Foot of perpendicular, Image of a point and Reflexive properties Questions in English

(A) The midpoint $M$ of the line segment $AB$ is given by $(\frac{1+5}{2}, \frac{1+7}{2}) = (3, 4)$.
The slope of the line segment $AB$ is $m_{AB} = \frac{7-1}{5-1} = \frac{6}{4} = \frac{3}{2}$.
The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{3/2} = -\frac{2}{3}$.
The equation of the line passing through $(3, 4)$ with slope $-\frac{2}{3}$ is given by $y - 4 = -\frac{2}{3}(x - 3)$.
Multiplying by $3$,we get $3y - 12 = -2x + 6$,which simplifies to $2x + 3y = 18$.

(C) Given points are $A(-1, 5)$,$B(0, 0)$,and $C(2, 2)$.
$D$ is the midpoint of $BC$,so $D = (\frac{0+2}{2}, \frac{0+2}{2}) = (1, 1)$.
The slope of line $AD$ is $m_{AD} = \frac{1-5}{1-(-1)} = \frac{-4}{2} = -2$.
The equation of line $AD$ is $y - 5 = -2(x + 1)$,which simplifies to $2x + y - 3 = 0$.
The line perpendicular to $AD$ has a slope $m = -\frac{1}{m_{AD}} = \frac{1}{2}$.
Since this line passes through $B(0, 0)$,its equation is $y - 0 = \frac{1}{2}(x - 0)$,which simplifies to $x - 2y = 0$.

(A) The slope of the line segment joining $(a, b)$ and $(a', b')$ is $m_1 = \frac{b' - b}{a' - a}$.
Since the required line is the perpendicular bisector,its slope $m$ is $-\frac{1}{m_1} = \frac{a - a'}{b' - b} = \frac{a' - a}{b - b'}$.
The midpoint of the segment is $\left( \frac{a + a'}{2}, \frac{b + b'}{2} \right)$.
The equation of the line is $y - \frac{b + b'}{2} = \frac{a' - a}{b - b'} \left( x - \frac{a + a'}{2} \right)$.
Multiplying by $2(b - b')$,we get $2(b - b')y - (b^2 - b'^2) = 2(a' - a)x - (a'^2 - a^2)$.
Rearranging terms: $2(a - a')x + 2(b - b')y = a^2 + b^2 - a'^2 - b'^2$.

(A) The given line is $x \sec \theta + y \csc \theta = a$,which can be written as $x/\cos \theta + y/\sin \theta = a$.
Its slope $m_1 = -(\sin \theta / \cos \theta) = -\tan \theta$.
Since the required line is perpendicular to this line,its slope $m_2 = -1/m_1 = \cot \theta = \cos \theta / \sin \theta$.
The equation of the line passing through $(a \cos^3 \theta, a \sin^3 \theta)$ with slope $m_2$ is:
$y - a \sin^3 \theta = (\cos \theta / \sin \theta)(x - a \cos^3 \theta)$
$y \sin \theta - a \sin^4 \theta = x \cos \theta - a \cos^4 \theta$
$x \cos \theta - y \sin \theta = a \cos^4 \theta - a \sin^4 \theta$
$x \cos \theta - y \sin \theta = a (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta)$
$x \cos \theta - y \sin \theta = a \cos 2\theta$.

(C) The midpoint $M$ of the line segment joining $(7, 4)$ and $(-1, -2)$ is given by $M = (\frac{7-1}{2}, \frac{4-2}{2}) = (3, 1)$.
The slope of the line segment joining $(7, 4)$ and $(-1, -2)$ is $m = \frac{-2-4}{-1-7} = \frac{-6}{-8} = \frac{3}{4}$.
The slope of the perpendicular bisector is $m' = -\frac{1}{m} = -\frac{4}{3}$.
The equation of the line passing through $(3, 1)$ with slope $-\frac{4}{3}$ is:
$y - 1 = -\frac{4}{3}(x - 3)$
$3(y - 1) = -4(x - 3)$
$3y - 3 = -4x + 12$
$4x + 3y = 15$.

(A) Step $1$: 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 $(-1, -1)$.
Step $2$: Find the equation of the line perpendicular to $3x - 5y + 11 = 0$.
The general form of a line perpendicular to $ax + by + c = 0$ is $bx - ay + k = 0$.
Thus,the line is $5x + 3y + k = 0$.
Step $3$: Since the line passes through $(-1, -1)$,substitute these coordinates into the equation:
$5(-1) + 3(-1) + k = 0$ $\Rightarrow -5 - 3 + k = 0$ $\Rightarrow k = 8$.
Therefore,the required equation is $5x + 3y + 8 = 0$.

(A) The midpoint of the segment joining $(-4, 6)$ and $(8, 8)$ is $(\frac{-4+8}{2}, \frac{6+8}{2}) = (2, 7)$.
The slope of the line joining $(-4, 6)$ and $(8, 8)$ is $m = \frac{8-6}{8-(-4)} = \frac{2}{12} = \frac{1}{6}$.
The slope of the perpendicular bisector is $m' = -\frac{1}{m} = -6$.
The equation of the line passing through $(2, 7)$ with slope $-6$ is $y - 7 = -6(x - 2)$.
$y - 7 = -6x + 12$.
$6x + y - 19 = 0$.

(A) The equation of a line perpendicular to $ax + by + c = 0$ is of the form $bx - ay + \lambda = 0$.
Given the line $3x - 5y + 7 = 0$,the perpendicular line is $5x + 3y + \lambda = 0$ $(i)$.
Since the line passes through $(1, -2)$,we substitute these coordinates into equation $(i)$:
$5(1) + 3(-2) + \lambda = 0$
$5 - 6 + \lambda = 0$
$-1 + \lambda = 0$
$\lambda = 1$
Substituting $\lambda = 1$ into equation $(i)$,we get the required equation: $5x + 3y + 1 = 0$.

