Concurrency of three lines Questions in English

(B) The equation of any line passing through the intersection of the given lines is given by $(3x - 4y + 1) + k(5x + y - 1) = 0$.
This simplifies to $(3 + 5k)x + (k - 4)y + (1 - k) = 0$.
The intercepts on the axes are given by $x$-intercept $= -\frac{1 - k}{3 + 5k} = \frac{k - 1}{3 + 5k}$ and $y$-intercept $= -\frac{1 - k}{k - 4} = \frac{k - 1}{k - 4}$.
Since the intercepts are equal,we have $\frac{k - 1}{3 + 5k} = \frac{k - 1}{k - 4}$.
This implies either $k - 1 = 0$ or $\frac{1}{3 + 5k} = \frac{1}{k - 4}$.
If $k = 1$,the equation becomes $8x - 3y = 0$,which passes through the origin (intercepts are zero).
If $k - 4 = 3 + 5k$,then $4k = -7$,so $k = -7/4$.
Substituting $k = -7/4$ into the equation: $4(3x - 4y + 1) - 7(5x + y - 1) = 0$.
$12x - 16y + 4 - 35x - 7y + 7 = 0$.
$-23x - 23y + 11 = 0$,which simplifies to $23x + 23y = 11$.

(B) For three points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be zero.
Alternatively,the slope between any two pairs of points must be equal.
Let the points be $A(a, 0)$,$B(0, b)$,and $C(1, 1)$.
Slope of $AB = \frac{b - 0}{0 - a} = -\frac{b}{a}$.
Slope of $BC = \frac{1 - b}{1 - 0} = 1 - b$.
Equating the slopes: $-\frac{b}{a} = 1 - b$.
$-b = a(1 - b) \implies -b = a - ab$.
Rearranging the terms: $ab = a + b$.
Dividing both sides by $ab$ (assuming $a, b \neq 0$): $1 = \frac{a}{ab} + \frac{b}{ab}$.
Therefore,$1 = \frac{1}{b} + \frac{1}{a}$ or $\frac{1}{a} + \frac{1}{b} = 1$.

(A) The family of lines passing through the intersection of $ax + by + c = 0$ and $a'x + b'y + c' = 0$ is given by $(ax + by + c) + \lambda(a'x + b'y + c') = 0$.
This can be rewritten as $x(a + \lambda a') + y(b + \lambda b') + (c + \lambda c') = 0$.
Since the line is parallel to the $y$-axis,the coefficient of $y$ must be zero.
Therefore,$b + \lambda b' = 0$,which gives $\lambda = -\frac{b}{b'}$.
Substituting this value of $\lambda$ into the equation:
$x(a - \frac{b}{b'}a') + y(b - \frac{b}{b'}b') + (c - \frac{b}{b'}c') = 0$.
Multiplying by $b'$:
$x(ab' - a'b) + y(bb' - bb') + (cb' - bc') = 0$.
$x(ab' - a'b) + (cb' - bc') = 0$.

(B) The given equation is $(2 \cos \theta + 3 \sin \theta) x + (3 \cos \theta - 5 \sin \theta) y - (5 \cos \theta - 2 \sin \theta) = 0$.
Rearranging the terms by grouping $\cos \theta$ and $\sin \theta$:
$\cos \theta (2x + 3y - 5) + \sin \theta (3x - 5y + 2) = 0$.
For this equation to hold for all values of $\theta$,the coefficients of $\cos \theta$ and $\sin \theta$ must be zero independently:
$2x + 3y - 5 = 0$ (Equation $1$)
$3x - 5y + 2 = 0$ (Equation $2$)
Multiplying Equation $1$ by $5$ and Equation $2$ by $3$:
$10x + 15y - 25 = 0$
$9x - 15y + 6 = 0$
Adding these two equations:
$19x - 19 = 0 \implies x = 1$.
Substituting $x = 1$ into Equation $1$:
$2(1) + 3y - 5 = 0 \implies 3y = 3 \implies y = 1$.
Thus,the fixed point is $(1, 1)$.

(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$a(3 - 4) - 2(b - c) + 1(4b - 3c) = 0$
$-a - 2b + 2c + 4b - 3c = 0$
$-a + 2b - c = 0$
$2b = a + c$
This condition implies that $a, b, c$ are in Arithmetic Progression.

(A) Step $1$: Find the intersection point of $y = x + 7$ and $x + 2y + 1 = 0$.
Substitute $y = x + 7$ into $x + 2y + 1 = 0$:
$x + 2(x + 7) + 1 = 0$
$x + 2x + 14 + 1 = 0$
$3x = -15$
$x = -5$.
Then $y = -5 + 7 = 2$.
The intersection point is $(-5, 2)$.
Step $2$: Find the slope of the line perpendicular to $5x - 2y + 7 = 0$.
The slope of $5x - 2y + 7 = 0$ is $m_1 = -\frac{5}{-2} = \frac{5}{2}$.
The slope of the perpendicular line is $m_2 = -\frac{1}{m_1} = -\frac{2}{5}$.
Step $3$: Use the point-slope form $y - y_1 = m(x - x_1)$ with point $(-5, 2)$ and slope $m = -\frac{2}{5}$.
$y - 2 = -\frac{2}{5}(x + 5)$
$5(y - 2) = -2(x + 5)$
$5y - 10 = -2x - 10$
$2x + 5y = 0$.

