Mix Examples-Straight Line Questions in English

(C) The original line is $x-2y-4=0$. The slope of this line is $m_1 = 1/2$.
When a line is shifted parallel to itself,its slope remains unchanged. Therefore,the slope of the line after shifting remains $m_1 = 1/2 = \tan \alpha$,where $\alpha = \tan^{-1}(1/2)$.
After rotating the line by an angle of $30^{\circ}$ in the anti-clockwise direction,the new slope $m$ is given by the formula for the tangent of the sum of angles:
$m = \tan(\alpha + 30^{\circ}) = \frac{\tan \alpha + \tan 30^{\circ}}{1 - \tan \alpha \tan 30^{\circ}}$
Substituting the values $\tan \alpha = 1/2$ and $\tan 30^{\circ} = 1/\sqrt{3}$:
$m = \frac{1/2 + 1/\sqrt{3}}{1 - (1/2)(1/\sqrt{3})} = \frac{(\sqrt{3}+2)/(2\sqrt{3})}{(\sqrt{3}-1)/(2\sqrt{3})} = \frac{\sqrt{3}+2}{\sqrt{3}-1}$
Rationalizing the denominator:
$m = \frac{(\sqrt{3}+2)(\sqrt{3}+1)}{3-1} = \frac{3+\sqrt{3}+2\sqrt{3}+2}{2} = \frac{5+3\sqrt{3}}{2}$
Using $\sqrt{3} \approx 1.732$:
$m \approx \frac{5+3(1.732)}{2} = \frac{5+5.196}{2} = \frac{10.196}{2} = 5.098$
The integral part of $m$ is $5$.

(C) Given $A=(-5,-4)$ and the slope of line $L$ is $\tan \theta$. The parametric form of the line is $x = -5 + r \cos \theta$ and $y = -4 + r \sin \theta$.
Point $B$ lies on $x+3y+2=0$,so $(-5+r_1 \cos \theta) + 3(-4+r_1 \sin \theta) + 2 = 0$.
$-5 + r_1 \cos \theta - 12 + 3r_1 \sin \theta + 2 = 0 \implies r_1(\cos \theta + 3 \sin \theta) = 15 \implies AB = r_1 = \frac{15}{\cos \theta + 3 \sin \theta}$.
Point $C$ lies on $2x+y+4=0$,so $2(-5+r_2 \cos \theta) + (-4+r_2 \sin \theta) + 4 = 0$.
$-10 + 2r_2 \cos \theta - 4 + r_2 \sin \theta + 4 = 0 \implies r_2(2 \cos \theta + \sin \theta) = 10 \implies AC = r_2 = \frac{10}{2 \cos \theta + \sin \theta}$.
Substituting into the given equation $\frac{100}{AC^2} - \frac{225}{AB^2} = 4 \cos 2\theta + \sin 2\theta$:
$(2 \cos \theta + \sin \theta)^2 - (\cos \theta + 3 \sin \theta)^2 = 4 \cos 2\theta + \sin 2\theta$.
$(4 \cos^2 \theta + \sin^2 \theta + 4 \sin \theta \cos \theta) - (\cos^2 \theta + 9 \sin^2 \theta + 6 \sin \theta \cos \theta) = 4(\cos^2 \theta - \sin^2 \theta) + 2 \sin \theta \cos \theta$.
$3 \cos^2 \theta - 8 \sin^2 \theta - 2 \sin \theta \cos \theta = 4 \cos^2 \theta - 4 \sin^2 \theta + 2 \sin \theta \cos \theta$.
$4 \sin^2 \theta + 4 \sin \theta \cos \theta + \cos^2 \theta = 0$.
$(2 \sin \theta + \cos \theta)^2 = 0 \implies 2 \sin \theta = -\cos \theta \implies \tan \theta = -\frac{1}{2}$.