(C) Let the vertices of the square be $A(-1, 1)$ and $B(5, 3)$.
Since $A$ and $B$ are opposite vertices,the midpoint of $AB$ is $M = (\frac{-1+5}{2}, \frac{1+3}{2}) = (2, 2)$.
The slope of diagonal $AB$ is $m_{AB} = \frac{3-1}{5-(-1)} = \frac{2}{6} = \frac{1}{3}$.
The other diagonal of a square is the perpendicular bisector of the diagonal $AB$.
Therefore,the slope of the other diagonal is $m = -\frac{1}{m_{AB}} = -3$.
The equation of the other diagonal passing through $M(2, 2)$ with slope $-3$ is:
$y - 2 = -3(x - 2)$
$y - 2 = -3x + 6$
$3x + y - 8 = 0$.

(A) The equation of a line perpendicular to $ax + by + c = 0$ is of the form $bx - ay + \lambda = 0$.....$(i)$
Since the line passes through $(a, b)$,we substitute these coordinates into equation $(i)$:
$b(a) - a(b) + \lambda = 0$
$ab - ab + \lambda = 0$
$\lambda = 0$
Substituting $\lambda = 0$ back into equation $(i)$,we get the required equation:
$bx - ay = 0$

(B) The given line is $3x - y + 4 = 0$.
Let the foot of the perpendicular from $P(2, 3)$ be $Q(h, k)$.
The slope of the line is $m_1 = 3$.
The slope of the perpendicular line $PQ$ is $m_2 = -\frac{1}{3}$.
The equation of line $PQ$ passing through $(2, 3)$ is $y - 3 = -\frac{1}{3}(x - 2)$,which simplifies to $3y - 9 = -x + 2$,or $x + 3y = 11$.
Solving the system of equations $3x - y = -4$ and $x + 3y = 11$:
Multiply the first equation by $3$: $9x - 3y = -12$.
Adding this to $x + 3y = 11$ gives $10x = -1$,so $x = -\frac{1}{10}$.
Substituting $x$ into $x + 3y = 11$: $-\frac{1}{10} + 3y = 11 \implies 3y = 11 + \frac{1}{10} = \frac{111}{10} \implies y = \frac{37}{10}$.
Thus,the foot of the perpendicular is $\left( - \frac{1}{10}, \frac{37}{10} \right)$.

(B) The line passes through $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$.
Its equation is $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = a \cos \frac{\alpha - \beta}{2}$.
The foot of the perpendicular from the origin $(0,0)$ to the line $x \cos \theta + y \sin \theta = p$ is $(p \cos \theta, p \sin \theta)$.
Here,$p = a \cos \frac{\alpha - \beta}{2}$ and $\theta = \frac{\alpha + \beta}{2}$.
Thus,the foot of the perpendicular is $\left( a \cos \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2}, a \cos \frac{\alpha - \beta}{2} \sin \frac{\alpha + \beta}{2} \right)$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ and $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we get:
$x = \frac{a}{2} [\cos \alpha + \cos \beta]$ and $y = \frac{a}{2} [\sin \alpha + \sin \beta]$.
Therefore,the correct option is $B$.

(A) Let the coordinates of the foot of the perpendicular be $(h, k)$.
Since $(h, k)$ lies on the line $ax + by + c = 0$,we have $ah + bk + c = 0$.
The slope of the line $ax + by + c = 0$ is $m_1 = -\frac{a}{b}$.
The slope of the line joining $(x_1, y_1)$ and $(h, k)$ is $m_2 = \frac{k - y_1}{h - x_1}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$,which implies $\left(-\frac{a}{b}\right) \times \left(\frac{k - y_1}{h - x_1}\right) = -1$,so $\frac{k - y_1}{h - x_1} = \frac{b}{a}$.
Using the property $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$,we solve for $h$ and $k$:
$h = x_1 - a \left( \frac{ax_1 + by_1 + c}{a^2 + b^2} \right) = \frac{x_1(a^2 + b^2) - a^2x_1 - aby_1 - ac}{a^2 + b^2} = \frac{b^2x_1 - aby_1 - ac}{a^2 + b^2}$.
$k = y_1 - b \left( \frac{ax_1 + by_1 + c}{a^2 + b^2} \right) = \frac{y_1(a^2 + b^2) - abx_1 - b^2y_1 - bc}{a^2 + b^2} = \frac{a^2y_1 - abx_1 - bc}{a^2 + b^2}$.
Thus,the correct option is $A$.

(B) Let the foot of the perpendicular from $P(x_1, y_1) = (2, 4)$ to the line $ax + by + c = 0$ be $(h, k)$.
Here,$a = 1, b = 1, c = -1$.
The formula for the foot of the perpendicular is $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$.
Substituting the values: $\frac{h - 2}{1} = \frac{k - 4}{1} = -\frac{1(2) + 1(4) - 1}{1^2 + 1^2}$.
$\frac{h - 2}{1} = \frac{k - 4}{1} = -\frac{5}{2}$.
$h - 2 = -2.5 \implies h = -0.5 = -\frac{1}{2}$.
$k - 4 = -2.5 \implies k = 1.5 = \frac{3}{2}$.
Thus,the foot of the perpendicular is $\left( -\frac{1}{2}, \frac{3}{2} \right)$.

(B) Let the given point be $P(2, 3)$ and the line be $L: x + y - 11 = 0$.
The slope of the line $L$ is $m_1 = -1$.
The slope of the line perpendicular to $L$ is $m_2 = -\frac{1}{m_1} = 1$.
The equation of the line passing through $(2, 3)$ with slope $1$ is $y - 3 = 1(x - 2)$,which simplifies to $x - y + 1 = 0$.
The foot of the perpendicular is the intersection of $x + y = 11$ and $x - y = -1$.
Adding the two equations: $(x + y) + (x - y) = 11 - 1 \implies 2x = 10 \implies x = 5$.
Substituting $x = 5$ into $x + y = 11$: $5 + y = 11 \implies y = 6$.
Thus,the foot of the perpendicular is $(5, 6)$.

(B) Let $O$ be the origin $(0, 0)$. The perpendicular distance $OD$ from the origin to the line $3x + 4y + 15 = 0$ is given by $OD = \frac{|3(0) + 4(0) + 15|}{\sqrt{3^2 + 4^2}} = \frac{15}{5} = 3 \text{ units}$.
In the right-angled triangle $ODA$,we have $OA = 9$ and $OD = 3$.
Using the Pythagorean theorem,$AD = \sqrt{OA^2 - OD^2} = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2} \text{ units}$.
Since $OD$ is perpendicular to $AB$,$D$ is the midpoint of $AB$,so $AB = 2AD = 2(6\sqrt{2}) = 12\sqrt{2} \text{ units}$.
The area of triangle $OAB$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OD = \frac{1}{2} \times 12\sqrt{2} \times 3 = 18\sqrt{2} \text{ sq. units}$.