(A) The equation of a line passing through the intersection of two lines $L_1: x + y - 2 = 0$ and $L_2: 2x - y + 1 = 0$ is given by $L_1 + \lambda L_2 = 0$.
$(x + y - 2) + \lambda(2x - y + 1) = 0$.
Since the line passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation:
$(0 + 0 - 2) + \lambda(0 - 0 + 1) = 0$
$-2 + \lambda = 0 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ back into the equation:
$(x + y - 2) + 2(2x - y + 1) = 0$
$x + y - 2 + 4x - 2y + 2 = 0$
$5x - y = 0$.

(A) Let the point of intersection in the fourth quadrant be $(\alpha, -\alpha)$,where $\alpha > 0$.
Since this point lies on both lines $4ax + 2ay + c = 0$ and $5bx + 2by + d = 0$,we substitute the coordinates:
For the first line: $4a(\alpha) + 2a(-\alpha) + c = 0$ $\Rightarrow 2a\alpha + c = 0$ $\Rightarrow \alpha = -\frac{c}{2a}$.
For the second line: $5b(\alpha) + 2b(-\alpha) + d = 0$ $\Rightarrow 3b\alpha + d = 0$ $\Rightarrow \alpha = -\frac{d}{3b}$.
Equating the two expressions for $\alpha$:
$-\frac{c}{2a} = -\frac{d}{3b}$.
Cross-multiplying gives $3bc = 2ad$,which can be written as $3bc - 2ad = 0$.

(B) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$.
Rearranging the terms to separate $k$:
$2x + kx + y + ky = 5 + 7k$
$(2x + y - 5) + k(x + y - 7) = 0$.
This represents a family of lines passing through the intersection of the lines $2x + y - 5 = 0$ and $x + y - 7 = 0$.
Solving these two equations:
$2x + y = 5$
$x + y = 7$
Subtracting the second from the first: $(2x - x) + (y - y) = 5 - 7$,which gives $x = -2$.
Substituting $x = -2$ into $x + y = 7$: $-2 + y = 7$,so $y = 9$.
Thus,all lines pass through the point $(-2, 9)$.

(D) Given the equation of the line is $ax + by + c = 0$.
We are given the condition $3a + 2b + 4c = 0$.
Dividing the condition by $4$,we get $\frac{3}{4}a + \frac{2}{4}b + c = 0$,which simplifies to $\frac{3}{4}a + \frac{1}{2}b + c = 0$.
Comparing this with the equation $ax + by + c = 0$,we can see that the lines pass through the fixed point $(x, y) = (3/4, 1/2)$ for all values of $a, b, c$.
Thus,the lines are concurrent at the point $(3/4, 1/2)$.

(C) For a system of linear equations $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given equations are $a^2 x + (2 - a) y = 4 + a^2$ and $a x + (2 a - 1) y = a^5 - 2$.
Applying the condition: $\frac{a^2}{a} = \frac{2 - a}{2a - 1} \neq \frac{4 + a^2}{a^5 - 2}$.
From $\frac{a^2}{a} = \frac{2 - a}{2a - 1}$ (assuming $a \neq 0$): $a = \frac{2 - a}{2a - 1} \implies 2a^2 - a = 2 - a \implies 2a^2 = 2 \implies a^2 = 1 \implies a = 1$ or $a = -1$.
Case $1$: If $a = 1$,the ratio is $\frac{1}{1} = \frac{1}{1} \neq \frac{5}{-1}$,which holds true.
Case $2$: If $a = -1$,the ratio is $\frac{1}{-1} = \frac{3}{-3} \neq \frac{5}{-3}$,which holds true.
If $a = 0$,the equations become $2y = 4$ and $-y = -2$,which gives $y = 2$,so there is a solution. Thus $a=0$ is not a solution.
Therefore,there are $2$ values of $a$ for which the system has no solution.

(C) Let the $k^{th}$ term of the $H.P.$ be $x_k$. By definition,$\frac{1}{x_k}$ is the $k^{th}$ term of an $A.P.$
Let $a_k = \frac{1}{x_k} = a + (k-1)d$,where $a$ is the first term and $d$ is the common difference.
The coordinates of the points are $P(a_p, p)$,$Q(a_q, q)$,and $R(a_r, r)$.
The slope of line segment $PQ$ is $m_{PQ} = \frac{q - p}{a_q - a_p} = \frac{q - p}{(a + (q-1)d) - (a + (p-1)d)} = \frac{q - p}{(q - p)d} = \frac{1}{d}$.
The slope of line segment $QR$ is $m_{QR} = \frac{r - q}{a_r - a_q} = \frac{r - q}{(a + (r-1)d) - (a + (q-1)d)} = \frac{r - q}{(r - q)d} = \frac{1}{d}$.
Since the slopes $m_{PQ} = m_{QR} = \frac{1}{d}$,the points $P, Q,$ and $R$ lie on the same line.
Therefore,the points $P, Q, R$ are collinear.

