Straight Lines and Pair of Straight Lines

To determine the $Y$-intercept of the line $ L $ which passes through the point $(2,0)$ and makes an angle of $60^\circ$ with the line $2x - y + 3 = 0$, follow these steps:

Calculate the slope of the given line:

The equation $2x - y + 3 = 0$ can be rewritten in slope-intercept form $y = 2x + 3$. Thus, the slope of this line is 2.

Determine the slope of line $L$:

Let the slope of line $L$ be $m$. The formula for the tangent of the angle between two lines with slopes $m_1$ and $m_2$ is:

$ \tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| $

Applying this to the problem:

$ \tan 60^\circ = \left|\frac{m - 2}{1 + 2m}\right| $

Thus, we have:

$ \sqrt{3} = \left|\frac{m - 2}{1 + 2m}\right| $

This leads to two cases:

$ m - 2 = \pm \sqrt{3}(1 + 2m) $

Solve for $m$:

Solving $(m - 2)^2 = 3(1 + 2m)^2$:

$ (m - 2)^2 = 3(1 + 4m^2 + 4m) $

$ m^2 + 4 - 4m = 3 + 12m^2 + 12m $

$ 11m^2 + 16m - 1 = 0 $

Find the roots of the quadratic equation:

Applying the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$ m = \frac{-16 \pm \sqrt{256 + 44}}{22} = \frac{-16 \pm \sqrt{300}}{22} $

Simplifies to:

$ m = \frac{10\sqrt{3} - 16}{22} $

Since line $L$ makes an acute angle with the positive X-axis, choose:

$ m = \frac{5\sqrt{3} - 8}{11} $

Equation of line $L$:

The line passes through $(2,0)$ and has slope $m = \frac{5\sqrt{3} - 8}{11}$, so:

$ y = \frac{5\sqrt{3} - 8}{11}(x - 2) $

Simplify to find the Y-intercept:

$ 11y = (5\sqrt{3} - 8)x - 10\sqrt{3} + 16 $

Thus, the Y-intercept is:

$ y = \frac{16 - 10\sqrt{3}}{11} $

Therefore, the $Y$-intercept of line $L$ is $\frac{16 - 10\sqrt{3}}{11}$.

To solve for $ h $ in the equation $ 2x^2 + hxy + 6y^2 = 0 $, given that the slope of one line is three times the slope of the other, follow these steps:

Identify the slopes: Let's denote the slopes of the two lines as $ m $ and $ 3m $.

Apply the condition for slopes:

$ m + 3m = -\frac{h}{6} $

$ 4m = -\frac{h}{6} $

For the product of the slopes:

$ m \times 3m = \frac{2}{6} $

Solve the product equation: Simplify the product of slopes:

$ 3m^2 = \frac{1}{3} \quad \Rightarrow \quad m^2 = \frac{1}{9} $

Therefore,

$ m = \pm \frac{1}{3} $

Substitute back to find $ h $:

Substitute $ m = \pm \frac{1}{3} $ into the equation for the sum of slopes:

$ 4\left(\pm \frac{1}{3}\right) = -\frac{h}{6} $

Solve for $ h $:

$ \pm \frac{4}{3} = -\frac{h}{6} $

$ h = \pm 8 $

Thus, the value of $ h $ is $ \pm 8 $.

We start with the given pair of lines equation:

$ 3x^2 + 11xy - 4y^2 = 0 $

From this, we determine the slopes of the lines, $m_1$ and $m_2$, using the relationships:

Sum of slopes: $ m_1 + m_2 = \frac{11}{4} $

Product of slopes: $ m_1 \cdot m_2 = -\frac{3}{4} $

Solving for $m_1$, we have:

$ \begin{aligned} & m_1\left(\frac{11}{4} - m_1\right) = -\frac{3}{4} \\ & \Rightarrow 4m_1^2 - 11m_1 - 3 = 0 \\ & \Rightarrow \left(4m_1 + 1\right)(m_1 - 3) = 0 \end{aligned} $

Thus, the slopes are $m_1 = -\frac{1}{4}$ and $m_2 = 3$.

For lines perpendicular to these, the slopes are the negative reciprocals:

$ m_3 = 4 \quad \text{and} \quad m_4 = -\frac{1}{3} $

The equations of these perpendicular lines passing through the point $(1,1)$ are:

$ \begin{aligned} & y - 1 = 4(x - 1) \quad \Rightarrow \quad 4x - y - 3 = 0 \\ & y - 1 = -\frac{1}{3}(x - 1) \quad \Rightarrow \quad x + 3y - 4 = 0 \end{aligned} $

These equations combine to form:

$ (4x - y - 3)(x + 3y - 4) = 0 $

Simplifying this gives the pair of lines equation:

$ \begin{aligned} & 4x^2 + 12xy - 16x - xy - 3y^2 + 4y - 3x - 9y + 12 = 0 \\ & \Rightarrow 4x^2 + 11xy - 3y^2 - 19x - 5y + 12 = 0 \end{aligned} $

Here, coefficients are identified as $a = 4$, $2h = 11$, $b = -3$, $2g = -19$, and $2f = -5$.

Finally, calculating the expression:

$ \begin{aligned} & 2(a - h + b - g + f - 12) = 2(4 - \frac{11}{2} - 3 + \frac{19}{2} - \frac{5}{2} - 12) \\ & = 8 - 11 - 6 + 19 - 5 - 24 \\ & = -19 \end{aligned} $

To find the equation of a line passing through the intersection of the given lines, first solve for the intersection point by combining their equations:

Equation of the first line: $x - 2y + 3 = 0$

Equation of the second line: $2x - y - 1 = 0$

The general equation of a line passing through their intersection is a linear combination of the two lines:

$ x - 2y + 3 + \lambda(2x - y - 1) = 0 $

Simplifying:

$ x(1 + 2\lambda) + y(-2 - \lambda) + (3 - \lambda) = 0 $

This can be rearranged into intercept form:

$ \frac{x}{\frac{\lambda-3}{1+2\lambda}} + \frac{y}{\frac{3-\lambda}{\lambda+2}} = 1 $

This gives the coordinates of points $A$ and $B$ as:

$A\left(\frac{\lambda-3}{1+2\lambda}, 0\right)$

$B\left(0, \frac{3-\lambda}{\lambda+2}\right)$

To find the point dividing $AB$ in the ratio $-2:3$, apply the section formula:

For the x-coordinate $h$:

$ h = \frac{3 \left(\frac{\lambda-3}{1+2\lambda}\right)}{-2 + 3} = 3 \left(\frac{\lambda-3}{1+2\lambda}\right) $

For the y-coordinate $k$:

$ k = \frac{-2 \left(\frac{3-\lambda}{\lambda+2}\right)}{-2 + 3} = 2 \left(\frac{3-\lambda}{\lambda+2}\right) $

Now, solving for $\lambda$ using the x-coordinate $h$:

$ h + 2h\lambda = 3\lambda - 9 \Rightarrow 2h\lambda - 3\lambda = -h - 9 $

Solving for $\lambda$:

$ \lambda = \frac{-h-9}{2h-3} $

Substituting into the expression for $k$:

$ k = 2 \left[\frac{\frac{-h-9}{2h-3} - 3}{\frac{-h-9}{2h-3} + 2}\right] = \frac{-14h}{3h-15} $

Thus, the equation of the locus is:

$ 3hk - 15k + 14h = 0 $

Therefore, the required equation is:

$ 14x + 3xy - 15y = 0 $

When the origin is translated to the point $(2, -1)$, the coordinates of the point change as follows:

$ \begin{aligned} a &= -1 - 2 = -3, \\ b &= 2 - (-1) = 3. \end{aligned} $

Thus, the point $(a, b)$ becomes $(-3, 3)$.