(C) Let the coordinate of vertex $A$ be $(h, k)$. Since the triangle is equilateral,the median $AD$ is perpendicular to the side $BC$. The centroid $O(0, 0)$ lies on the median $AD$. Thus,the line $OA$ is perpendicular to $BC$ $(x + y - 2 = 0)$.
The slope of $BC$ is $-1$,so the slope of $OA$ must be $1$.
$\frac{k - 0}{h - 0} = 1 \Rightarrow k = h$ ... $(i)$
Let $D(\alpha, \beta)$ be the midpoint of $BC$. Since $O$ is the centroid,it divides the median $AD$ in the ratio $2:1$. Thus,$O = \left(\frac{2\alpha + h}{3}, \frac{2\beta + k}{3}\right) = (0, 0)$.
This gives $2\alpha + h = 0 \Rightarrow \alpha = -h/2$ and $2\beta + k = 0 \Rightarrow \beta = -k/2$.
Since $D(\alpha, \beta)$ lies on $x + y - 2 = 0$,we have $\alpha + \beta - 2 = 0$.
Substituting $\alpha$ and $\beta$: $-h/2 - k/2 - 2 = 0 \Rightarrow h + k = -4$.
Using $(i)$,$h + h = -4$ $\Rightarrow 2h = -4$ $\Rightarrow h = -2$.
Thus,$k = -2$. The vertex $A$ is $(-2, -2)$.

(A) Let the foot of the perpendicular from the origin $(0, 0)$ to the line $3x + 4y - 5 = 0$ be $(h, k)$.
The slope of the given line $3x + 4y - 5 = 0$ is $m_1 = -\frac{3}{4}$.
The slope of the perpendicular line passing through the origin $(0, 0)$ and $(h, k)$ is $m_2 = \frac{k - 0}{h - 0} = \frac{k}{h}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$,so $\left( -\frac{3}{4} \right) \times \left( \frac{k}{h} \right) = -1$,which implies $\frac{k}{h} = \frac{4}{3}$,or $3k = 4h \implies k = \frac{4h}{3}$.
Since the point $(h, k)$ lies on the line $3x + 4y - 5 = 0$,we have $3h + 4k = 5$.
Substituting $k = \frac{4h}{3}$ into the equation: $3h + 4\left( \frac{4h}{3} \right) = 5$.
$3h + \frac{16h}{3} = 5 \implies \frac{9h + 16h}{3} = 5 \implies 25h = 15 \implies h = \frac{15}{25} = \frac{3}{5}$.
Now,$k = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}$.
Thus,the foot of the perpendicular is $\left( \frac{3}{5}, \frac{4}{5} \right)$.

(A) Let the image of point $A(3, 8)$ be $A'(x_1, y_1)$.
The line passing through $A$ and perpendicular to $x + 3y - 7 = 0$ has the form $3x - y + k = 0$.
Since it passes through $(3, 8)$,we have $3(3) - 8 + k = 0$,which gives $k = -1$. So,the line is $3x - y - 1 = 0$.
To find the intersection point of $x + 3y = 7$ and $3x - y = 1$,we solve the system:
$x + 3y = 7$ $(i)$
$9x - 3y = 3$ (ii)
Adding $(i)$ and (ii),we get $10x = 10$,so $x = 1$. Substituting $x=1$ into $(i)$,$1 + 3y = 7$,so $3y = 6$,$y = 2$. The intersection point is $(1, 2)$.
Since $(1, 2)$ is the midpoint of $AA'$,we have:
$\frac{x_1 + 3}{2} = 1 \Rightarrow x_1 = -1$
$\frac{y_1 + 8}{2} = 2 \Rightarrow y_1 = -4$
Thus,the image is $(-1, -4)$.

(A) Let $Q(a, b)$ be the reflection of $P(4, -13)$ in the line $5x + y + 6 = 0$.
The mid-point $R\left(\frac{a + 4}{2}, \frac{b - 13}{2}\right)$ lies on the line $5x + y + 6 = 0$.
Substituting the coordinates of $R$ into the line equation:
$5\left(\frac{a + 4}{2}\right) + \left(\frac{b - 13}{2}\right) + 6 = 0$
$5(a + 4) + (b - 13) + 12 = 0$
$5a + 20 + b - 13 + 12 = 0$
$5a + b + 19 = 0$ ......$(i)$
Also,the line segment $PQ$ is perpendicular to the line $5x + y + 6 = 0$.
Since the slope of the line is $-5$,the slope of $PQ$ must be $\frac{1}{5}$.
$\frac{b - (-13)}{a - 4} = \frac{1}{5}$
$5(b + 13) = a - 4$
$5b + 65 = a - 4$
$a - 5b - 69 = 0$ ......$(ii)$
Solving equations $(i)$ and $(ii)$:
From $(i)$,$b = -5a - 19$.
Substituting into $(ii)$:
$a - 5(-5a - 19) - 69 = 0$
$a + 25a + 95 - 69 = 0$
$26a + 26 = 0$
$a = -1$
Substituting $a = -1$ into $b = -5a - 19$:
$b = -5(-1) - 19 = 5 - 19 = -14$.
Thus,the reflection is $(-1, -14)$.

(D) The image of a point $(x, y)$ with respect to the line $y = x$ is given by the point $(y, x)$.
Given the point $(4, -3)$,we substitute $x = 4$ and $y = -3$ into the transformation rule $(y, x)$.
Thus,the image is $(-3, 4)$.

