \right|$.
This gives two cases:
Case $1$: $\frac{3m - 5}{3 + 5m} = 1$ $\Rightarrow 3m - 5 = 3 + 5m$ $\Rightarrow 2m = -8$ $\Rightarrow m = -4$.
The equation of the side passing through $(1,-2)$ with slope $-4$ is $y - (-2) = -4(x - 1)$ $\Rightarrow y + 2 = -4x + 4$ $\Rightarrow 4x + y - 2 = 0$.
Case $2$: $\frac{3m - 5}{3 + 5m} = -1$ $\Rightarrow 3m - 5 = -3 - 5m$ $\Rightarrow 8m = 2$ $\Rightarrow m = \frac{1}{4}$.
The equation of the side passing through $(1,-2)$ with slope $\frac{1}{4}$ is $y - (-2) = \frac{1}{4}(x - 1)$ $\Rightarrow 4y + 8 = x - 1$ $\Rightarrow x - 4y - 9 = 0$.
Comparing with the options,$x - 4y - 9 = 0$ is present.

(D) To check if the lines are concurrent,we solve the first two equations for their point of intersection:
$x+y = -4$ $(1)$
$x-2y = 4$ $(2)$
Subtracting $(2)$ from $(1)$: $(x+y) - (x-2y) = -4 - 4 \implies 3y = -8 \implies y = -8/3$.
Substituting $y = -8/3$ into $(1)$: $x - 8/3 = -4 \implies x = -4 + 8/3 = -4/3$.
The intersection point is $(-4/3, -8/3)$.
Now,check if this point satisfies the third equation $3x+4y-2=0$:
$3(-4/3) + 4(-8/3) - 2 = -4 - 32/3 - 2 = -6 - 32/3 = -18/3 - 32/3 = -50/3 \neq 0$.
Since the point does not satisfy the third equation,the lines are not concurrent.
Next,find the slopes of the lines:
$L_1: x+y+4=0 \implies m_1 = -1$.
$L_2: x-2y-4=0 \implies m_2 = 1/2$.
$L_3: 3x+4y-2=0 \implies m_3 = -3/4$.
Since no two slopes are equal,the lines are not parallel.
Since $m_1 \times m_2 = -1/2 \neq -1$,$m_2 \times m_3 = -3/8 \neq -1$,and $m_1 \times m_3 = 3/4 \neq -1$,no two lines are perpendicular.
Calculate the lengths of the sides of the triangle formed by the intersection points:
$P_1 (L_1 \cap L_2) = (-4/3, -8/3)$.
$P_2 (L_2 \cap L_3)$: $x-2y=4$ and $3x+4y=2$. Solving gives $x=2, y=-1$. $P_2 = (2, -1)$.
$P_3 (L_3 \cap L_1)$: $3x+4y=2$ and $x+y=-4$. Solving gives $x=18, y=-22$. $P_3 = (18, -22)$.
Calculate side lengths:
$d_{12} = \sqrt{(2 - (-4/3))^2 + (-1 - (-8/3))^2} = \sqrt{(10/3)^2 + (5/3)^2} = \sqrt{125/9} = 5\sqrt{5}/3$.
$d_{23} = \sqrt{(18 - 2)^2 + (-22 - (-1))^2} = \sqrt{16^2 + (-21)^2} = \sqrt{256 + 441} = \sqrt{697}$.
$d_{31} = \sqrt{(18 - (-4/3))^2 + (-22 - (-8/3))^2} = \sqrt{(58/3)^2 + (-58/3)^2} = 58\sqrt{2}/3$.
Since all side lengths are distinct,the triangle is scalene.