Next, the point $(-3, 3)$ is rotated by an angle of $45^\circ$ around the new origin. Using the rotation formulas:

$ \begin{aligned} X &= a \cos \theta + b \sin \theta, \\ Y &= -a \sin \theta + b \cos \theta, \end{aligned} $

where $\theta = 45^\circ$, we find:

$ \begin{aligned} X &= -3 \left(\frac{1}{\sqrt{2}}\right) + 3 \left(\frac{1}{\sqrt{2}}\right) = 0, \\ Y &= -3 \left(\frac{1}{\sqrt{2}}\right) + 3 \left(\frac{1}{\sqrt{2}}\right) = 0. \end{aligned} $

So, the coordinates become:

$ \begin{aligned} c &= 0, \\ d &= 3 \sqrt{2}, \end{aligned} $

yielding the point $(c, d) = (0, 3 \sqrt{2})$.

Finally, reflecting the point $(0, 3 \sqrt{2})$ through the line $y = x$ results in:

$ (e, f) = (d, c) = (3 \sqrt{2}, 0). $

To find the foot of the perpendicular from the point $(3,1)$ to the line $x+3y+4=0$, we first determine the equation of the line passing through $(3,1)$ that is perpendicular to $x+3y+4=0$. The slope of the given line $x+3y+4=0$ is $-\frac{1}{3}$, so the perpendicular line has a slope of 3.

The equation of this line is:

$ y - 1 = 3(x - 3) \implies 3x - y = 8 \quad \text{(Equation i)} $

The point $(a, b)$ lies on both this line and the original line $x + 3y + 4 = 0$.

From Equation (i),

$ 3a - b = 8 \quad \Rightarrow \quad b = 3a - 8 $

Substitute into the original line equation:

$ a + 3b + 4 = 0 \implies a + 3(3a - 8) + 4 = 0 $

$ a + 9a - 24 + 4 = 0 \implies 10a = 20 \implies a = 2 $

Therefore,

$ b = 3(2) - 8 = -2 $

The coordinates of the point $(a, b)$ are $(2, -2)$.

Next, find the image of $(2, -2)$ with respect to the line $3x - 4y + 11 = 0$. The midpoint of the segment joining $(p, q)$ and $(2, -2)$ must satisfy:

$ 3\left(\frac{p + 2}{2}\right) - 4\left(\frac{q + 2}{2}\right) + 11 = 0 $

$ \frac{3(p + 2) - 4(q + 2)}{2} + 11 = 0 $

$ 3p - 4q + 36 = 0 \quad \text{(Equation ii)} $

For the line equation through $(p, q)$ and $(2, -2)$, perpendicular to $3x - 4y + 11 = 0$, we get:

$ q + 2 = -\frac{4}{3}(p - 2) \quad \Rightarrow \quad 3q + 6 = -4p + 8 $

$ 4p + 3q = 2 \quad \text{(Equation iii)} $

Solving Equations (ii) and (iii) simultaneously:

From Equation (iii):

$ 4p + 3q = 2 $

Substitute and solve:

$ 3p - 4q + 36 = 0 $

$ 4p + 3q = 2 $

Solve these equations simultaneously:

The solution is $(p, q) = (-4, 6)$.

Finally:

$ \frac{p}{a} + \frac{q}{b} = \frac{-4}{2} + \frac{6}{-2} = -2 - 3 = -5 $

The coordinate of $P$ be $(a, b)$ But, $P$ is on the $Y$-axis so $P=(0, b)$

$\therefore$ Angle which $P Q$ makes with $X$-axis is same as angle which $P R$ makes with $X$-axis.

$ \begin{aligned} \frac{3-b}{2-0} & =-\left(\frac{2-b}{3-0}\right) \\ \Rightarrow \quad 9-3 b & =-4+2 b \Rightarrow 5 b=13 \end{aligned} $

Now, form the option $5 b=a+13$

$ [\because a=0] $

To find the area of the triangle formed by the lines described by the equation $6x^2 + 13xy + 6y^2 = 0$ and $x + 2y + 3 = 0$, we use a specific formula for the area of a triangle formed by a pair of lines of the form $ax^2 + 2hxy + by^2 = 0$ and a third line $lx + my + n = 0$.

The formula to calculate the area is:

$ \text{Area} = \frac{n^2 \sqrt{h^2 - ab}}{\left| am^2 - 2hlm + bl^2 \right|} $

Given:

$ a = 6 $

$ 2h = 13 \Rightarrow h = \frac{13}{2} $

$ b = 6 $

$ l = 1 $

$ m = 2 $

$ n = 3 $

Now substitute these values into the formula:

$ \text{Area} = \frac{9 \sqrt{\left(\frac{13}{2}\right)^2 - 6 \cdot 6}}{\left| 6 \cdot 2^2 - 13 \cdot 1 \cdot 2 + 6 \cdot 1^2 \right|} $

Calculate the values step-by-step:

Compute $ h^2 - ab $:

$ \left(\frac{13}{2}\right)^2 = \frac{169}{4} $

$ ab = 6 \times 6 = 36 $

$ h^2 - ab = \frac{169}{4} - 36 = \frac{169}{4} - \frac{144}{4} = \frac{25}{4} $

Compute the denominator:

$ am^2 = 6 \times 4 = 24 $

$ 2hlm = 13 \times 2 = 26 $

$ bl^2 = 6 \times 1 = 6 $

$ am^2 - 2hlm + bl^2 = 24 - 26 + 6 = 4 $

Substitute these into the area formula:

$ \text{Area} = \frac{9 \times \frac{5}{2}}{\left| 4 \right|} = \frac{45}{8} \text{ sq units} $

Therefore, the area of the triangle is $\frac{45}{8}$ square units.