(B) Let the points be $A(1, -1)$ and $B(2, 3)$.
The vector along the line $AB$ is $\vec{AB} = (2-1)i + (3 - (-1))j = i + 4j$.
$A$ vector perpendicular to $\vec{AB} = i + 4j$ is of the form $\lambda(4i - j)$.
The unit normal vector is $\pm \frac{4i - j}{\sqrt{4^2 + (-1)^2}} = \pm \frac{4i - j}{\sqrt{17}}$.
The line equation passing through $(1, -1)$ with direction vector $i + 4j$ is $\frac{x-1}{1} = \frac{y+1}{4}$,which simplifies to $4x - y - 5 = 0$.
The normal vector to the line $ax + by + c = 0$ is $ai + bj$. Here,the normal vector is $4i - j$.
To check the direction towards the origin $(0, 0)$,we test the point $(0, 0)$ in the expression $4x - y - 5$. Since $4(0) - 0 - 5 = -5 < 0$,the vector pointing towards the origin must be the negative of the normal vector $4i - j$.
Thus,the required unit normal vector is $-\frac{4i - j}{\sqrt{17}} = \frac{-4i + j}{\sqrt{17}}$.

(A) The line $L$ is given by $2x - y = 0$.
Let the point be $P(0, 1)$. The formula for the reflection $(x', y')$ of a point $(x_0, y_0)$ in the line $ax + by + c = 0$ is $\frac{x' - x_0}{a} = \frac{y' - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.
Substituting $a = 2, b = -1, c = 0$ and $(x_0, y_0) = (0, 1)$:
$\frac{x' - 0}{2} = \frac{y' - 1}{-1} = -2 \frac{2(0) - 1(1) + 0}{2^2 + (-1)^2} = -2 \frac{-1}{5} = \frac{2}{5}$.
So,$x' = 2 \times \frac{2}{5} = \frac{4}{5}$ and $y' - 1 = -1 \times \frac{2}{5} = -\frac{2}{5} \implies y' = 1 - \frac{2}{5} = \frac{3}{5}$.
Thus,Statement-$1$ is true.
Statement-$2$ describes the definition of reflection,which is that the line $L$ is the perpendicular bisector of the segment joining the point and its reflection. Therefore,the points are equidistant from the line and lie on opposite sides. Thus,Statement-$2$ is true and is the correct explanation for Statement-$1$.

(A) The reflection of a point $(h, k)$ with respect to the $y$-axis is given by $(-h, k)$.
Applying this rule to the point $(1, -2)$:
$h = 1, k = -2$.
Therefore,the reflected point is $(-1, -2)$.

(A) Let the line perpendicular to $2x - 3y - 3 = 0$ be $3x + 2y + K = 0$.
Since it passes through $P(-5, 13)$,we have $3(-5) + 2(13) + K = 0$,which gives $K = -11$.
Thus,the equation of the line $PQ$ is $3x + 2y - 11 = 0$.
The intersection point $R$ of the lines $2x - 3y - 3 = 0$ and $3x + 2y - 11 = 0$ is the midpoint of $PQ$.
Solving the system:
$2x - 3y = 3$ (multiply by $2$): $4x - 6y = 6$
$3x + 2y = 11$ (multiply by $3$): $9x + 6y = 33$
Adding gives $13x = 39$,so $x = 3$.
Substituting $x = 3$ into $3(3) + 2y = 11$ gives $2y = 2$,so $y = 1$.
Thus,$R = (3, 1)$.
Let $Q = (\alpha, \beta)$. Since $R$ is the midpoint of $PQ$:
$\frac{\alpha - 5}{2} = 3$ $\Rightarrow \alpha - 5 = 6$ $\Rightarrow \alpha = 11$
$\frac{\beta + 13}{2} = 1$ $\Rightarrow \beta + 13 = 2$ $\Rightarrow \beta = -11$
Therefore,the coordinates of $Q$ are $(11, -11)$.

(D) Let the point be $P(-1, 3)$ and the line be $L: x - y + 1 = 0$. Let the image of $P$ be $P'(x', y')$.
Using the formula for the image of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$:
$\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$
Here,$a = 1, b = -1, c = 1, x_1 = -1, y_1 = 3$.
$\frac{x' - (-1)}{1} = \frac{y' - 3}{-1} = -2 \frac{1(-1) + (-1)(3) + 1}{1^2 + (-1)^2}$
$\frac{x' + 1}{1} = \frac{y' - 3}{-1} = -2 \frac{-1 - 3 + 1}{1 + 1}$
$\frac{x' + 1}{1} = \frac{y' - 3}{-1} = -2 \frac{-3}{2} = 3$
Now,$x' + 1 = 3 \implies x' = 2$.
And $\frac{y' - 3}{-1} = 3 \implies y' - 3 = -3 \implies y' = 0$.
Thus,the image point is $(2, 0)$.

(C) The rule for finding the reflection of a point $(h, k)$ across the line $y = x$ is to swap the coordinates,resulting in the point $(k, h)$.
Given the point $(4, -5)$,we have $h = 4$ and $k = -5$.
Applying the rule,the reflected point is $(-5, 4)$.

(A) The incident ray is along the line $x = 1$,which is a vertical line. The reflecting line is $x + y = 1$,which has a slope of $-1$ (making an angle of $135^{\circ}$ with the positive $x$-axis).
The angle between the incident ray ($x = 1$,slope $m_1 = \infty$) and the reflecting line ($x + y = 1$,slope $m_2 = -1$) is $45^{\circ}$.
By the law of reflection,the angle of incidence equals the angle of reflection. The reflected ray must make an angle of $45^{\circ}$ with the reflecting line.
Since the incident ray is vertical,the reflected ray must be horizontal. $A$ horizontal line has a slope of $m = 0$.
The point of intersection of $x = 1$ and $x + y = 1$ is $(1, 0)$.
The equation of the reflected ray passing through $(1, 0)$ with slope $m = 0$ is $y - 0 = 0(x - 1)$,which simplifies to $y = 0$.

(C) The given line is $ax + by + c = 0$.
The slope of this line is $m_1 = -\frac{a}{b}$.
Any line perpendicular to this line will have a slope $m_2$ such that $m_1 \times m_2 = -1$.
So,$m_2 = \frac{b}{a}$.
The equation of a line with slope $m_2 = \frac{b}{a}$ passing through $(a, b)$ is given by $y - b = \frac{b}{a}(x - a)$.
Multiplying by $a$,we get $a(y - b) = b(x - a)$.
$ay - ab = bx - ab$.
$bx - ay = 0$.