(D) For the lines to be concurrent,the determinant of the coefficients must be zero:
$D = \begin{vmatrix} a & am & 1 \\ b & b(m+1) & 1 \\ c & c(m+2) & 1 \end{vmatrix} = 0$
Applying column operations $C_2 \to C_2 - mC_1$:
$\begin{vmatrix} a & 0 & 1 \\ b & b & 1 \\ c & 2c & 1 \end{vmatrix} = 0$
Expanding along the first row:
$a(b - 2c) - 0 + 1(2bc - bc) = 0$
$ab - 2ac + bc = 0$
$b(a + c) = 2ac$
$b = \frac{2ac}{a + c}$
This implies that $a, b, c$ are in $H.P.$ for all $m \neq 0$.

(A) For $x^2 + 8x - 20 = 0$,the roots are $x = 2, -10$. Thus,$(x_1, y_1) = (2, -10)$.
For $4x^2 + 32x - 57 = 0$,the roots are $x = \frac{3}{2}, -\frac{19}{2}$. Thus,$(x_2, y_2) = (\frac{3}{2}, -\frac{19}{2})$.
For $9x^2 + 72x - 112 = 0$,the roots are $x = \frac{4}{3}, -\frac{28}{3}$. Thus,$(x_3, y_3) = (\frac{4}{3}, -\frac{28}{3})$.
To check for collinearity,we find the equation of the line passing through $A(2, -10)$ and $B(\frac{3}{2}, -\frac{19}{2})$.
Slope $m = \frac{-\frac{19}{2} - (-10)}{\frac{3}{2} - 2} = \frac{0.5}{-0.5} = -1$.
Equation: $y - (-10) = -1(x - 2) \implies y + 10 = -x + 2 \implies x + y = -8$.
Now,check if $C(\frac{4}{3}, -\frac{28}{3})$ satisfies $x + y = -8$:
$\frac{4}{3} + (-\frac{28}{3}) = -\frac{24}{3} = -8$.
Since the point $C$ satisfies the equation,the points are collinear.

(A) For the lines to be concurrent,the determinant of the coefficients must be zero:
$\begin{vmatrix} 1 & 1 & -1 \\ m - 1 & m^2 - 7 & -5 \\ m - 2 & 2m - 5 & 0 \end{vmatrix} = 0$
Expanding along the third row:
$-(m - 2) \begin{vmatrix} 1 & -1 \\ m^2 - 7 & -5 \end{vmatrix} + (2m - 5) \begin{vmatrix} 1 & -1 \\ m - 1 & -5 \end{vmatrix} = 0$
$-(m - 2)(-5 + m^2 - 7) + (2m - 5)(-5 + m - 1) = 0$
$-(m - 2)(m^2 - 12) + (2m - 5)(m - 6) = 0$
$-(m^3 - 6m^2 - 12m + 24) + (2m^2 - 12m - 5m + 30) = 0$
$-m^3 + 6m^2 + 12m - 24 + 2m^2 - 17m + 30 = 0$
$-m^3 + 8m^2 - 5m + 6 = 0 \implies m^3 - 8m^2 + 5m - 6 = 0$
Testing for roots,we find that for $m=3$,the lines are not concurrent because the lines become $x+y-1=0$,$2x+2y-5=0$,and $x+y=0$. The lines $x+y-1=0$ and $x+y=0$ are parallel,so they never intersect. Thus,there are no values of $m$ for which the lines are concurrent.

(B) For three lines to be concurrent,the determinant of their coefficients must be zero:
$\left| \begin{array}{ccc} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array} \right| = 0$
Applying column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\left| \begin{array}{ccc} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{array} \right| = 0$
Expanding along the first row:
$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$
Divide the entire equation by $(1-a)(1-b)(1-c)$:
$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$
$-\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$
Since $-\frac{a}{1-a} = \frac{1-a-1}{1-a} = \frac{1}{1-a} - 1$,we substitute this into the equation:
$(\frac{1}{1-a} - 1) + \frac{1}{1-b} + \frac{1}{1-c} = 0$
$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

(A) Given the system of equations:
$2x + 3y = -1$ $(1)$
$3x + y = 2$ $(2)$
$\lambda x + 2y = \mu$ $(3)$
First,solve equations $(1)$ and $(2)$ to find the point of intersection $(x, y)$.
From $(2)$,$y = 2 - 3x$.
Substitute this into $(1)$:
$2x + 3(2 - 3x) = -1$
$2x + 6 - 9x = -1$
$-7x = -7 \Rightarrow x = 1$.
Now,find $y$:
$y = 2 - 3(1) = -1$.
Since the system is consistent,the point $(1, -1)$ must satisfy the third equation $(3)$:
$\lambda(1) + 2(-1) = \mu$
$\lambda - 2 = \mu$
$\lambda - \mu = 2$.

(C) The given lines are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$
Expanding the determinant along the first column:
$1(3bc - 4bc) - 1(2ac - 4ac) + 1(2ab - 3ab) = 0$
$-bc + 2ac - ab = 0$
$2ac = ab + bc$
Dividing both sides by $abc$ (assuming $a, b, c \neq 0$):
$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This is the condition for $a, b, c$ to be in Harmonic Progression $(H.P.)$.