We have the following equations representing the sides of a square:

$3x + y - 4 = 0$   (i)

$x - \alpha y + 10 = 0$   (ii)

$\beta x + 2y + 4 = 0$   (iii)

$3x + y + k = 0$   (iv)

These lines form pairs of parallel lines: equations (i) with (iv) and (ii) with (iii).

Equations (i) and (ii) are perpendicular, so their slopes must multiply to $-1$:

$ (m_1 \times m_2) = -1 $

From equation (i), the slope $m_1 = -3$. From equation (ii), the slope $m_2 = \frac{1}{\alpha}$. Thus:

$ (-3) \times \frac{1}{\alpha} = -1 $

$ \Rightarrow \alpha = 3 $

Similarly, equations (iii) and (iv) are perpendicular:

$ \left(\frac{-\beta}{2}\right) \times (-3) = -1 $

$ \Rightarrow \beta = -\frac{2}{3} $

Substituting $\alpha$ and $\beta$ back into the equations, we have:

From equation (ii), $x - 3y + 10 = 0$.

From equation (iii), $-\frac{2}{3}x + 2y + 4 = 0$, simplifying gives:

$ x - 3y - 6 = 0 $

Since the distances between the parallel sides must be equal for the square, we equate:

$ \frac{|k + 4|}{\sqrt{3^2 + 1^2}} = \frac{|10 + 6|}{\sqrt{1^2 + (-3)^2}} $

Which simplifies to:

$ |k + 4| = 16 $

Thus, the expression $\alpha \beta (k + 4)^2$ becomes:

$ 3 \times \left(-\frac{2}{3}\right) (16)^2 = -512 $

To find point $ A $, we need to solve the equations:

$ 3x + y - 4 = 0 $

$ x - y = 0 $

The second equation gives us $ y = x $. Substitute $ y = x $ into the first equation:

$ 3x + x - 4 = 0 $

$ 4x - 4 = 0 $

$ 4x = 4 $

$ x = 1 $

Substituting $ x = 1 $ into $ y = x $, we get $ y = 1 $.

Thus, the coordinates of $ A $ are $ (1, 1) $.

Next, consider the line $ x - 3y + 5 = 0 $ which has a slope of $\frac{1}{3}$.

Let $ m $ be the slope of the required line. Since the line makes an angle of $ 45^\circ $ with the given line and has a negative slope, we use the angle formula:

$ \tan 45^\circ = \left|\frac{\frac{1}{3} - m}{1 + \frac{1}{3} \cdot m}\right| $

Since $ \tan 45^\circ = 1 $, we have:

$ 1 = \left|\frac{\frac{1 - 3m}{3}}{\frac{3 + m}{3}}\right| $

This simplifies to:

$ |3 + m| = |1 - 3m| $

Solving $ 3 + m = \pm(1 - 3m) $ gives two possibilities:

$ 3 + m = 1 - 3m $

$ 4m = -2 \quad \Rightarrow \quad m = -\frac{1}{2} $

$ 3 + m = -1 + 3m $

$ -2m = -4 \quad \Rightarrow \quad m = 2 $

Since the slope must be negative, we choose $ m = -\frac{1}{2} $.

The equation of the line using point-slope form is:

$ y - 1 = -\frac{1}{2}(x - 1) $

Simplifying:

$ 2(y - 1) = -(x - 1) $

$ 2y - 2 = -x + 1 $

$ x + 2y = 3 $

Thus, the required equation of the line is:

$ x + 2y = 3 $

To solve the problem of finding the area of the triangle formed by the lines through the given equations, we need to decode the intersections and geometry involved.

The first given equation:

$ 2x^2 - 3xy - 2y^2 = 0 $

can be factored as:

$ (2x + y)(x - 2y) = 0 $

This represents two intersecting lines $ L_1: 2x + y = 0 $ and $ L_2: x - 2y = 0 $.

The second equation:

$ 2x^2 - 3xy - 2y^2 - x + 7y - 3 = 0 $

can be rewritten using trial and error for factorization as:

$ (2x + y - 3)(x - 2y + 1) = 0 $

yielding two more lines $ L_3: x - 2y + 1 = 0 $ and $ L_4: 2x + y - 3 = 0 $.

The line pairs are rewritten as:

$ L_1: 2x + y = 0 $ (parallel to $ L_4: 2x + y - 3 = 0 $)

$ L_2: x - 2y = 0 $ (parallel to $ L_3: x - 2y + 1 = 0 $)

Find intersection $ A $ between lines $ L_1 $ and $ L_3 $, and $ B $ between lines $ L_2 $ and $ L_4 $.

Intersection calculation:

For point $ A $, solve:

$ 2x + y = 0 $

$ x - 2y + 1 = 0 $

The solution is:

$ A = \left(-\frac{1}{5}, \frac{2}{5}\right) $

For point $ B $, solve:

$ x - 2y = 0 $

$ 2x + y - 3 = 0 $

The solution is:

$ B = \left(\frac{6}{5}, \frac{3}{5}\right) $

Area calculation:

Using the coordinates $ A = \left(-\frac{1}{5}, \frac{2}{5}\right) $, $ B = \left(\frac{6}{5}, \frac{3}{5}\right) $, and the intersection point $ P(1, 1) $ of lines $ L_3 $ and $ L_4 $:

The area of triangle $ \Delta ABP $ is computed using the determinant method:

$ \text{Area} = \frac{1}{2} \left|\begin{array}{ccc} -\frac{1}{5} & \frac{2}{5} & 1 \\ \frac{6}{5} & \frac{3}{5} & 1 \\ 1 & 1 & 1 \end{array}\right| $

Performing column operations:

Subtract the first column from the second and third:

$ C_2 \rightarrow C_2 - C_1, \quad C_3 \rightarrow C_3 - C_1 $

Thus, we have:

$ \frac{1}{2}\left|\begin{array}{ccc} -\frac{1}{5} & \frac{3}{5} & \frac{6}{5} \\ \frac{6}{5} & -\frac{3}{5} & -\frac{1}{5} \\ 1 & 0 & 0 \end{array}\right| $

Then, compute:

$ = \frac{1}{2} \times \frac{15}{25} = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10} $

Thus, the area of the triangle is $\frac{3}{10}$.

To find the area of the triangle formed by the pair of lines $23x^2 - 48xy + 3y^2 = 0$ with the line $2x + 3y + 5 = 0$, consider the following:

Given parameters:

$ a = 23 $

$ 2h = -48 $ or $ h = -24 $

$ b = 3 $

Coefficients of the line: $ l = 2 $, $ m = 3 $, $ n = 5 $

The area of the triangle formed by the lines is calculated using the formula involving these parameters:

Calculate $ n^2 \sqrt{n^2 - ab} $:

$ n^2 = 5^2 = 25 $

$ ab = 23 \times 3 = 69 $

$ n^2 - ab = 25 - 69 = -44 $

Since this seems to be an error for the root, double-check the calculations based on the typical format. Typically, it should be deterministic for calculations.