(A) Given vertices are $A(1,7)$,$B(-5,-1)$,and $C(7,4)$.
Let the internal angle bisector of $\angle ABC$ meet the side $AC$ at point $D$.
By the Angle Bisector Theorem,the ratio in which $D$ divides $AC$ is $\frac{AD}{DC} = \frac{BA}{BC}$.
Calculating the lengths of sides $BA$ and $BC$:
$BA = \sqrt{(1 - (-5))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
$BC = \sqrt{(7 - (-5))^2 + (4 - (-1))^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
Thus,$\frac{AD}{DC} = \frac{10}{13}$.
Using the section formula,the coordinates of $D$ are:
$D = \left( \frac{10(7) + 13(1)}{10 + 13}, \frac{10(4) + 13(7)}{10 + 13} \right) = \left( \frac{70 + 13}{23}, \frac{40 + 91}{23} \right) = \left( \frac{83}{23}, \frac{131}{23} \right)$.
The equation of the line passing through $B(-5, -1)$ and $D\left(\frac{83}{23}, \frac{131}{23}\right)$ is:
$y - (-1) = \frac{\frac{131}{23} - (-1)}{\frac{83}{23} - (-5)} (x - (-5))$
$y + 1 = \frac{\frac{131 + 23}{23}}{\frac{83 + 115}{23}} (x + 5)$
$y + 1 = \frac{154}{198} (x + 5)$
$y + 1 = \frac{7}{9} (x + 5)$
$9y + 9 = 7x + 35$
$7x - 9y + 26 = 0$.

(C) Let the sides of the triangle be $a, b, c$. Given $a = 17 \text{ cm}$ and the perimeter $P = a + b + c = 40 \text{ cm}$.
Given the sum of two sides is $a + b = 35 \text{ cm}$.
Substituting $a = 17$, we get $17 + b = 35$, so $b = 18 \text{ cm}$.
Since $a + b + c = 40$, we have $17 + 18 + c = 40$, which gives $c = 5 \text{ cm}$.
The semi-perimeter $s = \frac{40}{2} = 20 \text{ cm}$.
Using Heron's formula, the area $A = \sqrt{s(s-a)(s-b)(s-c)}$.
$A = \sqrt{20(20-17)(20-18)(20-5)} = \sqrt{20 \times 3 \times 2 \times 15}$.
$A = \sqrt{1800} = \sqrt{900 \times 2} = 30 \sqrt{2} \text{ cm}^2$.
Thus, the correct option is $C$.

(D) Let $G$ be the centroid of $\triangle ABC$. Since $AD$ and $BE$ are medians,$G$ divides $AD$ and $BE$ in the ratio $2:1$.
Thus,$AG = \frac{2}{3} AD = \frac{2}{3} \times 4 = \frac{8}{3}$ and $BG = \frac{2}{3} BE$.
In $\triangle ABG$,by the Sine Rule: $\frac{AG}{\sin(\angle ABG)} = \frac{BG}{\sin(\angle BAG)}$.
Given $\angle BAG = \frac{\pi}{6}$ and $\angle ABG = \frac{\pi}{3}$,we have $\frac{8/3}{\sin(\pi/3)} = \frac{BG}{\sin(\pi/6)}$.
$BG = \frac{8}{3} \times \frac{\sin(\pi/6)}{\sin(\pi/3)} = \frac{8}{3} \times \frac{1/2}{\sqrt{3}/2} = \frac{8}{3 \sqrt{3}}$.
Area of $\triangle ABG = \frac{1}{2} \times AG \times BG \times \sin(\angle AGB)$.
Since $\angle AGB = \pi - (\pi/6 + \pi/3) = \pi/2$,$\sin(\angle AGB) = 1$.
Area of $\triangle ABG = \frac{1}{2} \times \frac{8}{3} \times \frac{8}{3 \sqrt{3}} = \frac{32}{9 \sqrt{3}}$.
Since the centroid divides the triangle into three triangles of equal area,Area of $\triangle ABC = 3 \times \text{Area of } \triangle ABG = 3 \times \frac{32}{9 \sqrt{3}} = \frac{32}{3 \sqrt{3}}$.