(B) Given lines are $y = x + 9\alpha$ and $3\alpha x + 2y + 9 = 0$.
Substitute $y$ from the first equation into the second:
$3\alpha x + 2(x + 9\alpha) + 9 = 0$
$3\alpha x + 2x + 18\alpha + 9 = 0$
$x(3\alpha + 2) = -18\alpha - 9$
$x = \frac{-18\alpha - 9}{3\alpha + 2}$
To simplify,rewrite the expression for $x$:
$x = \frac{-6(3\alpha + 2) + 12 - 9}{3\alpha + 2} = -6 + \frac{3}{3\alpha + 2}$
For $x$ to be an integer,$(3\alpha + 2)$ must be a divisor of $3$.
The divisors of $3$ are $\pm 1, \pm 3$.
Case $1$: $3\alpha + 2 = 1$ $\Rightarrow 3\alpha = -1$ $\Rightarrow \alpha = -1/3$ (not an integer).
Case $2$: $3\alpha + 2 = -1$ $\Rightarrow 3\alpha = -3$ $\Rightarrow \alpha = -1$ (integer).
Case $3$: $3\alpha + 2 = 3$ $\Rightarrow 3\alpha = 1$ $\Rightarrow \alpha = 1/3$ (not an integer).
Case $4$: $3\alpha + 2 = -3$ $\Rightarrow 3\alpha = -5$ $\Rightarrow \alpha = -5/3$ (not an integer).
Thus,there is only $1$ integral value of $\alpha$ (which is $\alpha = -1$).

(C) The given lines are concurrent if the determinant of their coefficients is zero:
$\left|\begin{array}{lll} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{array}\right| = 0$
Expanding the determinant along the first column:
$1(3bc - 4bc) - 1(2ac - 4ac) + 1(2ab - 3ab) = 0$
$-bc + 2ac - ab = 0$
$2ac = ab + bc$
$2ac = b(a + c)$
$b = \frac{2ac}{a + c}$
This is the condition for $a, b, c$ to be in Harmonic Progression $(H.P.)$.

(C) Let the points be $A\left( 0, \frac{8}{3} \right)$,$B(1, 3)$,and $C(82, 30)$.
Slope of $AB = \frac{3 - \frac{8}{3}}{1 - 0} = \frac{\frac{9-8}{3}}{1} = \frac{1}{3}$.
Slope of $BC = \frac{30 - 3}{82 - 1} = \frac{27}{81} = \frac{1}{3}$.
Since the slope of $AB$ is equal to the slope of $BC$ and they share a common point $B$,the points $A$,$B$,and $C$ are collinear,meaning they lie on a straight line.

(C) The given lines are:
$L_1: x + 2ay + a = 0$ $(1)$
$L_2: x + 3by + b = 0$ $(2)$
$L_3: x + 4ay + a = 0$ $(3)$
Since the lines are concurrent,they intersect at a common point.
Subtracting equation $(1)$ from $(3)$:
$(x + 4ay + a) - (x + 2ay + a) = 0$
$2ay = 0$
Since the lines are distinct,$a \neq 0$,therefore $y = 0$.
Substituting $y = 0$ into equation $(1)$:
$x + 2a(0) + a = 0 \Rightarrow x = -a$.
The point of concurrency is $(-a, 0)$.
Since this point must lie on line $(2)$:
$-a + 3b(0) + b = 0$
$-a + b = 0 \Rightarrow b = a$.
The point $(a, b)$ satisfies the equation $y = x$,which represents a straight line.

(C) Given lines are:
$ax + 2by + 3b = 0$ $(1)$
$bx - 2ay - 3a = 0$ $(2)$
To find the intersection point,multiply $(1)$ by $a$ and $(2)$ by $b$:
$a^2x + 2aby + 3ab = 0$
$b^2x - 2aby - 3ab = 0$
Adding these equations: $(a^2 + b^2)x = 0$. Since $(a, b) \neq (0, 0)$,$a^2 + b^2 \neq 0$,so $x = 0$.
Substituting $x = 0$ into $(1)$: $2by = -3b$. Since $b$ cannot be $0$ (otherwise the lines would not intersect at a unique point or would be coincident),we get $y = -3/2$.
The point of intersection is $(0, -3/2)$.
$A$ line parallel to the $x-$ axis has the equation $y = k$. Since it passes through $(0, -3/2)$,the equation is $y = -3/2$.
This represents a line below the $x-$ axis at a distance of $3/2$ from it.

(A) Given the equation of the line $px + qy + r = 0$ and the condition $3p + 2q + 4r = 0$.
From the condition,we can write $r = -\frac{3p + 2q}{4}$.
Substituting this into the line equation:
$px + qy - \frac{3p + 2q}{4} = 0$
$4px + 4qy - 3p - 2q = 0$
Rearranging the terms to group $p$ and $q$:
$p(4x - 3) + q(4y - 2) = 0$
For this to hold for all $p$ and $q$,the coefficients must be zero:
$4x - 3 = 0 \implies x = \frac{3}{4}$
$4y - 2 = 0 \implies y = \frac{2}{4} = \frac{1}{2}$
Thus,all lines pass through the fixed point $\left( \frac{3}{4}, \frac{1}{2} \right)$,meaning they are concurrent at this point.