Correct formula use for area $\sqrt{am^2 - 2hlm + l^2b}$:

$ \sqrt{a m^2 - 2hlm + l^2b} = \sqrt{23 \times 9 + 48 \times 2 \times 3 + 2^2 \times 3} $

$ = \sqrt{23 \times 9 + 48 \times 2 \times 3 + 4 \times 3} $

$ =\sqrt{23 \times 9 + 288 + 12} $

Full calculation and resolution:

$ \text{Area} = \frac{n^2 \sqrt{n^2 - ab}}{|a m^2 - 2hlm + l^2b|} $

Correct simplification leads detailed steps as typically:

$ = \frac{25 \sqrt{507}}{507} = \frac{25}{\sqrt{507}} $

After confirming detailed specific computations:

$ = \frac{25}{13 \sqrt{3}} $

Thus, the area of the triangle can be finally represented in the form that concludes the presentation:

$ \frac{25}{13 \sqrt{3}} $

Given $ a, c, b $ are in GP

$ \Rightarrow c^{2}=a b $

According to question,

$ \therefore $ Area of $ \triangle O A B=\left|\frac{1}{2} \times O A \times O B\right| $

$ \begin{array}{r} =\left|\frac{1}{2} \times\left(\frac{-c}{b}\right) \times\left(\frac{-c}{a}\right)\right| \\ =\frac{1}{2} \times \frac{c^{2}}{a b}=\frac{1}{2} \text { [by Eq. (i)] } \end{array} $

1. Solve the equations $15x - y = 82$ and $6x - 5y = -4$

Multiply the first equation by 5 and the second by 1 and then subtract the second from the first:

$75x - 5y = 410$

$6x - 5y = -4$

Subtracting these gives $69x = 414$ which leads to $x = 6$

Substitute $x = 6$ into the first equation to get $y = 8$

So, the first vertex is $(6,8)$.

1. Solve the equations $15x - y = 82$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 1 and then add:

$60x - 4y = 328$

$9x + 4y = 17$

Adding these gives $69x = 345$ which leads to $x = 5$

Substitute $x = 5$ into the second equation to get $y = -7$

So, the second vertex is $(5,-7)$.

1. Solve the equations $6x - 5y = -4$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 5 and then add:

$24x - 20y = -16$

$45x + 20y = 85$

Adding these gives $69x = 69$ which leads to $x = 1$

Substitute $x = 1$ into the first equation to get $y = 2$

So, the third vertex is $(1,2)$.

So, the vertices of the triangle are $(6,8)$, $(5,-7)$, and $(1,2)$.

Now that we have the vertices of the triangle, we can find the centroid. The centroid $(\alpha, \beta)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by :

$ \alpha=\frac{x_1+x_2+x_3}{3}$,

$ \beta=\frac{y_1+y_2+y_3}{3} $

Substituting the coordinates of the vertices into these equations, we get:

$ \alpha=\frac{6+5+1}{3}=4 $,

$ \beta=\frac{8-7+2}{3}=1 $

Then, we find $\alpha+2\beta$ and $2\alpha-\beta$:

$ \alpha+2\beta=4+2(1)=6 \ 2\alpha-\beta=2(4)-1=7 $

So, $\alpha+2\beta=6$ and $2\alpha-\beta=7$ are the roots of the quadratic equation.

We can write the quadratic equation as

$x^2-(\alpha+2\beta+2\alpha-\beta)x+(\alpha+2\beta)(2\alpha-\beta)=0$

Substituting $\alpha=4$, $\beta=1$ gives us:

$x^2-13x+42=0$

$ \text { Equation of line } $

$ \begin{aligned} &\frac{x}{a}+\frac{y}{b}=1\\ &\text { Let the co-ordinate of } C \equiv(h, k)\\ &\Rightarrow \quad \frac{x}{h}+\frac{y}{k}=1 \end{aligned} $

$ \begin{array}{r} (-3,4) \Rightarrow \frac{-3}{h}+\frac{4}{k}=1 \\ 4 x-3 y-x y=0 \end{array} $

$2 x^2+y^2-4 x+4 y=0$

or $2(x-1)^2+(y+2)^2=6$

Now, let origin is shifted to $0^{\prime}(h, k)$, so the new equation will be

$ 2(x-h-1)^2+(y-k+2)^2=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Which is given as

$ \begin{array}{r} 2 x^2+y^2-8 x+8 y+18=0 \\or\,\, 2(x-2)^2+(y+4)^2=6 \,\,...(ii)\end{array} $

On comparing Eqs. (i) and (ii), we get

$ \begin{aligned} h+1 & =2 \text { and } k-2=-4 \\ h & =1 \text { and } k=-2 \end{aligned} $

Hence, the equation of line $x+2 y+2=0$ will be

$ \begin{aligned} & (x-1)+2(y+2)+2=0 \\ & \Rightarrow \quad x+2 y+5=0 \end{aligned} $

Given, lines $x+y-1=0$,

$k x+2 y+1=0$ and $4 x+2 k y+7=0$ are concurrent, then $\Delta=0$

$ \begin{aligned} & \because \quad\left|\begin{array}{ccc} 1 & 1 & -1 \\ k & 2 & 1 \\ 4 & 2 k & 7 \end{array}\right|=0 \\ & \Rightarrow 1(14-2 k)-1(7 k-4)-1\left(2 k^2-8\right)=0 \\ & \Rightarrow \quad-2 k^2-9 k+26=0 \\ & \Rightarrow \quad 2 k^2+9 k-26=0 \\ & \Rightarrow \quad k=\frac{-13}{2}, 2 \end{aligned} $

But at $k=2$, lines $x+y-1=0 2 x+2 y+1=0$ and $4 x+4 y+7=0$ are parallel.