(B) Given that the perimeter $2s = 16 \text{ cm}$,so the semi-perimeter $s = 8 \text{ cm}$.
Let the sides be $a, b, c$. Given $a = 6 \text{ cm}$ and area $\Delta = 12 \text{ cm}^2$.
Using Heron's formula: $\Delta^2 = s(s-a)(s-b)(s-c)$.
$144 = 8(8-6)(8-b)(8-c)$.
$144 = 8(2)(8-b)(8-c) \Rightarrow 144 = 16(8-b)(8-c)$.
$9 = (8-b)(8-c)$.
Since $a+b+c = 16$ and $a=6$,we have $b+c = 10$,so $c = 10-b$.
Substituting $c$: $9 = (8-b)(8-(10-b)) = (8-b)(b-2)$.
$9 = 8b - 16 - b^2 + 2b \Rightarrow b^2 - 10b + 25 = 0$.
$(b-5)^2 = 0 \Rightarrow b = 5 \text{ cm}$.
Then $c = 10 - 5 = 5 \text{ cm}$.
Since two sides are equal $(b=c=5 \text{ cm})$,the triangle is isosceles.

(D) Let the vertices be $A(a, 1)$,$B(1, b)$,and $D(-1, d)$. Since $AD$ is parallel to $y=2x$,the slope of $AD$ is $2$. Thus,$\frac{d-1}{-1-a} = 2$ $\Rightarrow d-1 = -2-2a$ $\Rightarrow d = -1-2a$. Since $AB$ is perpendicular to $AD$,the slope of $AB$ is $-\frac{1}{2}$. Thus,$\frac{b-1}{1-a} = -\frac{1}{2}$ $\Rightarrow 2b-2 = a-1$ $\Rightarrow b = \frac{a+1}{2}$. Since $ABCD$ is a rectangle,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$. Midpoint of $BD = (\frac{1-1}{2}, \frac{b+d}{2}) = (0, \frac{b+d}{2})$. Midpoint of $AC = (\frac{a+x_c}{2}, \frac{1+y_c}{2})$. Equating these,we get $x_c = -a$ and $y_c = b+d-1 = \frac{a+1}{2} - 1 - 2a - 1 = \frac{a+1-2-4a-2}{2} = \frac{-3a-3}{2}$. Thus,$C$ is $(-a, \frac{-3(a+1)}{2})$. Checking the options,none of the given coordinates match this form for any real $a$.

(B) Given the three sides of a triangle are:
$x+8y-22=0$ $(i)$
$5x+2y-34=0$ $(ii)$
$2x-3y+13=0$ $(iii)$
Solving $(i)$ and $(ii)$:
$x=6, y=2$. Vertex $A = (6, 2)$.
Solving $(ii)$ and $(iii)$:
$x=4, y=7$. Vertex $B = (4, 7)$.
Solving $(i)$ and $(iii)$:
$x=-2, y=3$. Vertex $C = (-2, 3)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
$\text{Area} = \frac{1}{2} |6(7-3) + 4(3-2) + (-2)(2-7)|$
$\text{Area} = \frac{1}{2} |6(4) + 4(1) - 2(-5)|$
$\text{Area} = \frac{1}{2} |24 + 4 + 10|$
$\text{Area} = \frac{1}{2} |38| = 19$ sq units.

(A) The given equations are $y = \sqrt{3}x$,$y = -\sqrt{3}x$,and $y = 1$.
These can be written as $y = \tan(60^{\circ})x$,$y = \tan(120^{\circ})x$,and $y = 1$.
The lines $y = \sqrt{3}x$ and $y = -\sqrt{3}x$ pass through the origin $(0,0)$ and make angles of $60^{\circ}$ and $120^{\circ}$ with the positive $x$-axis,respectively.
The angle between these two lines is $120^{\circ} - 60^{\circ} = 60^{\circ}$.
The line $y = 1$ intersects $y = \sqrt{3}x$ at $(\frac{1}{\sqrt{3}}, 1)$ and $y = -\sqrt{3}x$ at $(-\frac{1}{\sqrt{3}}, 1)$.
The length of the base on the line $y = 1$ is $\frac{1}{\sqrt{3}} - (-\frac{1}{\sqrt{3}}) = \frac{2}{\sqrt{3}}$.
The lengths of the other two sides are $\sqrt{(\frac{1}{\sqrt{3}})^2 + 1^2} = \sqrt{\frac{1}{3} + 1} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Since all three sides are equal to $\frac{2}{\sqrt{3}}$,the triangle is an equilateral triangle.