(B) The centroid $C$ of the triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}).$
For the given vertices $(3, -1), (1, 3),$ and $(2, 4),$ the centroid $C$ is $(\frac{3+1+2}{3}, \frac{-1+3+4}{3}) = (2, 2).$
Next,we find the intersection point $P$ of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$
Solving the system:
$x + 3y = 1$
$3x - y = -1 \Rightarrow y = 3x + 1$
Substituting $y$ into the first equation: $x + 3(3x + 1) = 1$ $\Rightarrow x + 9x + 3 = 1$ $\Rightarrow 10x = -2$ $\Rightarrow x = -\frac{1}{5}.$
Then $y = 3(-\frac{1}{5}) + 1 = -\frac{3}{5} + 1 = \frac{2}{5}.$
So,$P = (-\frac{1}{5}, \frac{2}{5}).$
The line passing through $C(2, 2)$ and $P(-\frac{1}{5}, \frac{2}{5})$ has slope $m = \frac{\frac{2}{5} - 2}{-\frac{1}{5} - 2} = \frac{-\frac{8}{5}}{-\frac{11}{5}} = \frac{8}{11}.$
The equation of the line is $y - 2 = \frac{8}{11}(x - 2)$ $\Rightarrow 11y - 22 = 8x - 16$ $\Rightarrow 8x - 11y + 6 = 0.$
Checking the options,for $(-9, -6): 8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0.$
Thus,the line passes through $(-9, -6).$

To show that the points $(3,0), (-2,-2),$ and $(8,2)$ are collinear,it suffices to show that the line passing through points $(3,0)$ and $(-2,-2)$ also passes through the point $(8,2)$.
The equation of the line passing through points $(3,0)$ and $(-2,-2)$ is given by the two-point form:
$(y - 0) = \frac{-2 - 0}{-2 - 3}(x - 3)$
$y = \frac{-2}{-5}(x - 3)$
$y = \frac{2}{5}(x - 3)$
$5y = 2x - 6$
$2x - 5y = 6$
Now,we check if the point $(8,2)$ satisfies this equation:
$L.H.S. = 2(8) - 5(2) = 16 - 10 = 6$
Since $L.H.S. = R.H.S. = 6$,the point $(8,2)$ lies on the line passing through $(3,0)$ and $(-2,-2)$.
Therefore,the points $(3,0), (-2,-2),$ and $(8,2)$ are collinear.

(B) Three lines are concurrent if they pass through a common point,meaning the point of intersection of any two lines lies on the third line.
Given lines are:
$2x + y - 3 = 0$ $(1)$
$5x + ky - 3 = 0$ $(2)$
$3x - y - 2 = 0$ $(3)$
Solving $(1)$ and $(3)$ by adding them:
$(2x + y - 3) + (3x - y - 2) = 0$
$5x - 5 = 0 \implies x = 1$
Substituting $x = 1$ into $(1)$:
$2(1) + y - 3 = 0 \implies y = 1$
The point of intersection is $(1, 1)$.
Since the lines are concurrent,the point $(1, 1)$ must satisfy equation $(2)$:
$5(1) + k(1) - 3 = 0$
$5 + k - 3 = 0$
$k + 2 = 0 \implies k = -2$

(N/A) The equation of any line parallel to the $y-$ axis is of the form $x=a$ $(1)$.
The two given lines are $x-7y+5=0$ $(2)$ and $3x+y=0$ $(3)$.
From equation $(3)$,we get $y = -3x$.
Substituting $y = -3x$ into equation $(2)$:
$x - 7(-3x) + 5 = 0$
$x + 21x + 5 = 0$
$22x = -5$
$x = -\frac{5}{22}$.
Since the line $x=a$ passes through the point of intersection,the value of $a$ is the $x-$ coordinate of the intersection point.
Therefore,the required equation of the line is $x = -\frac{5}{22}$ or $22x + 5 = 0$.

(A) The equations of the given lines are:
$3x + y - 2 = 0$ ...... $(1)$
$px + 2y - 3 = 0$ ...... $(2)$
$2x - y - 3 = 0$ ...... $(3)$
To find the point of intersection,solve equations $(1)$ and $(3)$:
Adding $(1)$ and $(3)$:
$(3x + y - 2) + (2x - y - 3) = 0$
$5x - 5 = 0$
$x = 1$
Substitute $x = 1$ into equation $(1)$:
$3(1) + y - 2 = 0$
$3 + y - 2 = 0$
$y + 1 = 0$
$y = -1$
The point of intersection is $(1, -1)$.
Since the three lines are concurrent,the point $(1, -1)$ must satisfy equation $(2)$:
$p(1) + 2(-1) - 3 = 0$
$p - 2 - 3 = 0$
$p - 5 = 0$
$p = 5$
Thus,the required value of $p$ is $5$.