(C) Given equations are $x^2-3|x|+2=0$ and $y^2-3y+2=0$.
Solving $x^2-3|x|+2=0$: $(|x|-2)(|x|-1)=0$,so $|x|=2$ or $|x|=1$,which gives $x=\pm 2, \pm 1$.
Solving $y^2-3y+2=0$: $(y-2)(y-1)=0$,so $y=2, 1$.
The lines form two squares with vertices:
Square $1$: $A(-1, 1), B(-1, 2), C(-2, 2), D(-2, 1)$.
Square $2$: $A'(2, 1), B'(2, 2), C'(1, 2), D'(1, 1)$.
The diagonals of these squares are lines like $x+y=k$ and $x-y=k$.
For the square $ABCD$ formed by these diagonals,the sides are parallel to the axes or at $45^\circ$.
Checking the options,the equations $x+y=3$ and $x-y=-3$ represent lines that can form the sides of such a square.
Thus,the correct option is $C$.

(B) Given lines are:
$2x + 3y - 1 = 0$ $(i)$
$3x + 4y - 6 = 0$ $(ii)$
Multiplying $(i)$ by $3$ and $(ii)$ by $2$:
$6x + 9y - 3 = 0$
$6x + 8y - 12 = 0$
Subtracting the equations: $y + 9 = 0 \Rightarrow y = -9$.
Substituting $y = -9$ in $(i)$: $2x + 3(-9) - 1 = 0$ $\Rightarrow 2x - 27 - 1 = 0$ $\Rightarrow 2x = 28$ $\Rightarrow x = 14$.
The point of intersection is $(14, -9)$.
The line perpendicular to $5x - 2y = 7$ is of the form $2x + 5y + k = 0$.
Since it passes through $(14, -9)$:
$2(14) + 5(-9) + k = 0$
$28 - 45 + k = 0$
$-17 + k = 0 \Rightarrow k = 17$.
Thus,the equation is $2x + 5y + 17 = 0$.

(C) First,find the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$.
Multiplying the second equation by $3$,we get $9x + 6y + 15 = 0$.
Adding this to the first equation: $(5x - 6y - 1) + (9x + 6y + 15) = 0$ $\Rightarrow 14x + 14 = 0$ $\Rightarrow x = -1$.
Substituting $x = -1$ into $3x + 2y + 5 = 0$: $3(-1) + 2y + 5 = 0$ $\Rightarrow -3 + 2y + 5 = 0$ $\Rightarrow 2y = -2$ $\Rightarrow y = -1$.
The point of intersection is $(-1, -1)$.
The slope of the line $3x - 5y + 11 = 0$ is $m_1 = \frac{3}{5}$.
The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{5}{3}$.
The equation of the line passing through $(-1, -1)$ with slope $m_2 = -\frac{5}{3}$ is:
$(y - (-1)) = -\frac{5}{3}(x - (-1))
$ $\Rightarrow 3(y + 1) = -5(x + 1)
$ $\Rightarrow 3y + 3 = -5x - 5
$ $\Rightarrow 5x + 3y + 8 = 0$.

(A) The given line is $5x - 2y = 7$. The slope of this line is $m_1 = \frac{5}{2}$.
Any line perpendicular to this line will have a slope $m_2 = -\frac{2}{5}$.
Thus,the equation of the required line is of the form $2x + 5y = \lambda$.
Now,we find the point of intersection of the lines $2x + 3y = 1$ $(i)$ and $3x + 4y = 6$ (ii).
Multiplying $(i)$ by $3$ and (ii) by $2$,we get:
$6x + 9y = 3$
$6x + 8y = 12$
Subtracting the two equations,we get $y = -9$.
Substituting $y = -9$ into $(i)$: $2x + 3(-9) = 1$ $\Rightarrow 2x - 27 = 1$ $\Rightarrow 2x = 28$ $\Rightarrow x = 14$.
The point of intersection is $(14, -9)$.
Since the required line passes through $(14, -9)$,we substitute these values into $2x + 5y = \lambda$:
$2(14) + 5(-9) = \lambda$
$28 - 45 = \lambda$
$\lambda = -17$.
Therefore,the equation of the line is $2x + 5y = -17$,which can be written as $2x + 5y + 17 = 0$.