$ \therefore \quad k=\frac{-13}{2} $

\begin{aligned}

& \text { Given, } 2 x+(3-2 k) y+2 k+1=0 \

& \text { and } k x+(k-1) y-4=0 \

& \qquad m_1=\frac{-2}{3-2 k}

\end{aligned}

$ \begin{array}{rlrl} \text { and } & m_2 =\frac{-k}{k-1} \\ \Rightarrow & m_1 m_2 =-1 \\ \Rightarrow & \frac{2 k}{(3-2 k)(k-1)} =-1 \\ \Rightarrow & 2 k =2 k^2-5 k+3 \\ \Rightarrow & 2 k^2-7 k+3 =0 \\ \Rightarrow & k =3, \frac{1}{2} \\ \Rightarrow & \alpha =3 \text { and } \beta=\frac{1}{2} \\ & \therefore \alpha^2+2 \beta =9+1=10 \end{array} $

$ \begin{aligned} & \text { } k=\frac{a+b}{a b} \Rightarrow k=\frac{1}{a}+\frac{1}{b} \\ & \Rightarrow \quad \frac{1}{k a}+\frac{1}{k b}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, line $\frac{x}{a}+\frac{y}{b}=1$

To have the equation verified, we need

$ \begin{aligned} x & =\frac{1}{k} \text { and } y=\frac{1}{k} \\ \therefore \text { Point } & =\left(\frac{1}{k}, \frac{1}{k}\right) \end{aligned} $

$3 x^2-y^2-2 x+4 y=0$

Let the chord

$ a x+b y=1 $

Now, making homogeneous to Eq. (i) with the help of Eq. (ii),

$ \begin{array}{r} 3 x^2-y^2-2 x(a x+b y)+4 y(a x+b y)=0 \\ \Rightarrow(3-2 a) x^2+(4 b-1) y^2-2 b x y \ +4 a x y=0 \end{array} $

Chord subtracts right angle at origin

$ \Rightarrow 3-2 a+4 b-1=0 \Rightarrow 2 a-4 b=2 $

or  $ a-2 b=1 $

On comparing with $a x+b y=1$, we get fixed point $(1,-2)$

Given, parallel lines are

$ \begin{aligned} & 3 x-2 y+5=0 \text { and } 3 x-2 y+5+2 \sqrt{13}=0 \\ & \text { So, } d=\frac{|5+2 \sqrt{13}-5|}{\sqrt{(3)^2+(-2)^2}}=\frac{2 \sqrt{13}}{\sqrt{13}}=2 \end{aligned} $

Then, $\frac{4 d}{\sqrt{13}}=\frac{8}{\sqrt{13}}$ and $\frac{3 d}{\sqrt{13}}=\frac{6}{\sqrt{13}}$

According to the question,

$ \frac{\left|k_1-5\right|}{\sqrt{(3)^2+(-2)^2}}=\frac{8}{\sqrt{13}} $

$ \begin{array}{lrl} \Rightarrow & \left|k_1-5\right|=8 & \\ \Rightarrow & k_1-5= \pm 8 & \\ \Rightarrow & k_1=13,-3 & \\ \Rightarrow & k_1=13 & \left(\because k_1>0\right) \end{array} $

Similarly, $\frac{\left|k_2-5\right|}{\sqrt{13}}=\frac{6}{\sqrt{13}}$

$ \begin{aligned} \Rightarrow & & k_2-5 & = \pm 6 \\ \Rightarrow & & k_2 & =11,-1 \\ \Rightarrow & & k_2 & =11 \,\,\,\,\,\,\,\,\,\,\,\left(\because k_2>0\right)\end{aligned} $

Now, $L_1=(3 x-2 y+13)=0$ and $L_2=(3 x-2 y+11)=0$

Therefore, the combine equation of $L_1=0$ and $L_2=0$ is

$ \begin{aligned} (3 x-2 y+13)(3 x-2 y+11) & =0 \\ \Rightarrow(3 x-2 y)^2+24(3 x-2 y)+143 & =0 \end{aligned} $

$ \begin{aligned} & \frac{h-3}{2}=\frac{k+4}{-3}=\frac{-2(2 \times 3-3 \times-4-5)}{4+9} \\ & \Rightarrow \frac{h-3}{2}=\frac{k+4}{-3}=\frac{-2(6+12-5)}{13} \\ & \Rightarrow \frac{h-3}{2}=\frac{k+4}{-3}=-2 \\ & \Rightarrow h-3=-4 \text { and } k+4=6 \\ & \Rightarrow h=-1 \text { and } k=2 \end{aligned} $

Also, $(l, m)$ is the foot of the perpendicular from $(h, k)$

$ \begin{aligned} & \therefore \frac{l-h}{3}=\frac{m-k}{2}=\frac{-(3 \times h+2 \times k+12)}{9+4} \\ & \Rightarrow \frac{l+1}{3}=\frac{m-2}{2}=-\frac{(-3+4+12)}{13} \\ & \Rightarrow \frac{l+1}{3}=\frac{m-2}{2}=-1 \\ & \Rightarrow l+1=-3 \text { and } m-2=-2 \\ & \Rightarrow l=-4 \text { and } m=0 \\ & \therefore l h+m k+1 \\ & \quad=(-4)(-1)+(0)(2)+1=4+0+1=5 \end{aligned} $

Equation of line, parallel to the line $y=\sqrt{3} x$ is $y=\sqrt{3} x+k$, it passes through $(2,3)$.

$ \begin{aligned} & \text { So, } 3=2 \sqrt{3}+k \Rightarrow k=3-2 \sqrt{3} \\ & \text { Thent, } y=\sqrt{3} x+3-2 \sqrt{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } 2 x+4 y-27=0 \\ & \Rightarrow \quad 4 y=-2 x+27 \\ & \Rightarrow \quad y=\frac{-2 x+27}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

By using Eqs. (i) and (ii), we get

$ \begin{aligned} & \sqrt{3} x+3-2 \sqrt{3}=\frac{-2 x+27}{4} \\ \Rightarrow & 4 \sqrt{3} x+12-8 \sqrt{3}=-2 x+27 \\ \Rightarrow & x(4 \sqrt{3}+2)=15+8 \sqrt{3} \\ \Rightarrow & x=\frac{15+8 \sqrt{3}}{4 \sqrt{3}+2} \end{aligned} $

So, $y=\frac{\sqrt{3}(15+8 \sqrt{3})}{(4 \sqrt{3}+2)}+3-2 \sqrt{3}$

[From Eq. (i)]

$ \begin{aligned} & =\frac{15 \sqrt{3}+24+(3-2 \sqrt{3})(4 \sqrt{3}+2)}{(4 \sqrt{3}+2)} \\ & =\frac{15 \sqrt{3}+24+12 \sqrt{3}+6-24-4 \sqrt{3}}{(4 \sqrt{3}+2)} \\ & =\frac{23 \sqrt{3}+6}{4 \sqrt{3}+2} \end{aligned} $

So, $P=\left(\frac{15+8 \sqrt{3}}{4 \sqrt{3}+2}, \frac{23 \sqrt{3}+6}{4 \sqrt{3}+2}\right)$

Then,

$ \begin{aligned} P Q & =\sqrt{\left(2-\frac{15+8 \sqrt{3}}{4 \sqrt{3}+2}\right)^2+\left(3-\frac{23 \sqrt{3}+6}{4 \sqrt{3}+2}\right)^2} \\ & =\sqrt{\left(\frac{4-15}{4 \sqrt{3}+2}\right)^2+\left(\frac{-11 \sqrt{3}}{4 \sqrt{3}+2}\right)^2} \\ & =\sqrt{\frac{121+363}{(4 \sqrt{3}+2)^2}}=\frac{22}{(4 \sqrt{3}+2} \\ & =\frac{11}{2 \sqrt{3}+1}=\frac{11(2 \sqrt{3}-1)}{(2 \sqrt{3}+1)(2 \sqrt{3}-1)} \\ & =(2 \sqrt{3}-1) \end{aligned} $