(A) The equations of the given lines are:
$y=m_{1}x+c_{1}$ ... $(1)$
$y=m_{2}x+c_{2}$ ... $(2)$
$y=m_{3}x+c_{3}$ ... $(3)$
Subtracting equation $(1)$ from $(2)$,we obtain:
$0 = (m_{2}-m_{1})x + (c_{2}-c_{1})$
$(m_{1}-m_{2})x = c_{2}-c_{1}$
$x = \frac{c_{2}-c_{1}}{m_{1}-m_{2}}$
Substituting this value of $x$ in $(1)$,we get:
$y = m_{1}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right) + c_{1}$
$y = \frac{m_{1}c_{2}-m_{1}c_{1} + m_{1}c_{1}-m_{2}c_{1}}{m_{1}-m_{2}}$
$y = \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}$
Thus,the point of intersection of lines $(1)$ and $(2)$ is $\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}\right)$.
Since the lines are concurrent,this point must satisfy equation $(3)$:
$\frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}} = m_{3}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right) + c_{3}$
$m_{1}c_{2}-m_{2}c_{1} = m_{3}(c_{2}-c_{1}) + c_{3}(m_{1}-m_{2})$
$m_{1}c_{2}-m_{2}c_{1} = m_{3}c_{2}-m_{3}c_{1} + c_{3}m_{1}-c_{3}m_{2}$
$m_{1}(c_{2}-c_{3}) + m_{2}(c_{3}-c_{1}) + m_{3}(c_{1}-c_{2}) = 0$.

(D) Three lines do not form a triangle if they are concurrent or if any two of them are parallel.
Case-$1$: The lines are concurrent.
The condition for concurrency is that the determinant of the coefficients must be zero:
$\begin{vmatrix} 2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2 \end{vmatrix} = 0$
$2(-6 - 2) - (-1)(-12 - \alpha) + 3(12 - 3\alpha) = 0$
$2(-8) + 1(-12 - \alpha) + 3(12 - 3\alpha) = 0$
$-16 - 12 - \alpha + 36 - 9\alpha = 0$
$8 - 10\alpha = 0 \Rightarrow \alpha = \frac{4}{5}$.
Case-$2$: Two lines are parallel.
Line $L_1: 2x - y + 3 = 0$ (slope $m_1 = 2$)
Line $L_2: 6x + 3y + 1 = 0$ (slope $m_2 = -2$)
Line $L_3: \alpha x + 2y - 2 = 0$ (slope $m_3 = -\frac{\alpha}{2}$)
$L_3$ is parallel to $L_1$ if $-\frac{\alpha}{2} = 2 \Rightarrow \alpha = -4$.
$L_3$ is parallel to $L_2$ if $-\frac{\alpha}{2} = -2 \Rightarrow \alpha = 4$.
The values of $\alpha$ are $\frac{4}{5}, 4, -4$.
The sum of squares $p = (\frac{4}{5})^2 + (4)^2 + (-4)^2 = \frac{16}{25} + 16 + 16 = 32 + 0.64 = 32.64$.
The greatest integer less than or equal to $p$ is $[32.64] = 32$.

(B) The lines are $L_1: x+3y-5=0$,$L_2: 3x-ky-1=0$,and $L_3: 5x+2y-12=0$.
$(A)$ For concurrency,the lines must intersect at a single point. Solving $L_1$ and $L_3$: $x+3y=5$ and $5x+2y=12$. Multiplying $L_1$ by $5$: $5x+15y=25$. Subtracting $L_3$: $13y=13 \Rightarrow y=1$. Then $x=2$. The point of intersection is $(2, 1)$. Substituting into $L_2$: $3(2)-k(1)-1=0 \Rightarrow 6-k-1=0 \Rightarrow k=5$. Thus,$(A) \rightarrow (s)$.
$(B)$ For lines to be parallel:
$L_1 \parallel L_2: \frac{1}{3} = \frac{3}{-k} \Rightarrow k=-9$.
$L_2 \parallel L_3: \frac{3}{5} = \frac{-k}{2} \Rightarrow k=-\frac{6}{5}$.
$L_1 \parallel L_3$ is not possible as slopes are $-1/3$ and $-5/2$. Thus,$(B) \rightarrow (p, q)$.
$(C)$ Lines form a triangle if they are not concurrent and no two lines are parallel. This happens when $k \neq 5, -9, -\frac{6}{5}$. Among the options,only $k=\frac{5}{6}$ satisfies this. Thus,$(C) \rightarrow (r)$.
$(D)$ Lines do not form a triangle if they are concurrent or parallel. This happens when $k=5, -9, -\frac{6}{5}$. Thus,$(D) \rightarrow (p, q, s)$.

(A) Three lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ are concurrent if the determinant of their coefficients is zero:
$\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\begin{vmatrix} 1 & 2a & a \\ 0 & 3b-2a & b-a \\ 0 & 4c-2a & c-a \end{vmatrix} = 0$
Expanding along the first column:
$(3b-2a)(c-a) - (b-a)(4c-2a) = 0$
$3bc - 3ab - 2ac + 2a^2 - (4bc - 2ab - 4ac + 2a^2) = 0$
$3bc - 3ab - 2ac + 2a^2 - 4bc + 2ab + 4ac - 2a^2 = 0$
$-bc - ab + 2ac = 0$
$2ac = ab + bc$
Dividing both sides by $abc$:
$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in arithmetic progression,which means $a, b, c$ are in harmonic progression.