(D) The equation of the line is $3x - 4y + 5 = 0$,which can be written as $3x - 4y = -5$. Dividing by $-5$,we get $-\frac{3}{5}x + \frac{4}{5}y = 1$.
Comparing this with the normal form $x \cos \alpha + y \sin \alpha = p$,we have $\cos \alpha = -\frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$.
Thus,$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4/5}{-3/5} = -\frac{4}{3}$.
The line $ax + by = 1$ passes through $(1, -1)$ and has slope $m = \tan \alpha = -\frac{4}{3}$.
The equation of the line is $y - (-1) = -\frac{4}{3}(x - 1)$.
$3(y + 1) = -4(x - 1)$ $\Rightarrow 3y + 3 = -4x + 4$ $\Rightarrow 4x + 3y = 1$.
Comparing $4x + 3y = 1$ with $ax + by = 1$,we get $a = 4$ and $b = 3$.
Therefore,$a + ab + b = 4 + (4 \times 3) + 3 = 4 + 12 + 3 = 19$.

(D) The point of intersection $P$ of lines $L_1: x-y-7=0$ and $L_2: x+y-5=0$ is $(6, -1)$.
For point $A(x_1, y_1)$ on $L_1$ with $PA=3\sqrt{2}$,the symmetric form is $\frac{x-6}{\cos \theta_1} = \frac{y+1}{\sin \theta_1} = \pm 3\sqrt{2}$,where $\tan \theta_1 = 1$ (slope of $L_1$ is $1$).
Thus,$x_1 = 6 \pm 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 6 \pm 3$ and $y_1 = -1 \pm 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = -1 \pm 3$.
Since $x_1, y_1 \geq 0$,we choose the positive sign: $A = (9, 2)$.
For point $B(x_2, y_2)$ on $L_2$ with $PB=\sqrt{2}$,the symmetric form is $\frac{x-6}{\cos \theta_2} = \frac{y+1}{\sin \theta_2} = \pm \sqrt{2}$,where $\tan \theta_2 = -1$ (slope of $L_2$ is $-1$).
Thus,$x_2 = 6 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 6 \pm 1$ and $y_2 = -1 \pm \sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) = -1 \mp 1$.
Since $x_2, y_2 \geq 0$,we choose $x_2 = 6-1=5$ and $y_2 = -1+1=0$,so $B = (5, 0)$.
Let $O$ be the origin $(0, 0)$. The vectors are $\vec{OA} = (9, 2)$ and $\vec{OB} = (5, 0)$.
The angle $\theta$ at the origin is given by $\cos \theta = \frac{\vec{OA} \cdot \vec{OB}}{|\vec{OA}| |\vec{OB}|} = \frac{9 \cdot 5 + 2 \cdot 0}{\sqrt{9^2+2^2} \sqrt{5^2+0^2}} = \frac{45}{\sqrt{85} \cdot 5} = \frac{9}{\sqrt{85}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{9}{\sqrt{85}}\right)$.