(A) Let the foot of the perpendicular be $P(h, k)$.
Given the line $2x + 3y - 4 = 0$,the slope of this line is $m_1 = -\frac{2}{3}$.
The slope of the perpendicular line passing through $(7, 8)$ and $(h, k)$ is $m_2 = \frac{k - 8}{h - 7}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$,so $(-\frac{2}{3}) \times (\frac{k - 8}{h - 7}) = -1$,which simplifies to $2(k - 8) = 3(h - 7)$,or $3h - 2k = 5$.
Since $P(h, k)$ lies on the line $2x + 3y - 4 = 0$,we have $2h + 3k = 4$.
Solving the system of equations:
$3h - 2k = 5$ (multiply by $3$): $9h - 6k = 15$
$2h + 3k = 4$ (multiply by $2$): $4h + 6k = 8$
Adding these gives $13h = 23$,so $h = \frac{23}{13}$.
Substituting $h$ into $2h + 3k = 4$: $2(\frac{23}{13}) + 3k = 4 \implies \frac{46}{13} + 3k = 4 \implies 3k = 4 - \frac{46}{13} = \frac{52 - 46}{13} = \frac{6}{13} \implies k = \frac{2}{13}$.
The foot of the perpendicular is $\left( \frac{23}{13}, \frac{2}{13} \right)$.

(A) Let the origin be $O(0, 0)$ and the foot of the perpendicular be $P(3, -4)$.
The slope of the line segment $OP$ is $m_{OP} = \frac{-4 - 0}{3 - 0} = -\frac{4}{3}$.
Since the line is perpendicular to $OP$,the slope of the required line $m$ is the negative reciprocal of $m_{OP}$,so $m = -\frac{1}{m_{OP}} = \frac{3}{4}$.
The required line passes through the point $P(3, -4)$ and has a slope $m = \frac{3}{4}$.
Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - (-4) = \frac{3}{4}(x - 3)$
$y + 4 = \frac{3}{4}(x - 3)$
$4(y + 4) = 3(x - 3)$
$4y + 16 = 3x - 9$
$3x - 4y = 25$.

(B) The given line is $x + \sqrt{3}y = \sqrt{3}$,which can be written as $y = -\frac{1}{\sqrt{3}}x + 1$.
The slope of this line is $m_1 = -\frac{1}{\sqrt{3}}$,so the angle it makes with the $x$-axis is $\theta = 150^{\circ}$ or $-30^{\circ}$.
The ray hits the $x$-axis at the point where $y = 0$,which is $x = \sqrt{3}$. So the point of incidence is $(\sqrt{3}, 0)$.
According to the law of reflection,the reflected ray makes an angle of $+30^{\circ}$ with the $x$-axis. Thus,its slope is $m_2 = \tan(30^{\circ}) = \frac{1}{\sqrt{3}}$.
The equation of the reflected line passing through $(\sqrt{3}, 0)$ with slope $m_2 = \frac{1}{\sqrt{3}}$ is:
$y - 0 = \frac{1}{\sqrt{3}}(x - \sqrt{3})$
$\sqrt{3}y = x - \sqrt{3}$.

(D) Let the coordinates of the foot of the perpendicular be $(h, k)$.
Given the line equation $ax + by + c = 0$ and point $(x_1, y_1)$,the formula for the foot of the perpendicular $(h, k)$ is given by:
$\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$
Here,$a = 3, b = -4, c = -5$ and $(x_1, y_1) = (0, 5)$.
Substituting these values into the formula:
$\frac{h - 0}{3} = \frac{k - 5}{-4} = -\frac{3(0) - 4(5) - 5}{3^2 + (-4)^2}$
$\frac{h}{3} = \frac{k - 5}{-4} = -\frac{0 - 20 - 5}{9 + 16}$
$\frac{h}{3} = \frac{k - 5}{-4} = -\frac{-25}{25}$
$\frac{h}{3} = \frac{k - 5}{-4} = 1$
From $\frac{h}{3} = 1$,we get $h = 3$.
From $\frac{k - 5}{-4} = 1$,we get $k - 5 = -4$,so $k = 1$.
Thus,the coordinates of the foot of the perpendicular are $(3, 1)$.

(D) The slope of $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.
The slope of the perpendicular bisector is $m_{\perp} = -(k-1) = 1-k$.
The midpoint of $PQ$ is $M = \left(\frac{k+1}{2}, \frac{7}{2}\right)$.
The equation of the perpendicular bisector is $y - \frac{7}{2} = (1-k)\left(x - \frac{k+1}{2}\right)$.
Setting $x=0$ for the $y$-intercept,we get $y - \frac{7}{2} = (1-k)\left(-\frac{k+1}{2}\right)$.
Given the $y$-intercept is $-4$,we have $-4 - \frac{7}{2} = -\frac{(1-k)(k+1)}{2}$.
$-\frac{15}{2} = -\frac{1-k^2}{2} \implies 15 = 1-k^2$.
$k^2 = 1 - 15 = -14$. Since $k$ must be real,let us re-evaluate the slope calculation.
Slope of perpendicular bisector is $m_{\perp} = k-1$. Midpoint is $(\frac{k+1}{2}, 3.5)$.
Equation: $y - 3.5 = (k-1)(x - \frac{k+1}{2})$.
At $x=0$,$y = -4$: $-4 - 3.5 = (k-1)(-\frac{k+1}{2})$.
$-7.5 = -\frac{k^2-1}{2} \implies 15 = k^2-1 \implies k^2 = 16 \implies k = \pm 4$.
Given the options,$k = -4$ is the correct choice.

(D) The point $P$ makes an angle $\theta$ with the positive $x$-axis.
The point $Q$ makes an angle $(\alpha - \theta)$ with the positive $x$-axis.
The angle between $OP$ and $OQ$ is $\angle POQ = \theta - (\alpha - \theta) = 2\theta - \alpha$ is incorrect; rather,the angle between the vectors is $\theta - (\alpha - \theta)$ is not the approach.
Let us consider the rotation: $A$ rotation of $P(\cos \theta, \sin \theta)$ by an angle $\phi$ clockwise results in $(\cos(\theta - \phi), \sin(\theta - \phi))$.
Setting $\theta - \phi = \alpha - \theta$,we get $\phi = 2\theta - \alpha$,which depends on $\theta$.
Re-evaluating: The reflection of $P(\cos \theta, \sin \theta)$ in a line $y = x \tan(\alpha/2)$ (which makes an angle $\alpha/2$ with the $x$-axis) is given by the rotation formula for reflection: $x' = \cos(2\phi - \theta), y' = \sin(2\phi - \theta)$ where $\phi = \alpha/2$.
Thus,$x' = \cos(\alpha - \theta)$ and $y' = \sin(\alpha - \theta)$.
Therefore,$Q$ is the reflection of $P$ in the line passing through the origin with slope $\tan(\alpha/2)$.