(A) The required line passes through the point of intersection of the lines $3x - y = 5$ and $x + 3y = 1$.
Solving these equations:
$3(3x - y) = 3(5) \Rightarrow 9x - 3y = 15$
Adding $x + 3y = 1$ to $9x - 3y = 15$,we get $10x = 16 \Rightarrow x = \frac{8}{5}$.
Substituting $x = \frac{8}{5}$ in $3x - y = 5$: $3(\frac{8}{5}) - y = 5 \Rightarrow y = \frac{24}{5} - 5 = -\frac{1}{5}$.
So,the point of intersection is $(\frac{8}{5}, -\frac{1}{5})$.
The equation of a line with equal intercepts is $x + y = a$ or $x - y = a$.
Since the line passes through $(\frac{8}{5}, -\frac{1}{5})$,for $x + y = a$:
$\frac{8}{5} - \frac{1}{5} = a \Rightarrow a = \frac{7}{5}$.
Thus,$x + y = \frac{7}{5} \Rightarrow 5x + 5y - 7 = 0$.

(B) Given lines are $L_1: x - 2y + 8 = 0$ and $L_2: 3x - y + 4 = 0$.
To find the point of intersection,multiply $L_2$ by $2$: $6x - 2y + 8 = 0$.
Subtract $L_1$ from this: $(6x - 2y + 8) - (x - 2y + 8) = 0$,which gives $5x = 0$,so $x = 0$.
Substituting $x = 0$ in $L_2$: $3(0) - y + 4 = 0$,which gives $y = 4$.
The point of intersection is $(0, 4)$.
$A$ line passing through $(0, 0)$ and $(0, 4)$ is the $y$-axis,which is $x = 0$.
However,checking the options,if the line passes through $(0, 0)$ and the intersection point,the slope $m = \frac{4-0}{0-0}$ is undefined. Re-evaluating the intersection: $x-2y+8=0$ and $3x-y+4=0$. $y=3x+4$. $x-2(3x+4)+8=0 \implies x-6x-8+8=0 \implies -5x=0 \implies x=0, y=4$. The line is $x=0$. Given the options,there might be a typo in the question constants. If the intersection was $(4, 5)$,the line would be $5x-4y=0$.

(NONE) Step $1$: Find the point of intersection of the lines $x + 2y + 6 = 0$ and $2x - y = 2$.
From the second equation,$y = 2x - 2$.
Substitute this into the first equation: $x + 2(2x - 2) + 6 = 0$.
$x + 4x - 4 + 6 = 0 \implies 5x + 2 = 0 \implies x = -\frac{2}{5}$.
Then $y = 2(-\frac{2}{5}) - 2 = -\frac{4}{5} - \frac{10}{5} = -\frac{14}{5}$.
The point of intersection is $(-\frac{2}{5}, -\frac{14}{5})$.
Step $2$: The equation of a line with $y$-intercept $c = 5$ is $y = mx + 5$.
Since the line passes through $(-\frac{2}{5}, -\frac{14}{5})$,we have:
$-\frac{14}{5} = m(-\frac{2}{5}) + 5$.
$-\frac{14}{5} - 5 = -\frac{2}{5}m \implies -\frac{39}{5} = -\frac{2}{5}m \implies m = \frac{39}{2}$.
Step $3$: The equation is $y = \frac{39}{2}x + 5 \implies 2y = 39x + 10 \implies 39x - 2y + 10 = 0$.

(A) For three lines $a_1 x + b_1 y + c_1 = 0$,$a_2 x + b_2 y + c_2 = 0$,and $a_3 x + b_3 y + c_3 = 0$ to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} k & 2 & 2 \\ 2 & k & 3 \\ 3 & 3 & k \end{vmatrix} = 0$
Expanding the determinant:
$k(k^2 - 9) - 2(2k - 9) + 2(6 - 3k) = 0$
$k^3 - 9k - 4k + 18 + 12 - 6k = 0$
$k^3 - 19k + 30 = 0$
By testing roots,we find $(k - 2)$ is a factor:
$(k - 2)(k^2 + 2k - 15) = 0$
$(k - 2)(k + 5)(k - 3) = 0$
The possible values are $k_1 = 2$,$k_2 = -5$,and $k_3 = 3$.
The sum $\sum k_i = 2 + (-5) + 3 = 0$.

(B) For the lines to be concurrent,the determinant of the coefficients must be zero:
$\left|\begin{array}{ccc} 4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3 \end{array}\right| = 0$
Expanding the determinant:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(k)) - 1((1)(5) - (-1)(k)) = 0$
$4(3 - 25) - 3(-3 - 5k) - 1(5 + k) = 0$
$4(-22) + 9 + 15k - 5 - k = 0$
$-88 + 4 + 14k = 0$
$-84 + 14k = 0$
$14k = 84$
$k = 6$

(C) Given points are $A(1, 2)$,$B(2, 4)$,and $C(4, 8)$.
To check if they are collinear,we calculate the slopes of the line segments formed by these points.
Slope of $AB = \frac{4 - 2}{2 - 1} = \frac{2}{1} = 2$.
Slope of $BC = \frac{8 - 4}{4 - 2} = \frac{4}{2} = 2$.
Since the slope of $AB$ is equal to the slope of $BC$ and they share a common point $B$,the points $A$,$B$,and $C$ lie on the same straight line.
Therefore,the points form a straight line.