(A) The slope of the line $3x - 4y + k = 0$ is $m = \frac{3}{4}$.
Since $L_1$ is perpendicular to this line,its slope is $m_1 = -\frac{4}{3}$.
The equation of line $L_1$ passing through $P(4, -3)$ is $y - (-3) = -\frac{4}{3}(x - 4)$,which simplifies to $3y + 9 = -4x + 16$,or $4x + 3y = 7$.
We need the distance of $P(4, -3)$ from the line $L: 5x - 3y - 2 = 0$ measured along $L_1$. This is the distance between $P(4, -3)$ and the intersection point of $L_1$ and $L$.
Solving the system:
$4x + 3y = 7$
$5x - 3y = 2$
Adding the equations: $9x = 9 \Rightarrow x = 1$.
Substituting $x = 1$ into $4x + 3y = 7$: $4(1) + 3y = 7$ $\Rightarrow 3y = 3$ $\Rightarrow y = 1$.
The intersection point is $(1, 1)$.
The distance $d$ between $P(4, -3)$ and $(1, 1)$ is $\sqrt{(4 - 1)^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(B) Let the point on the line $4x - y - 2 = 0$ be $P(x, y)$.
Let $A = (-5, 6)$ and $B = (3, 2)$.
Since $P$ is equidistant from $A$ and $B$,we have $AP = PB$,which implies $AP^2 = PB^2$.
Using the distance formula: $(x + 5)^2 + (y - 6)^2 = (x - 3)^2 + (y - 2)^2$.
Expanding both sides: $x^2 + 10x + 25 + y^2 - 12y + 36 = x^2 - 6x + 9 + y^2 - 4y + 4$.
Simplifying: $16x - 8y + 48 = 0$,which reduces to $2x - y + 6 = 0$ (Equation $2$).
We are given the line $4x - y - 2 = 0$ (Equation $1$).
Subtracting Equation $2$ from Equation $1$: $(4x - y - 2) - (2x - y + 6) = 0$.
$2x - 8 = 0 \Rightarrow x = 4$.
Substituting $x = 4$ into $4x - y - 2 = 0$: $4(4) - y - 2 = 0$ $\Rightarrow 16 - y - 2 = 0$ $\Rightarrow y = 14$.
Thus,the required point is $(4, 14)$.

(B) Let the line passing through the origin $O(0,0)$ be $L_1$. It intersects $L_2: 10x - 8y - 10 = 0$ (which simplifies to $5x - 4y - 5 = 0$) at point $P$ at right angles.
It also intersects $L_3: \frac{x}{4} - \frac{y}{5} + 1 = 0$ (which simplifies to $5x - 4y + 20 = 0$) at point $Q$ at right angles.
Since $L_2$ and $L_3$ have the same slope,they are parallel lines.
The perpendicular distance from the origin $O(0,0)$ to line $L_2$ is $OP = \frac{|5(0) - 4(0) - 5|}{\sqrt{5^2 + (-4)^2}} = \frac{5}{\sqrt{41}}$.
The perpendicular distance from the origin $O(0,0)$ to line $L_3$ is $OQ = \frac{|5(0) - 4(0) + 20|}{\sqrt{5^2 + (-4)^2}} = \frac{20}{\sqrt{41}}$.
Since $P$ and $Q$ lie on opposite sides of the origin (as the constant terms $-5$ and $+20$ have opposite signs),the origin $O$ divides the segment $PQ$ internally in the ratio $OP : OQ = \frac{5}{\sqrt{41}} : \frac{20}{\sqrt{41}} = 1 : 4$.

(B) The vertices are $O(0,0), A(-3,-1)$,and $B(-1,-3)$.
Slope of $OB = \frac{-3-0}{-1-0} = 3$.
Equation of line $OB$ is $y = 3x$,or $3x - y = 0$.
Since $AD \perp OB$,the slope of $AD$ is $-\frac{1}{3}$.
Equation of line $AD$ passing through $A(-3,-1)$ is $y - (-1) = -\frac{1}{3}(x - (-3))$,which simplifies to $y + 1 = -\frac{1}{3}(x + 3)$,or $x + 3y + 6 = 0$.
To find $D$,the intersection of $OB$ $(y=3x)$ and $AD$ $(x+3y+6=0)$:
$x + 3(3x) + 6 = 0$ $\Rightarrow 10x = -6$ $\Rightarrow x = -\frac{3}{5}$.
$y = 3(-\frac{3}{5}) = -\frac{9}{5}$. So,$D = (-\frac{3}{5}, -\frac{9}{5})$.
Point $P$ divides $AD$ in the ratio $3:4$. Using the section formula:
$P = \left( \frac{3(-\frac{3}{5}) + 4(-3)}{3+4}, \frac{3(-\frac{9}{5}) + 4(-1)}{3+4} \right) = \left( \frac{-\frac{9}{5} - 12}{7}, \frac{-\frac{27}{5} - 4}{7} \right) = \left( \frac{-69}{35}, \frac{-47}{35} \right)$.
Line $L$ is parallel to $OB$ $(y=3x)$,so its equation is $y = 3x + k$.
Since $L$ passes through $P(-\frac{69}{35}, -\frac{47}{35})$:
$-\frac{47}{35} = 3(-\frac{69}{35}) + k \Rightarrow k = \frac{-47 + 207}{35} = \frac{160}{35} = \frac{32}{7}$.
Equation $L: y = 3x + \frac{32}{7}$ $\Rightarrow 7y = 21x + 32$ $\Rightarrow 21x - 7y + 32 = 0$.