(C) The slope of $PQ$ is $m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$.
The slope of the perpendicular bisector is $m = -(k-1) = 1-k$.
The midpoint of $PQ$ is $M = \left(\frac{k+1}{2}, \frac{7}{2}\right)$.
The equation of the perpendicular bisector is $y - \frac{7}{2} = (1-k)(x - \frac{k+1}{2})$.
To find the $y-$intercept,set $x = 0$:
$y - \frac{7}{2} = (1-k)(0 - \frac{k+1}{2})$
$y = \frac{7}{2} - \frac{(1-k)(k+1)}{2}$
$y = \frac{7 - (1-k^2)}{2} = \frac{7 - 1 + k^2}{2} = \frac{6 + k^2}{2}$.
Given the $y-$intercept is $-4$:
$\frac{6 + k^2}{2} = -4$
$6 + k^2 = -8$
$k^2 = -14$ (This implies no real solution for $k$ based on the provided options).
Re-evaluating the slope calculation:
Slope of $PQ = \frac{3-4}{k-1} = \frac{-1}{k-1}$.
Slope of perpendicular bisector $= k-1$.
Equation: $y - \frac{7}{2} = (k-1)(x - \frac{k+1}{2})$.
At $x=0$,$y = \frac{7}{2} - (k-1)(\frac{k+1}{2}) = \frac{7 - (k^2-1)}{2} = \frac{8-k^2}{2}$.
Given $y-$intercept is $-4$:
$\frac{8-k^2}{2} = -4$ $\Rightarrow 8-k^2 = -8$ $\Rightarrow k^2 = 16$ $\Rightarrow k = \pm 4$.
Thus,a possible value of $k$ is $-4$.

(B) The equation of the incident ray is $x + \sqrt{3}y = \sqrt{3}$.
Find the point of incidence $A$ on the $x$-axis (where $y=0$): $x + \sqrt{3}(0) = \sqrt{3} \Rightarrow x = \sqrt{3}$. So,$A = (\sqrt{3}, 0)$.
Take a point $B$ on the incident ray,e.g.,$x=0$ $\Rightarrow \sqrt{3}y = \sqrt{3}$ $\Rightarrow y=1$. So,$B = (0, 1)$.
The reflected ray passes through $A(\sqrt{3}, 0)$ and the image of $B(0, 1)$ with respect to the $x$-axis,which is $B'(0, -1)$.
The slope of the reflected ray $AB'$ is $m = \frac{-1 - 0}{0 - \sqrt{3}} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}$.
The equation of the reflected ray is $y - 0 = \frac{1}{\sqrt{3}}(x - \sqrt{3})$.
$\sqrt{3}y = x - \sqrt{3}$.

(A) Let the image of point $A(3, 10)$ with respect to the line $2x + y - 6 = 0$ be $A'(x', y')$.
Using the formula $\frac{x' - 3}{2} = \frac{y' - 10}{1} = -2 \frac{2(3) + 10 - 6}{2^2 + 1^2} = -2 \frac{10}{5} = -4$.
So,$x' - 3 = -8 \Rightarrow x' = -5$ and $y' - 10 = -4 \Rightarrow y' = 6$.
Thus,$A' = (-5, 6)$.
The reflected beam passes through $B(5, 6)$ and $A'(-5, 6)$. Since both points have $y = 6$,the equation of the reflected beam is $y = 6$.
The incident beam passes through $A(3, 10)$ and the point of incidence $P$. The point $P$ is the intersection of $AB'$ (where $B'$ is the image of $B$) or simply the intersection of the line $A'B$ with the mirror line.
The line $A'B$ is $y = 6$. The intersection of $y = 6$ and $2x + y - 6 = 0$ is $2x + 6 - 6 = 0 \Rightarrow x = 0$. So $P = (0, 6)$.
The incident beam passes through $A(3, 10)$ and $P(0, 6)$.
The slope is $m = \frac{6 - 10}{0 - 3} = \frac{-4}{-3} = \frac{4}{3}$.
The equation is $y - 6 = \frac{4}{3}(x - 0)$ $\Rightarrow 3y - 18 = 4x$ $\Rightarrow 4x - 3y + 18 = 0$.

(D) Let the point $P = (\lambda^2 + 1, \lambda)$ and its image $P' = (\lambda, \lambda - 1)$ with respect to the line $3x + y - 6k = 0$.
$1$. The midpoint of $PP'$ must lie on the line $3x + y - 6k = 0$.
Midpoint $M = (\frac{\lambda^2 + 1 + \lambda}{2}, \frac{\lambda + \lambda - 1}{2}) = (\frac{\lambda^2 + \lambda + 1}{2}, \frac{2\lambda - 1}{2})$.
Substituting $M$ into the line equation: $3(\frac{\lambda^2 + \lambda + 1}{2}) + (\frac{2\lambda - 1}{2}) = 6k$.
$3\lambda^2 + 3\lambda + 3 + 2\lambda - 1 = 12k \implies 3\lambda^2 + 5\lambda + 2 = 12k$.
$2$. The line $PP'$ must be perpendicular to the line $3x + y - 6k = 0$.
The slope of $PP'$ is $m_{PP'} = \frac{(\lambda - 1) - \lambda}{(\lambda) - (\lambda^2 + 1)} = \frac{-1}{-\lambda^2 + \lambda - 1} = \frac{1}{\lambda^2 - \lambda + 1}$.
The slope of the line $3x + y - 6k = 0$ is $m_L = -3$.
Since $m_{PP'} \times m_L = -1$,we have $(\frac{1}{\lambda^2 - \lambda + 1}) \times (-3) = -1 \implies \lambda^2 - \lambda + 1 = 3 \implies \lambda^2 - \lambda - 2 = 0$.
Solving for $\lambda$: $(\lambda - 2)(\lambda + 1) = 0$,so $\lambda = 2$ or $\lambda = -1$.
$3$. Calculate $k$ for each $\lambda$:
If $\lambda = 2$: $12k = 3(2)^2 + 5(2) + 2 = 12 + 10 + 2 = 24 \implies k = 2$.
If $\lambda = -1$: $12k = 3(-1)^2 + 5(-1) + 2 = 3 - 5 + 2 = 0 \implies k = 0$.
Sum of values of $k = 2 + 0 = 2$.