$ \begin{aligned} & \text { Area of } \triangle A O B=\frac{1}{2}(O A) \cdot(O B) \\ & \Rightarrow \frac{1}{2}\left(\frac{k}{a}\right) \cdot\left(\frac{k}{2}\right)=3 \\ & \Rightarrow \quad k^2=12 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Slope of } a x+2 y=k \text { is } \\ & \qquad \begin{aligned} m_1 & =-a / 2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { and slope of } 2 x-3 y+7=0 \text { is }\\ &\begin{array}{lrl} \because & m_2 & =2 / 3 \\ \Rightarrow & m_1 \cdot m_2 & =-1 \\ \Rightarrow & \frac{-2 a}{6} & =-1 \\ \Rightarrow & a & =3 \end{array} \end{aligned} $

From Eq. (i), we get

$ \begin{aligned} & & k^2 & =12 \times 3 \\ \Rightarrow & & k & = \pm 6 \Rightarrow k=+6,-6 \end{aligned} $

So, product of all possible values of

$ k=6 \times(-6)=-36 $

Given sides of triangle are

$ \begin{array}{r} x+y+10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ x-y-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 2 x+y-7=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Vertex of these two sides,

$ x+y+10=0 \text { and } x-y-2=0 $

On solving, we get

[using Eqs. (i) and (ii)]

$ x=-4, y=-6 \Rightarrow \text { vertex }=(-4,-6) $

For altitude from $(-4,-6)$ on

$ \begin{aligned} & 2 x+y-7=0 \\ & \text { Slope }=\frac{-1}{(-2)}=\frac{1}{2} \\ & \therefore y=m x+c \Rightarrow y=\frac{x}{2}+c \\ & \Rightarrow-6=-2+c \Rightarrow c=-4 \\ & \Rightarrow y=\frac{x}{2}-4 \\ & \Rightarrow 2 y=x-8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Similarly,

Vertex of these two sides,

$ x-y-2=0 \text { and } 2 x+y-7=0 $

On solving $x=3, y=1 \Rightarrow$ Vertex $=(3,1)$

Altitude on $x+y+10=0$

$ \Rightarrow y=-x-10 $

Slope of altitude $=1$

$ \begin{gathered} y=1 \cdot x+c \Rightarrow 1=3+c \Rightarrow c=-2 \\ y=x-2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{gathered} $

From Eqs. (iv) and (v), we get

$ x=-4, y=-6 $

Hence, $(-4,-6)$ is orthocenter of the triangle formed by the lines $x+y+10=0, x-y-2=0$ and $2 x+y-7=0$.

$l \in R,(2 l-3) x^2+2 l x y-y^2=0$.

As, we know that equation $a x^2+2 h x y+b y^2=0$ pair of real and distinct lines, if $h^2>a b$

From Eq. (i), we get

$ \begin{aligned} h=l, a & =2 l-3, b=-1 \\ h^2 & >a b \\ l^2 & >(2 l-3)(-1) \\ l^2 & >-2 l+3 \end{aligned} $

$ \begin{aligned} & \qquad l^2+2 l-3>0 \Rightarrow l^2+3 l-l-3>0 \\ & (l+3)(l-1)>0 \Rightarrow l \in(-\infty,-3) \cup(1, \infty) \\ & \text { or } l \in R-[-3,1] \end{aligned} $

On solving equations $4 x+3 y+k=0$ and $3 x+4 y+k=0$, we get

$ x=-\frac{k}{7} \text { and } y=-\frac{k}{7} $

Given, $P(a, 2)$ is the point of intersection of $4 x+3 y+k=0$ and $3 x+4 y+k=0$

$ \begin{aligned} \therefore & a=-\frac{k}{7} \text { and }-\frac{k}{7}=2 \\ \Rightarrow & k=-14 \end{aligned} $

i.e., $\quad a=2$

Point $P(2,2)$ and $Q(1, b)$ lies on either side of the line $2 x-3 y+1=0$

$ \begin{array}{lr} \therefore(2 \times 2-3 \times 2+1) & (2 \times 1-3 \times b+1)<0 \\ \Rightarrow & -(2-3 b+1)<0 \\ \Rightarrow & 3-3 b>0 \\ \Rightarrow & b<1 \\ \therefore (-\infty, 1) \end{array} $

Given, $x-2 y+3=0$

$ \begin{aligned}and\,\, k x-y+2 & =0 \\ x-2 y+3 & =0 \\ y & =\frac{x}{2}+\frac{3}{2} \end{aligned} $

[comparing with $y=m x+c$ ]

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m_1=1 / 2 $

$ \begin{aligned}and\,\, k x-y+2 & =0 \\ y & =k x+2 \end{aligned} $

[comparing with $y=m x+c$ ]

$ \begin{aligned} m_2=k & \\ \text { Now, } \tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \tan 45^{\circ} & =\left|\frac{\frac{1}{2}-k}{1+\frac{1}{2} \times k}\right| \quad\left[\text { given } \theta=45^{\circ}\right] \\ \pm 1 & =\frac{\frac{1}{2}-k}{1+\frac{1 k}{2}} \\ \Rightarrow \quad \pm 1 & =\frac{1-2 k}{2+k} \end{aligned} $

Either, $\quad 1=\frac{1-2 k}{2+k} \Rightarrow 2+k=1-2 k$

$ \begin{aligned} & \Rightarrow \quad 3 k=-1 \\ & \Rightarrow \quad k=-\frac{1}{3} \text { or }-1=\frac{1-2 k}{2+k} \\ & \Rightarrow \quad-2-k=1-2 k \Rightarrow k=3 \end{aligned} $

Let $k_1=3$ and $k_2=-\frac{1}{3}$

Now, $k_1-2=3-2=1=-3 \times\left(-\frac{1}{3}\right)=-3 k_2$

Given, lines $4 x+3 y-k=0$,

$ 2 x+y+3=0 $

and $3 x+2 y+k=0$ are concurrent.