(A, C) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$C(a \cos(\alpha + 90^{\circ}), a \sin(\alpha + 90^{\circ})) = (-a \sin \alpha, a \cos \alpha)$,and $B(a \cos \alpha - a \sin \alpha, a \sin \alpha + a \cos \alpha)$.
The first diagonal passes through $O(0,0)$ and $B(a(\cos \alpha - \sin \alpha), a(\sin \alpha + \cos \alpha))$.
The slope of this diagonal is $m_1 = \frac{a(\sin \alpha + \cos \alpha)}{a(\cos \alpha - \sin \alpha)} = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha}$.
The equation is $y = m_1 x \Rightarrow y(\cos \alpha - \sin \alpha) = x(\sin \alpha + \cos \alpha)$.
The second diagonal passes through $A(a \cos \alpha, a \sin \alpha)$ and $C(-a \sin \alpha, a \cos \alpha)$.
The slope of this diagonal is $m_2 = \frac{a \cos \alpha - a \sin \alpha}{-a \sin \alpha - a \cos \alpha} = -\frac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}$.
The equation is $y - a \sin \alpha = m_2(x - a \cos \alpha)$.
$(y - a \sin \alpha)(\cos \alpha + \sin \alpha) = -(\cos \alpha - \sin \alpha)(x - a \cos \alpha)$.
$y(\cos \alpha + \sin \alpha) - a \sin \alpha \cos \alpha - a \sin^2 \alpha = -x(\cos \alpha - \sin \alpha) + a \cos^2 \alpha - a \sin \alpha \cos \alpha$.
$y(\cos \alpha + \sin \alpha) + x(\cos \alpha - \sin \alpha) = a(\cos^2 \alpha + \sin^2 \alpha) = a$.

(A) The given lines are $L_1: x+y=0$,$L_2: 5x+y=4$,and $L_3: x+5y=4$.
Solving $L_1$ and $L_2$: $x+y=0 \implies y=-x$. Substituting into $L_2$: $5x-x=4 \implies 4x=4 \implies x=1, y=-1$. So,$A = (1, -1)$.
Solving $L_1$ and $L_3$: $x+y=0 \implies x=-y$. Substituting into $L_3$: $-y+5y=4 \implies 4y=4 \implies y=1, x=-1$. So,$C = (-1, 1)$.
Solving $L_2$ and $L_3$: $5x+y=4$ and $x+5y=4$. Subtracting: $4x-4y=0 \implies x=y$. Substituting into $L_2$: $5x+x=4 \implies 6x=4 \implies x=2/3, y=2/3$. So,$B = (2/3, 2/3)$.
Calculating side lengths:
$AB = \sqrt{(2/3 - 1)^2 + (2/3 + 1)^2} = \sqrt{(-1/3)^2 + (5/3)^2} = \sqrt{1/9 + 25/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.
$BC = \sqrt{(-1 - 2/3)^2 + (1 - 2/3)^2} = \sqrt{(-5/3)^2 + (1/3)^2} = \sqrt{25/9 + 1/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.
$CA = \sqrt{(-1 - 1)^2 + (1 - (-1))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.
Since $AB = BC$,the triangle is an isosceles triangle.