(A) The given equation is $(\lambda + 2)x + (\lambda - 1)y - (8\lambda + 1) = 0$.
Rearranging the terms to isolate $\lambda$,we get $(2x - y - 1) + \lambda(x + y - 8) = 0$.
This represents a family of lines passing through the intersection of $2x - y - 1 = 0$ and $x + y - 8 = 0$.
Solving these equations: Adding them gives $3x - 9 = 0$,so $x = 3$. Substituting $x = 3$ into $x + y - 8 = 0$ gives $y = 5$.
Thus,the family of lines passes through the fixed point $P(3, 5)$.
The image of point $P(3, 5)$ in the line mirror $y = x$ is $P'(5, 3)$.
Any line passing through $P'(5, 3)$ can be represented as $a(x - 5) + b(y - 3) = 0$,which simplifies to $ax + by - (5a + 3b) = 0$.
Comparing this with the options,we check for the form $(\mu + 1)x - (2\mu + 1)y + \mu - 2 = 0$.
For this line to pass through $(5, 3)$,we substitute $x = 5$ and $y = 3$: $(\mu + 1)(5) - (2\mu + 1)(3) + \mu - 2 = 5\mu + 5 - 6\mu - 3 + \mu - 2 = 0$.
Since the result is $0$,the line passes through $(5, 3)$ for all $\mu$.

(B) Let the angle bisector of $B$ be $L: y = x$.
The reflection of vertex $A(3, 5)$ across the line $y = x$ will lie on the line $BC$.
To find the reflection $(x', y')$ of point $(x_0, y_0)$ across the line $y = x$,we swap the coordinates: $(x', y') = (y_0, x_0)$.
Thus,the reflection of $A(3, 5)$ is $(5, 3)$.
Therefore,the point $(5, 3)$ must lie on the line $BC$.

(D) The coordinates of the points are:
$P(\alpha, \beta)$,$Q(\beta, \alpha)$,$R(-\beta, \alpha)$,$S(-\beta, -\alpha)$,and $T(\beta, -\alpha)$.
The pentagon $PQRST$ can be divided into a rectangle with vertices $(-\beta, \alpha), (\beta, \alpha), (\beta, -\alpha), (-\beta, -\alpha)$ and a triangle with vertices $(\beta, \alpha), (\alpha, \beta), (\beta, -\alpha)$.
The area of the rectangle is $(2\beta) \times (2\alpha) = 4\alpha\beta$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2\alpha) \times (\alpha - \beta) = \alpha(\alpha - \beta)$.
Total area $= 4\alpha\beta + \alpha^2 - \alpha\beta = \alpha^2 + 3\alpha\beta = \alpha(\alpha + 3\beta) = 187$.
Since $187 = 11 \times 17$,we have $\alpha = 11$ and $\alpha + 3\beta = 17$.
Substituting $\alpha = 11$,we get $11 + 3\beta = 17$,so $3\beta = 6$,which gives $\beta = 2$.
We need to find $\alpha + \beta^2 = 11 + 2^2 = 11 + 4 = 15$.

(A) Let $P = (1, 2)$ be the point from which the ray originates and $Q = (5, 3)$ be the point through which the reflected ray passes.
Since the ray is reflected at point $A$ on the $x$-axis,the angle of incidence equals the angle of reflection.
Let $R$ be the reflection of point $P(1, 2)$ across the $x$-axis. The coordinates of $R$ are $(1, -2)$.
The reflected ray passes through $A$ and $Q$. Therefore,the points $R, A,$ and $Q$ are collinear.
The equation of the line passing through $R(1, -2)$ and $Q(5, 3)$ is given by:
$y - (-2) = \frac{3 - (-2)}{5 - 1}(x - 1)$
$y + 2 = \frac{5}{4}(x - 1)$
$4(y + 2) = 5(x - 1)$
$4y + 8 = 5x - 5$
$5x - 4y - 13 = 0$
Point $A$ lies on the $x$-axis,so its $y$-coordinate is $0$. Substituting $y = 0$ into the equation:
$5x - 4(0) - 13 = 0$
$5x = 13$
$x = \frac{13}{5}$
Thus,the coordinates of point $A$ are $\left( \frac{13}{5}, 0 \right)$.

(C) Let the point $P = (a, a - 1)$ and its image $Q = (a^2 + 1, a)$ with respect to the line $3x + y - 6a = 0$.
$1$. The midpoint $L$ of $PQ$ must lie on the line $3x + y = 6a$.
Midpoint $L = \left( \frac{a + a^2 + 1}{2}, \frac{a - 1 + a}{2} \right) = \left( \frac{a^2 + a + 1}{2}, \frac{2a - 1}{2} \right)$.
Substituting $L$ into the line equation: $3\left( \frac{a^2 + a + 1}{2} \right) + \left( \frac{2a - 1}{2} \right) = 6a$.
$3a^2 + 3a + 3 + 2a - 1 = 12a \Rightarrow 3a^2 - 7a + 2 = 0$.
$(3a - 1)(a - 2) = 0 \Rightarrow a = 2$ or $a = 1/3$.
$2$. The line $PQ$ must be perpendicular to the mirror line $3x + y = 6a$.
Slope of $PQ = \frac{a - (a - 1)}{a^2 + 1 - a} = \frac{1}{a^2 - a + 1}$.
Slope of mirror line is $-3$.
Since $PQ \perp \text{mirror}$,product of slopes is $-1$: $\left( \frac{1}{a^2 - a + 1} \right) \times (-3) = -1$.
$a^2 - a + 1 = 3 \Rightarrow a^2 - a - 2 = 0$.
$(a - 2)(a + 1) = 0 \Rightarrow a = 2$ or $a = -1$.
Comparing both conditions,the common value is $a = 2$.