(C) The point $(4,1)$ undergoes the following transformations:
$(i)$ Reflection in the line $x-y=0$: The reflection of a point $(x,y)$ in the line $y=x$ (or $x-y=0$) is $(y,x)$. Thus,the point $(4,1)$ becomes $(1,4)$.
(ii) Shifting through a distance of $2$ units along the positive $X$-axis: Adding $2$ to the $x$-coordinate,we get $(1+2, 4) = (3,4)$.
(iii) Projection on the $X$-axis: The projection of a point $(x,y)$ on the $X$-axis is $(x,0)$. Thus,the point $(3,4)$ becomes $(3,0)$.
Therefore,the final coordinates are $(3,0)$.

(D) The equation of the family of lines passing through the intersection of $x+2y-19=0$ and $x-2y-3=0$ is given by $(x+2y-19) + \lambda(x-2y-3) = 0$.
This simplifies to $(1+\lambda)x + (2-2\lambda)y - (19+3\lambda) = 0$.
The perpendicular distance from the point $(-2,4)$ to this line is $5$ units.
Using the distance formula $d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$,we have $5 = \frac{|(1+\lambda)(-2) + (2-2\lambda)(4) - (19+3\lambda)|}{\sqrt{(1+\lambda)^2 + (2-2\lambda)^2}}$.
$5 = \frac{|-2-2\lambda + 8-8\lambda - 19-3\lambda|}{\sqrt{1+2\lambda+\lambda^2 + 4-8\lambda+4\lambda^2}}$.
$5 = \frac{|-13\lambda - 13|}{\sqrt{5\lambda^2-6\lambda+5}}$.
Squaring both sides: $25(5\lambda^2-6\lambda+5) = (-13\lambda-13)^2$.
$125\lambda^2 - 150\lambda + 125 = 169\lambda^2 + 338\lambda + 169$.
$44\lambda^2 + 488\lambda + 44 = 0 \Rightarrow 11\lambda^2 + 122\lambda + 11 = 0$.
$(11\lambda+1)(\lambda+11) = 0$,so $\lambda = -1/11$ or $\lambda = -11$.
For $\lambda = -1/11$: $(1-1/11)x + (2+2/11)y - (19-3/11) = 0$ $\Rightarrow (10/11)x + (24/11)y - (206/11) = 0$ $\Rightarrow 5x+12y-103=0$.
Here $b=12, c=-103$,so $5+b+c = 5+12-103 = -86$.
For $\lambda = -11$: $(1-11)x + (2+22)y - (19-33) = 0$ $\Rightarrow -10x+24y+14=0$ $\Rightarrow 5x-12y-7=0$.
Here $b=-12, c=-7$,so $5+b+c = 5-12-7 = -14$.