(C) Let the given vertex be $V = (-4, 5)$ and the given diagonal be $L_1: 7x - y + 8 = 0$.
Note that the vertex $V(-4, 5)$ does not lie on the diagonal $L_1$ because $7(-4) - 5 + 8 = -28 - 5 + 8 = -25 \neq 0$.
In a square,the diagonals are perpendicular to each other and bisect each other.
Let the other diagonal be $L_2$. Since $L_2$ passes through the vertex $V(-4, 5)$ and is perpendicular to $L_1$,its slope must be the negative reciprocal of the slope of $L_1$.
The slope of $L_1$ is $m_1 = 7$.
Therefore,the slope of $L_2$ is $m_2 = -1/7$.
The equation of $L_2$ passing through $(-4, 5)$ is:
$y - 5 = -\frac{1}{7}(x + 4)$
$7(y - 5) = -(x + 4)$
$7y - 35 = -x - 4$
$x + 7y = 31$
Thus,the equation of the other diagonal is $x + 7y = 31$ or $7y + x = 31$.

(D) The reflection of point $P(3,6)$ on the line $y=x$ gives the point $Q(6,3)$.
The reflection of point $Q(6,3)$ on the line $y=-x$ gives the point $Q^{\prime}(-3,-6)$.
Now,the slope of $PQ = \frac{3-6}{6-3} = \frac{-3}{3} = -1$.
The slope of $QQ^{\prime} = \frac{-6-3}{-3-6} = \frac{-9}{-9} = 1$.
Since the product of the slopes is $(-1) \times (1) = -1$,the lines $PQ$ and $QQ^{\prime}$ are perpendicular.
Therefore,$\Delta PQQ^{\prime}$ is a right-angled triangle with the right angle at $Q$.
The circumcentre of a right-angled triangle is the midpoint of its hypotenuse.
The hypotenuse is $PQ^{\prime}$,with endpoints $P(3,6)$ and $Q^{\prime}(-3,-6)$.
Midpoint of $PQ^{\prime} = \left(\frac{3+(-3)}{2}, \frac{6+(-6)}{2}\right) = (0,0)$.

(A) Let the four points be $A(-a,-b)$,$B(a, b)$,$C(0,0)$,and $D(a^{2}, ab)$.
To check if $A, B$,and $C$ are collinear,we calculate the determinant:
$\left|\begin{array}{ccc}-a & -b & 1 \\ a & b & 1 \\ 0 & 0 & 1\end{array}\right| = -a(b-0) + b(a-0) + 1(0) = -ab + ab = 0$.
Since the determinant is $0$,points $A, B$,and $C$ are collinear.
Now,check if $B, C$,and $D$ are collinear:
$\left|\begin{array}{ccc}a & b & 1 \\ 0 & 0 & 1 \\ a^{2} & ab & 1\end{array}\right| = a(0-ab) - b(0-a^{2}) + 1(0) = -a^{2}b + a^{2}b = 0$.
Since the determinant is $0$,points $B, C$,and $D$ are collinear.
Since $A, B, C$ are collinear and $B, C, D$ are collinear,all four points $A, B, C$,and $D$ must lie on the same line.

(B) The given equations of the lines are:
$x-y=7$ ...$(i)$
$x+4y=2$ ...(ii)
Solving equations $(i)$ and (ii),we find the intersection point $B(6,-1)$.
Let the slope of line $AC$ be $m$. Since $A(2,-7)$ lies on $x-y=7$,line $AC$ passes through $A(2,-7)$.
The angle between line $AC$ and line $AB$ (which is $x-y=7$,slope $m_1=1$) must be equal to the angle between line $AC$ and line $BC$ (which is $x+4y=2$,slope $m_2=-1/4$) because $AB=AC$ implies $\triangle ABC$ is isosceles with $\angle ABC = \angle ACB$.
Using the formula for the angle between two lines $\tan \theta = \left| \frac{m_1-m_2}{1+m_1m_2} \right|$,we equate the tangents of the angles:
$\left| \frac{m-1}{1+m} \right| = \left| \frac{m-(-1/4)}{1+m(-1/4)} \right| = \left| \frac{4m+1}{4-m} \right|$.
Solving this gives $m=1$ or $m=-23/7$.
For $m=1$,the line equation is $y-(-7)=1(x-2) \Rightarrow x-y-9=0$.
For $m=-23/7$,the line equation is $y-(-7)=-\frac{23}{7}(x-2) \Rightarrow 23x+7y+3=0$.