(C) The reflection of a point $(x, y)$ about the line $y = x$ is $(y, x)$.
Given point $A(1, 2)$,its reflection $B$ about the line $y = x$ is $(2, 1)$.
The reflection of a point $(x, y)$ about the $x$-axis $(y = 0)$ is $(x, -y)$.
Therefore,the reflection of $B(2, 1)$ about $y = 0$ is $(2, -1)$.
Comparing this with $(\alpha, \beta)$,we get $\alpha = 2$ and $\beta = -1$.

(A) $1$. The mirror image of point $(x, y)$ in the line $y = x$ is $(y, x)$. Thus,for $A(1, 4)$,$B = (4, 1)$.
$2$. The mirror image of point $(x, y)$ in the line $y = -x$ is $(-y, -x)$. Thus,for $B(4, 1)$,$C = (-1, -4)$.
$3$. The mirror image of point $(x, y)$ in the $x$-axis is $(x, -y)$. Thus,for $C(-1, -4)$,$D = (-1, 4)$.
$4$. The coordinates are $A(1, 4)$,$B(4, 1)$,and $D(-1, 4)$.
$5$. The area of $\triangle ABD$ with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
$6$. Area $= \frac{1}{2} |1(1 - 4) + 4(4 - 4) + (-1)(4 - 1)| = \frac{1}{2} |1(-3) + 4(0) - 1(3)| = \frac{1}{2} |-3 - 3| = \frac{1}{2} |-6| = 3 \, sq. \, units$.

(B) Let the point $P = (\lambda, 2\lambda)$ and its image $Q = (\mu, 3\lambda)$ with respect to the line $x - y + 2 = 0$.
The midpoint $F$ of $PQ$ is $\left( \frac{\lambda + \mu}{2}, \frac{2\lambda + 3\lambda}{2} \right) = \left( \frac{\lambda + \mu}{2}, \frac{5\lambda}{2} \right)$.
Since $F$ lies on the line $x - y + 2 = 0$,we have $\frac{\lambda + \mu}{2} - \frac{5\lambda}{2} + 2 = 0$,which simplifies to $\frac{\mu - 4\lambda}{2} + 2 = 0$,or $\mu - 4\lambda + 4 = 0$ $(1)$.
The slope of $PQ$ is $m_{PQ} = \frac{3\lambda - 2\lambda}{\mu - \lambda} = \frac{\lambda}{\mu - \lambda}$.
Since $PQ$ is perpendicular to the line $x - y + 2 = 0$ (which has slope $1$),we have $m_{PQ} \times 1 = -1$,so $\frac{\lambda}{\mu - \lambda} = -1$,which implies $\lambda = -\mu + \lambda$,so $\mu = 0$.
Substituting $\mu = 0$ into $(1)$,we get $0 - 4\lambda + 4 = 0$,so $\lambda = 1$.
Thus,$\lambda = 1$ and $\mu = 0$. Checking the options:
$A) \mu - \lambda = 0 - 1 = -1 \neq 0$
$B) \lambda + \mu = 1 + 0 = 1$
$C) \lambda - \mu = 1 - 0 = 1 \neq -1$
$D) \lambda + \mu = 1 + 0 = 1 \neq 0$
The correct option is $B$.

(A) The equation of the line $3x + y = \lambda$ can be written in intercept form as $\frac{x}{\lambda/3} + \frac{y}{\lambda} = 1$.
Thus,the coordinates of $A$ are $(\frac{\lambda}{3}, 0)$ and $B$ are $(0, \lambda)$.
The foot of the perpendicular $P$ from the origin $(0,0)$ to the line $ax + by + c = 0$ is given by $(\frac{-ac}{a^2+b^2}, \frac{-bc}{a^2+b^2})$.
Here,$3x + y - \lambda = 0$,so $a=3, b=1, c=-\lambda$.
$P = (\frac{-3(-\lambda)}{3^2+1^2}, \frac{-1(-\lambda)}{3^2+1^2}) = (\frac{3\lambda}{10}, \frac{\lambda}{10})$.
Now,we calculate the distances $BP$ and $PA$:
$BP^2 = (\frac{3\lambda}{10} - 0)^2 + (\frac{\lambda}{10} - \lambda)^2 = \frac{9\lambda^2}{100} + \frac{81\lambda^2}{100} = \frac{90\lambda^2}{100} = \frac{9\lambda^2}{10}$.
$PA^2 = (\frac{\lambda}{3} - \frac{3\lambda}{10})^2 + (0 - \frac{\lambda}{10})^2 = (\frac{10\lambda - 9\lambda}{30})^2 + \frac{\lambda^2}{100} = \frac{\lambda^2}{900} + \frac{9\lambda^2}{900} = \frac{10\lambda^2}{900} = \frac{\lambda^2}{90}$.
Therefore,$\frac{BP^2}{PA^2} = \frac{9\lambda^2/10}{\lambda^2/90} = \frac{9}{10} \times 90 = 81$.
Thus,$\frac{BP}{PA} = \sqrt{81} = 9$.
So,the ratio $BP : PA$ is $9 : 1$.

(C) Let the slope of the incident ray be $m$. The line $7x - y + 1 = 0$ has a slope $m_1 = 7$. The reflected ray $y + 2x = 1$ has a slope $m_2 = -2$.
Since the angle of incidence equals the angle of reflection,the angle between the incident ray and the mirror line equals the angle between the reflected ray and the mirror line.
Using the formula for the angle between two lines $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\left| \frac{m - 7}{1 + 7m} \right| = \left| \frac{7 - (-2)}{1 + 7(-2)} \right| = \left| \frac{9}{1 - 14} \right| = \left| \frac{9}{-13} \right| = \frac{9}{13}$.
Case $1$: $\frac{m - 7}{1 + 7m} = \frac{9}{13}$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$. The line equation is $y - 1 = -2(x - 0) \Rightarrow 2x + y - 1 = 0$ (This is the reflected ray itself).
Case $2$: $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$ $\Rightarrow 13m - 91 = -9 - 63m$ $\Rightarrow 76m = 82$ $\Rightarrow m = \frac{41}{38}$.
The equation of the incident line is $y - 1 = \frac{41}{38}(x - 0)$ $\Rightarrow 38y - 38 = 41x$ $\Rightarrow 41x - 38y + 38 = 0$.