(B) Let the point be $P(x, y)$ on the line $3x + 4y = 5$.
Since $P$ is equidistant from $A(1, 2)$ and $B(3, 4)$,we have $PA^2 = PB^2$.
$(x - 1)^2 + (y - 2)^2 = (x - 3)^2 + (y - 4)^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16$
$-2x - 4y + 5 = -6x - 8y + 25$
$4x + 4y = 20$
$x + y = 5$
Now,solve the system of equations:
$3x + 4y = 5$
$x + y = 5 \Rightarrow x = 5 - y$
Substitute $x$ in the first equation:
$3(5 - y) + 4y = 5$
$15 - 3y + 4y = 5$
$y = -10$
$x = 5 - (-10) = 15$
Thus,the point is $(15, -10)$.

(A) The line $ax + 2y = k$ is perpendicular to $2x - 3y + 7 = 0$.
The slope of $ax + 2y = k$ is $m_1 = -a/2$.
The slope of $2x - 3y + 7 = 0$ is $m_2 = 2/3$.
Since they are perpendicular,$m_1 \times m_2 = -1$.
$(-a/2) \times (2/3) = -1 \implies -a/3 = -1 \implies a = 3$.
The equation of the line is $3x + 2y = k$.
Converting to intercept form: $\frac{x}{k/3} + \frac{y}{k/2} = 1$.
The area of the triangle formed with coordinate axes is $\frac{1}{2} \times |\frac{k}{3}| \times |\frac{k}{2}| = 3$.
$\frac{|k^2|}{12} = 3 \implies |k^2| = 36 \implies k^2 = 36 \implies k = \pm 6$.
The product of all possible values of $k$ is $6 \times (-6) = -36$.

(A) The sides of the rhombus are parallel to $x+y+c_1=0$ and $7x-y+c_2=0$.
Since the centre is $(1,3)$,the diagonals bisect the angles between the sides.
The lines passing through the centre $(1,3)$ parallel to the sides are $x+y-4=0$ and $7x-y-4=0$.
The diagonals of a rhombus are perpendicular bisectors of each other.
The slopes of the sides are $m_1 = -1$ and $m_2 = 7$.
The slopes of the diagonals are $m_d = \frac{1-m_1m_2 \pm \sqrt{(1-m_1^2)(1-m_2^2)}}{m_1+m_2}$ is not applicable here; rather,the diagonals bisect the angles between the sides.
The angle bisectors of $x+y-4=0$ and $7x-y-4=0$ are $\frac{x+y-4}{\sqrt{2}} = \pm \frac{7x-y-4}{\sqrt{50}}$.
Simplifying,$5(x+y-4) = \pm (7x-y-4)$.
Case $1$: $5x+5y-20 = 7x-y-4 \implies 2x-6y+16=0 \implies x-3y+8=0$.
Case $2$: $5x+5y-20 = -7x+y+4 \implies 12x+4y-24=0 \implies 3x+y-6=0$.
Vertex $A(\alpha, \beta)$ lies on $15x-5y=6$ and on one of the diagonals.
Solving $3x+y=6$ and $15x-5y=6$: $15x+5y=30$ and $15x-5y=6 \implies 30x=36 \implies x=1.2, y=2.4$. $\alpha+\beta = 3.6 = \frac{18}{5}$.
Solving $x-3y=-8$ and $15x-5y=6$: $5x-15y=-40$ and $45x-15y=18 \implies 40x=58 \implies x=1.45, y=3.15$. $\alpha+\beta = 4.6 = \frac{23}{5}$ (not in options).
Thus,the correct value is $\frac{18}{5}$.