(B) The side lengths are calculated as follows:
$AB = \sqrt{(5 - (-1))^2 + (1 - (-7))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$
$BC = \sqrt{(1 - 5)^2 + (4 - 1)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = 5$
By the Angle Bisector Theorem,the bisector of $\angle ABC$ divides the opposite side $AC$ in the ratio of the adjacent sides $AB:BC = 10:5 = 2:1$.
Let $P$ be the point on $AC$ that divides it in the ratio $2:1$. Using the section formula:
$P = \left( \frac{2(1) + 1(-1)}{2 + 1}, \frac{2(4) + 1(-7)}{2 + 1} \right) = \left( \frac{2 - 1}{3}, \frac{8 - 7}{3} \right) = \left( \frac{1}{3}, \frac{1}{3} \right)$
The equation of the line passing through $B(5, 1)$ and $P(\frac{1}{3}, \frac{1}{3})$ is given by:
$y - 1 = \frac{\frac{1}{3} - 1}{\frac{1}{3} - 5}(x - 5)$
$y - 1 = \frac{-2/3}{-14/3}(x - 5)$
$y - 1 = \frac{1}{7}(x - 5)$
$7y - 7 = x - 5$
$7y = x + 2$

(C) In a rhombus,the diagonals bisect each other at right angles. Let $O$ be the intersection of diagonals $AC$ and $BD$. The coordinates of $O$ are the midpoint of $AC$:
$O = \left(\frac{1-3}{2}, \frac{2-6}{2}\right) = (-1, -2)$.
Since $O$ is also the midpoint of $BD$,we have:
$\frac{\alpha+\gamma}{2} = -1 \implies \alpha+\gamma = -2$
$\frac{\beta+\delta}{2} = -2 \implies \beta+\delta = -4$
We need to find $|\alpha+\beta+\gamma+\delta|$.
$|\alpha+\beta+\gamma+\delta| = |(\alpha+\gamma) + (\beta+\delta)| = |-2 + (-4)| = |-6| = 6$.

(C) Let the side length of the equilateral triangle $ABC$ be $a$. The distance between the parallel lines $L_1$ and $L_2$ is $6 + 3 = 9$ units.
Let $\theta$ be the angle that the side $BC$ makes with the line $L_2$. Since $C$ lies on $L_2$,the vertical distance from $A$ to $L_2$ is $9$. In the right-angled triangle formed by the projection of $A$ onto $L_2$,we have $\sin \theta = \frac{3}{a}$ and $\sin(60^{\circ} + \theta) = \frac{9}{a}$.
Expanding $\sin(60^{\circ} + \theta) = \sin 60^{\circ} \cos \theta + \cos 60^{\circ} \sin \theta = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{9}{a}$.
Substituting $\sin \theta = \frac{3}{a}$ and $\cos \theta = \sqrt{1 - \frac{9}{a^2}}$,we get $\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \cdot \frac{3}{a} = \frac{9}{a}$.
$\frac{\sqrt{3}}{2} \sqrt{\frac{a^2 - 9}{a^2}} = \frac{9}{a} - \frac{3}{2a} = \frac{15}{2a}$.
$\sqrt{3} \sqrt{a^2 - 9} = 15 \implies 3(a^2 - 9) = 225 \implies a^2 - 9 = 75 \implies a^2 = 84$.
The area of the equilateral triangle is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 84 = 21 \sqrt{3}$ sq. units.