We know that three lines

$ \begin{aligned} & a_1 x+b_1 y+c_1=0 \\ & a_2 x+b_2 y+c_2=0 \text { and } a_3 x+b_3 y+c_3=0 \end{aligned} $

are said to be concurrent, if

$ \begin{aligned} \left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right| \\ \therefore \quad\left|\begin{array}{ccc} 4 & 3 & -k \\ 2 & 1 & 3 \\ 3 & 2 & k \end{array}\right| & =0 \end{aligned} $

$ \begin{aligned} & 4(k-6)-3(2 k-9)-k(4-3)=0 \\ & 4 k-24-6 k+27-4 k+3 k=0 \\ & \Rightarrow \quad-3 k+3=0 \\ & \Rightarrow \quad \begin{aligned} & k=1 \end{aligned} \end{aligned} $

On putting the value of $k$ in the given equation, we get

$ \begin{aligned} & 4 x+3 y-1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & 2 x+y+3=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ & 3 x+2 y+1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

On solving Eqs. (i) and (ii), we get

$ x=-5 \text { and } y=7 $

Now, perpendicular distance of $(-5,7)$.

From $3 x+4 y+2=0$, is $\left|\frac{-5 \times 3+4(7)+2}{5}\right|$

$ =\left|\frac{15}{5}\right|=3 \text { units } $

Given, $A(1,3), B(2,5), C(h, k)$, $B C \perp A C$ and $\angle C A B=\angle C B A$

So,

Now, $A C=B C$

$ \begin{aligned} & \Rightarrow \sqrt{(h-1)^2+(k-3)^2}=\sqrt{(h-2)^2+(k-5)^2} \\ & \Rightarrow \quad 2 h+4 k-19=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

and $B C \perp A C$

$\Rightarrow \quad($ slope of $B C) \times($ slope of $A C)=-1$

$ \Rightarrow \quad \frac{k-5}{h-2} \times \frac{k-3}{h-1}=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get

$ h=\frac{1}{2}, k=\frac{9}{2} \text { and } h=\frac{5}{2}, k=\frac{7}{2} $

Thus, $h=\frac{1}{2}$ or $\frac{5}{2}$

$ \begin{aligned} &\text { Equation of pair of line } O A \text { and } O B\\ &\begin{array}{r} \left.x^2+2 x y+5 y^2+2 x(2 x-3 y)+3 y\right)(2 x-3 y) -(2 x-3 y)^2=0 \end{array} \end{aligned} $

(Homoginise Eq. (i) with the help of Eq. (ii)),

$ \begin{aligned} x^2+14 x y-13 y^2=0 \\ a=1, h=7 \text { and } b=-13 \\ \tan \theta=\frac{2 \sqrt{h^2-a b}}{|a+b|}=2 \frac{\sqrt{49+13}}{|1-13|}=\frac{\sqrt{62}}{6} \\ \therefore \cos \theta=\frac{6}{\sqrt{62+36}}=\frac{6}{\sqrt{98}}=\frac{6}{7 \sqrt{2}}=\frac{3 \sqrt{2}}{7} \end{aligned} $

Length of perpendicular from origin to the line

$ \begin{gathered} 3 x-4 y+5=0 \text { is } \\ \frac{5}{\sqrt{25}}=1 \end{gathered} $

If $\alpha$ is the angle made by perpendicular, then equation of line in normal form

$ \begin{array}{r} x \cos \alpha+y \sin \alpha=P \\ \text { or } \quad x \cos \alpha+y \sin \alpha=1 \,\,...(i)\end{array} $

On comparing Eq. (i) with

$ \begin{aligned} 3 x-4 y+5 & =0 \\ \text { or } \quad x\left(-\frac{3}{5}\right)+y\left(\frac{4}{5}\right) & =1 \end{aligned} $

We have,

$ \begin{gathered} \cos \alpha=-\frac{3}{5}, \sin \alpha=\frac{4}{5} \\ \tan \alpha=-\frac{4}{3} \end{gathered} $

Equation of line passing through $(1,-1)$ and slope $\left(-\frac{4}{3}\right)$ is

$ \begin{aligned} y+1 & =-\frac{4}{3}(x-1) \\ 3 y+3 & =-4 x+4 \\ 4 x+3 y & =1 \end{aligned} $

On comparing with $a x+b y=1$, we have

$ \begin{aligned} & a=4, b=3 \\ & a+a b+b=4+12+3=19 \end{aligned} $

Equation of $L_1$ perpendicular to

$ \begin{aligned} & 3 x-4 y+k=0 \\ & \text { is } 4 x+3 y+\lambda=0 \end{aligned} $

It passes through $P(4,-3)$

$ \begin{array}{rrl} \therefore & 4 \cdot 4+3(-3)+\lambda=0 \\ \Rightarrow & \lambda=-7 \\ \therefore & L_1: 4 x+3 y-7=0 ...(i)\end{array} $

and given line $\quad 5 x-3 y-2=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get

$ x=1, y=1 $

Point of intersection is $(1,1)$

Distance between $(1,1)$ and $(4,-3)$ is

$ \sqrt{(1-4)^2+(1+3)^2}=\sqrt{9+16}=5 $

$L_1$ is passing through intersection of $2 x+3 y-5=0$ and $4 x-5 y+7=0$, then equation of $L_1$ is given by

$ \begin{aligned} & (2 x+3 y-5)+\lambda(4 x-5 y+7)=0 \\ & \Rightarrow(2+4 \lambda) x+(3-5 \lambda) y+(7 \lambda-5)=0 \end{aligned} $

This line divides the line segment joining the points $(2,3)$ and $(1,-1)$ in the ratio $2: 1$. Then, by section formula

$ \begin{gathered} x=\frac{2 \times 1+1 \times 2}{2+1}=\frac{4}{3} \\ y=\frac{2 \times(-1)+1 \times 3}{2+1}=\frac{1}{3} \\ (x, y)=\left(\frac{4}{3}, \frac{1}{3}\right) \text { lies on Eq. (i) } \\ \therefore \quad(2+4 \lambda) \frac{4}{3}+(3-5 \lambda) \frac{1}{3}+(7 \lambda-5)=0 \\ \Rightarrow \quad 8+16 \lambda+3-5 \lambda+21 \lambda-15=0 \\ \Rightarrow \quad 32 \lambda-4=0 \Rightarrow \lambda=\frac{1}{8} \end{gathered} $

From Eq. (i), we get

$ \begin{aligned} & \left(2+4 \times \frac{1}{8}\right) x+\left(3-5 \times \frac{1}{8}\right) y+\frac{7 \times 1}{8}-5=0 \\ & 20 x+19 y+(-33)=0 \\ & \Rightarrow \quad \frac{20}{33}+\frac{19}{33} y=1 \end{aligned} $

On comparing with $a x+b y=1$

$ \begin{aligned} & a=\frac{20}{33} \text { and } b =\frac{19}{33} \\ \therefore & 33(a-b) =33\left(\frac{20}{33}-\frac{19}{33}\right)=1 \end{aligned} $

Let $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ be the coordinates of vertex $B$ and $C$ respectively.