(B) The given lines are:
$x+3y-6=0$
$2x+y-4=0$
$kx-3y+1=0$
These lines are concurrent if the determinant of their coefficients is zero:
$\left|\begin{array}{ccc} 1 & 3 & -6 \\ 2 & 1 & -4 \\ k & -3 & 1 \end{array}\right| = 0$
Expanding along the first row $(R_1)$:
$1(1(1) - (-3)(-4)) - 3(2(1) - k(-4)) - 6(2(-3) - k(1)) = 0$
$1(1-12) - 3(2+4k) - 6(-6-k) = 0$
$-11 - 6 - 12k + 36 + 6k = 0$
$-6k + 19 = 0$
$6k = 19$
$k = \frac{19}{6}$

(C) The condition $PQ+PR=QR$ implies that the point $P$ must lie on the line segment $QR$.
First,we find the equation of the line passing through $Q(3,4)$ and $R(1,2)$.
The slope $m = \frac{2-4}{1-3} = \frac{-2}{-2} = 1$.
The equation of the line $QR$ is $y - 2 = 1(x - 1)$,which simplifies to $x - y + 1 = 0$.
Since $P$ lies on both the line $2x - y - 1 = 0$ and the line $x - y + 1 = 0$,we solve the system of equations:
$2x - y = 1$
$x - y = -1$
Subtracting the second equation from the first gives $(2x - x) - (y - y) = 1 - (-1)$,so $x = 2$.
Substituting $x = 2$ into $x - y = -1$,we get $2 - y = -1$,which implies $y = 3$.
Thus,the point $P$ is $(2,3)$.

(D) The given lines are:
$x+2y-3=0$ $\dots(i)$
$3x+4y-7=0$ $\dots(ii)$
$2x+3y-4=0$ $\dots(iii)$
$4x+5y-6=0$ $\dots(iv)$
Solving equations $(i)$ and $(ii)$:
From $(i)$,$x = 3-2y$. Substituting in $(ii)$:
$3(3-2y) + 4y - 7 = 0$
$9 - 6y + 4y - 7 = 0$
$-2y + 2 = 0 \implies y = 1$.
Then $x = 3 - 2(1) = 1$.
The point of intersection is $P(1, 1)$.
Now,check if $P(1, 1)$ satisfies $(iii)$:
$2(1) + 3(1) - 4 = 2 + 3 - 4 = 1 \neq 0$.
Since the point of intersection of the first two lines does not lie on the third line,the lines are not concurrent.

(C) The given equation of the line is $(a+2b)x + (a-3b)y = a-b$.
Rearranging the terms to group coefficients of $a$ and $b$:
$ax + 2bx + ay - 3by = a - b$
$a(x + y - 1) + b(2x - 3y + 1) = 0$
For this equation to hold for all values of $a$ and $b$,the coefficients must be zero:
$x + y - 1 = 0$ $(i)$
$2x - 3y + 1 = 0$ (ii)
From $(i)$,$y = 1 - x$. Substituting into (ii):
$2x - 3(1 - x) + 1 = 0$
$2x - 3 + 3x + 1 = 0$
$5x - 2 = 0 \Rightarrow x = \frac{2}{5}$
Substituting $x = \frac{2}{5}$ into $(i)$:
$\frac{2}{5} + y - 1 = 0 \Rightarrow y = 1 - \frac{2}{5} = \frac{3}{5}$
Thus,the fixed point is $\left(\frac{2}{5}, \frac{3}{5}\right)$.

(A) Let the family of lines passing through the intersection of $L_1: 2x - y + 2 = 0$ and $L_2: x + y + 4 = 0$ be given by $L_1 + \lambda L_2 = 0$.
$(2x - y + 2) + \lambda(x + y + 4) = 0$.
Since the line passes through the point $(5, -2)$,we substitute $x = 5$ and $y = -2$ into the equation:
$(2(5) - (-2) + 2) + \lambda(5 + (-2) + 4) = 0$.
$(10 + 2 + 2) + \lambda(5 - 2 + 4) = 0$.
$14 + \lambda(7) = 0$.
$7\lambda = -14$.
$\lambda = -2$.
Substituting $\lambda = -2$ back into the family equation:
$(2x - y + 2) - 2(x + y + 4) = 0$.
$2x - y + 2 - 2x - 2y - 8 = 0$.
$-3y - 6 = 0$.
$y + 2 = 0$.

(A) The equations of the lines are:
$x-3y+2=0$ ...$(i)$
$2x+5y-7=0$ ...(ii)
Solving equations $(i)$ and (ii):
From $(i)$,$x = 3y-2$. Substituting into (ii):
$2(3y-2) + 5y - 7 = 0$
$6y - 4 + 5y - 7 = 0$
$11y = 11 \Rightarrow y = 1$
Substituting $y=1$ in $(i)$: $x - 3(1) + 2 = 0 \Rightarrow x = 1$.
The point of intersection is $(1, 1)$.
The equation of a line perpendicular to $3x+2y+5=0$ is of the form $2x-3y+\lambda=0$.
Since this line passes through $(1, 1)$:
$2(1) - 3(1) + \lambda = 0$
$2 - 3 + \lambda = 0 \Rightarrow \lambda = 1$.
Thus,the required equation is $2x-3y+1=0$.