(B) Let the line pass through $A(1, 5)$ with slope $m = \tan \theta$. The equation of the line is $\frac{y-5}{x-1} = m$,or $y - 5 = m(x - 1)$.
Let the line intersect $5x - y - 4 = 0$ at $P_1(x_1, y_1)$ and $3x + 4y - 4 = 0$ at $P_2(x_2, y_2)$.
Since $(1, 5)$ is the midpoint,$P_1 = (1+r\cos\theta, 5+r\sin\theta)$ and $P_2 = (1-r\cos\theta, 5-r\sin\theta)$ for some distance $r$.
Substituting $P_1$ into $5x - y - 4 = 0$: $5(1+r\cos\theta) - (5+r\sin\theta) - 4 = 0$ $\Rightarrow 5r\cos\theta - r\sin\theta = 4$ $\Rightarrow r = \frac{4}{5\cos\theta - \sin\theta}$.
Substituting $P_2$ into $3x + 4y - 4 = 0$: $3(1-r\cos\theta) + 4(5-r\sin\theta) - 4 = 0$ $\Rightarrow 3 - 3r\cos\theta + 20 - 4r\sin\theta - 4 = 0$ $\Rightarrow 19 = r(3\cos\theta + 4\sin\theta)$ $\Rightarrow r = \frac{19}{3\cos\theta + 4\sin\theta}$.
Equating the two expressions for $r$: $\frac{4}{5\cos\theta - \sin\theta} = \frac{19}{3\cos\theta + 4\sin\theta}$.
$12\cos\theta + 16\sin\theta = 95\cos\theta - 19\sin\theta$.
$35\sin\theta = 83\cos\theta \Rightarrow \tan\theta = \frac{83}{35}$.
The equation of the line is $y - 5 = \frac{83}{35}(x - 1)$.
$35y - 175 = 83x - 83 \Rightarrow 83x - 35y + 92 = 0$.

(C) Let the point be $P(h, k)$.
Given the distance from the line $x-2y+1=0$ is $\sqrt{5}$,we have:
$\left|\frac{h-2k+1}{\sqrt{1^2+(-2)^2}}\right| = \sqrt{5} \Rightarrow |h-2k+1| = \sqrt{5} \times \sqrt{5} = 5$.
This gives two parallel lines: $h-2k+1 = 5$ or $h-2k+1 = -5$.
Given the distance from the line $2x+3y-1=0$ is $\sqrt{13}$,we have:
$\left|\frac{2h+3k-1}{\sqrt{2^2+3^2}}\right| = \sqrt{13} \Rightarrow |2h+3k-1| = \sqrt{13} \times \sqrt{13} = 13$.
This gives two parallel lines: $2h+3k-1 = 13$ or $2h+3k-1 = -13$.
Each pair of parallel lines intersects the other pair at $2 \times 2 = 4$ distinct points.
Thus,there are $4$ such points.

(D) Solution of statement - $1$:
Let the third vertex be $C(h, k)$.
The orthocentre $O$ is $(0, 0)$.
Since $AO \perp BC$,the slope of $AO$ is $m_{AO} = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$.
The slope of $BC$ is $m_{BC} = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
Since $m_{AO} \cdot m_{BC} = -1$,we have $(-\frac{1}{5}) \cdot (\frac{k - 3}{h + 2}) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ ....$(1)$
Since $BO \perp AC$,the slope of $BO$ is $m_{BO} = \frac{0 - 3}{0 - (-2)} = -\frac{3}{2}$.
The slope of $AC$ is $m_{AC} = \frac{k - (-1)}{h - 5} = \frac{k + 1}{h - 5}$.
Since $m_{BO} \cdot m_{AC} = -1$,we have $(-\frac{3}{2}) \cdot (\frac{k + 1}{h - 5}) = -1$ $\Rightarrow 3(k + 1) = 2(h - 5)$ $\Rightarrow 3k + 3 = 2h - 10$ $\Rightarrow 2h - 3k = 13$ ....$(2)$
Solving equations $(1)$ and $(2)$,we get $h = -4$ and $k = -7$.
So,the third vertex is $(-4, -7)$. Thus,statement $1$ is correct.
Solution of statement - $2$:
Given $2a, b, c$ are in $A.P.$,so $2b = 2a + c \Rightarrow 2a - 2b + c = 0$.
The line equation is $ax + by + c = 0$.
Comparing this with $2a - 2b + c = 0$,we can see that for $x = 2$ and $y = -2$,the equation $a(2) + b(-2) + c = 0$ holds true.
Thus,the lines $ax + by + c = 0$ are concurrent at $(2, -2)$. Thus,statement $2$ is correct.