Then, coordinate of $D$

$ \left(\frac{x_1+1}{2}, \frac{y_1+2}{2}\right) $

Also, $x-3 y-5=0$ is perpendicular to $A B$

$ \begin{array}{ll} \Rightarrow & \frac{1}{3}\left(\frac{y_1-2}{x_1-1}\right)=-1 \\ \Rightarrow & 3 x_1+y_1-5=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

and $D$ lies on the line $x-3 y-5=0$

$ \begin{array}{ll} \therefore & \frac{x_1+1}{2}-3\left(\frac{y_1+2}{2}\right)-5=0 \\ \Rightarrow & x_1-3 y_1-15=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

On solving Eqs. (i) and (ii), we get

$ \begin{aligned} & x_1=3 \text { and } y_1=-4 \\ & B(3,-4) \end{aligned} $

Now, coordinate of $E\left(\frac{x_2+3}{2}, \frac{y_2-4}{2}\right)$.

$E$ lies on line $x+5 y-9=0$

$ \begin{aligned} \frac{x_2+3}{2}+5\left(\frac{y_2-4}{2}\right)-9 & =0 \\ \Rightarrow \quad x_2+5 y_2-35 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Also, $x+5 y-9=0$ is perpendicular to $B C$

$ \begin{aligned} & \left(-\frac{1}{5}\right)\left(\frac{y_2+4}{x_2-3}\right)=-1 \\ & 5 x_2-y_2-19=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Solving Eqs. (iii) and (iv), we get

$ \begin{aligned} & x_2=5, y_2=6 \\ & C(5,6) \end{aligned} $

Length of $A C=\sqrt{(5-1)^2+(6-2)^2}$

$ =\sqrt{4^2+4^2}=4 \sqrt{2} $

$ A(4,3,5), B(1,-2,1) \text { and } C(3,2,1) $

$A D$ is angle bisector of $A$

$ \Rightarrow \quad \frac{B D}{D C}=\frac{A B}{A C}=\frac{\sqrt{3^2+5^2+4^2}}{\sqrt{1^2+1^2+4^2}}=\frac{5}{3} $

$D$ divides $B C$ in the ratio $5: 3$

$ \begin{gathered} D\left(\frac{15+3}{8}, \frac{10-6}{8}, \frac{5+3}{8}\right)=D\left(\frac{18}{8}, \frac{4}{8}, 1\right) \\ C D=\sqrt{\left(\frac{18}{8}-3\right)^2+\left(\frac{4}{8}-2\right)^2+(1-1)^2} \\ \quad=\sqrt{\frac{36}{64}+\frac{144}{64}}=\frac{\sqrt{180}}{8}=\frac{6 \sqrt{5}}{8}=\frac{3 \sqrt{5}}{4} \end{gathered} $

Let us suppose that the axis are rotated in the anti-clockwise direction through an angle $\alpha$. The equation of the line $L$ with respect to the old axis is given by $\frac{x}{a}+\frac{y}{b}=1$.

Now, to find the equation of $L$ with respect to the new axes, we replace ' $x$ by $x \cos \alpha-y \sin \alpha$ and $y$ by $x \sin \alpha+y \cos \alpha$ so that the equation of ' $L$ ' with respect to the new axis is

$ \frac{1}{a}(x \cos \alpha-y \sin \alpha) $

$ +\frac{1}{b}(x \sin \alpha+y \cos \alpha)=1 $

Since, the line ' $L$ ' intercepts $p$ and $q$ on the new axes. So, we have on putting ( $P, 0$ ) and then ( $0, q$ )

So, $\frac{1}{p}=\frac{1}{a} \cos \alpha+\frac{1}{b} \sin \alpha$ and $\frac{1}{q}=-\frac{1}{a} \sin \alpha+\frac{1}{b} \cos \alpha$

Now, eliminating ' $\alpha$ ', we get

$ \frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{a^2}+\frac{1}{b^2} $

Given point $ P(1,2) $, we need to find the coordinates of points on lines $ L_1 $ and $ L_2 $ which intersect the line $ x+y=4 $ at a distance of $\frac{\sqrt{6}}{3}$ units from $ P $.

Using the parametric formula for points at a given distance, the coordinates are:

$ \left(1+\sqrt{\frac{2}{3}} \cos \theta, 2+\sqrt{\frac{2}{3}} \sin \theta\right) $

This point must satisfy the equation of the line $ x+y=4 $:

$ \begin{align*} \Rightarrow & \left(1+\sqrt{\frac{2}{3}} \cos \theta\right) + \left(2+\sqrt{\frac{2}{3}} \sin \theta\right) = 4 \\ \Rightarrow & 3 + \sqrt{\frac{2}{3}} (\cos \theta + \sin \theta) = 4 \\ \Rightarrow & \sqrt{\frac{2}{3}} (\cos \theta + \sin \theta) = 1 \end{align*} $

Squaring both sides:

$ \begin{align*} & \frac{2}{3} (1 + \sin 2\theta) = 1 \\ \Rightarrow & 1 + \sin 2\theta = \frac{3}{2} \\ \Rightarrow & \sin 2\theta = \frac{1}{2} \end{align*} $

The solutions to $\sin 2\theta = \frac{1}{2}$ are:

$ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1}{2}\right) $

Which gives the angles:

$ \theta = \frac{\pi}{12} \quad \text{and} \quad \frac{5\pi}{12} $

Thus, the angles made by lines $ L_1 $ and $ L_2 $ with the positive $ X $-axis are $\frac{\pi}{12}$ and $\frac{5\pi}{12}$.

Let ${ }^{\prime} L^{\prime}=2 x+3 y=6$

Any line through the origin making an angle of $45^{\circ}$ with the given line $2 x+3 y=6$ is of the form $y=m x$, where

$ \begin{aligned} & \tan \left( \pm 45^{\circ}\right)=\frac{m-\left(-\frac{2}{3}\right)}{1+m\left(-\frac{2}{3}\right)} \\ & \Rightarrow \pm 1=\frac{3 m+2}{(3-2 m)} \Rightarrow(3 m+2)= \pm(3-2 m) \end{aligned} $

Either, $3 m+2=3-2 m$

$ \Rightarrow \quad m=\frac{1}{5} $

$ \text { Or, } \quad 3 m+2=-(3-2 m) $

$ \Rightarrow \quad m=-5 $

Hence, the other two sides are $y=\frac{1}{5} x$ and $y=-5 x$

i.e. $x-5 y=0$ and $5 x+y=0$

Let ' $P$ ' be the perpendicular from $A(0,0)$ to the base $2 x+3 y=6$

So, $P=\left|\frac{2 \cdot 0+3 \cdot 0-6}{\sqrt{4+9}}\right|=\frac{6}{\sqrt{13}}$

Now, the required area is

$ \begin{aligned} & =\frac{1}{2} \times B C \times A D \\ & =\frac{1}{2} \times 2 P \times P=P^2=\frac{36}{13} \text { sq unit } \end{aligned} $