(C) Let the midpoints be $D(\frac{5}{2}, 7)$,$E(\frac{5}{2}, 3)$,and $F(4, 5)$.
The vertices of $\triangle ABC$ are $A, B, C$. Using the property that the vertices of the original triangle are $A = E+F-D$,$B = D+F-E$,and $C = D+E-F$:
$A = (\frac{5}{2}+4-\frac{5}{2}, 3+5-7) = (4, 1)$
$B = (\frac{5}{2}+4-\frac{5}{2}, 7+5-3) = (4, 9)$
$C = (\frac{5}{2}+\frac{5}{2}-4, 7+3-5) = (1, 5)$
The side lengths are $a = BC = \sqrt{(4-1)^2 + (9-5)^2} = \sqrt{3^2+4^2} = 5$,$b = AC = \sqrt{(4-1)^2 + (1-5)^2} = \sqrt{3^2+(-4)^2} = 5$,and $c = AB = \sqrt{(4-4)^2 + (9-1)^2} = 8$.
Since $a=b=5$,the triangle is isosceles. The incentre $(h, k)$ is given by $(\frac{ax_A+bx_B+cx_C}{a+b+c}, \frac{ay_A+by_B+cy_C}{a+b+c})$.
$h = \frac{5(4)+5(4)+8(1)}{5+5+8} = \frac{20+20+8}{18} = \frac{48}{18} = \frac{8}{3}$.
$k = \frac{5(1)+5(9)+8(5)}{5+5+8} = \frac{5+45+40}{18} = \frac{90}{18} = 5$.
Thus,$3h + k = 3(\frac{8}{3}) + 5 = 8 + 5 = 13$.

(D) In an equilateral triangle,the orthocentre coincides with the centroid.
Let $P = (3, 5)$ and the line $QR$ be $x + y - 4 = 0$.
The altitude from $P$ to $QR$ is perpendicular to $x + y = 4$. The slope of $QR$ is $-1$,so the slope of the altitude is $1$.
The equation of the altitude passing through $(3, 5)$ is $y - 5 = 1(x - 3)$,which simplifies to $y = x + 2$.
The foot of the altitude $F$ is the intersection of $x + y = 4$ and $y = x + 2$. Substituting $y$,we get $x + (x + 2) = 4$,so $2x = 2$,$x = 1$. Then $y = 3$. Thus,$F = (1, 3)$.
The centroid $G(\alpha, \beta)$ divides the altitude $PF$ in the ratio $2:1$ from the vertex $P$.
Using the section formula,$G = \left( \frac{2(1) + 1(3)}{2 + 1}, \frac{2(3) + 1(5)}{2 + 1} \right) = \left( \frac{5}{3}, \frac{11}{3} \right)$.
Thus,$\alpha = 5/3$ and $\beta = 11/3$.
Therefore,$9(\alpha + \beta) = 9(5/3 + 11/3) = 9(16/3) = 48$.

(A) The given lines are $x - \sqrt{3}y = \alpha$ (for $y \ge 0$) and $x + \sqrt{3}y = \alpha$ (for $y < 0$).
The point of intersection $P$ is $(\alpha, 0)$.
The angle between the lines is $60^\circ$ because the slopes are $m_1 = 1/\sqrt{3}$ and $m_2 = -1/\sqrt{3}$.
Since $PA = PB = \alpha$ and the angle $\angle APB = 60^\circ$,$\triangle PAB$ is an equilateral triangle with side length $a = \alpha$.
The height $PQ$ of the equilateral triangle is given by $PQ = \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2}\alpha$.
Given $PQ = \frac{9}{2}$,we have $\frac{\sqrt{3}}{2}\alpha = \frac{9}{2}$,which implies $\alpha = 3\sqrt{3}$.
The circumradius $R$ of an equilateral triangle with side $a$ is $R = \frac{a}{\sqrt{3}}$.
Thus,$R = \frac{\alpha}{\sqrt{3}} = \frac{3\sqrt{3}}{\sqrt{3}} = 3$.
Finally,$\frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9$.