(4)

Equation of AC : $y - 3 = \frac{-4}{4} (x + 1)$

$\Rightarrow x + y = 2$

Equation AB : $y - 2 = \frac{1}{1} (x + 2)$

$\Rightarrow x - y + 4 = 0$

Equation BC : $y + 1 = \frac{-3}{5} (x - 3)$

$\Rightarrow 3x + 5y = 4$

Distance of AC from origin O(0, 0) = $\frac{|-2|}{\sqrt{2}} = \sqrt{2}$

Distance of AB from origin O(0, 0) = $\frac{\left|4\right|}{\sqrt{2}} = 2\sqrt{2}$

Distance of BC from origin O(0, 0) = $\frac{|-4|}{\sqrt{9 + 25}} = \frac{4}{\sqrt{34}}$

When each side of $∆ABC$ shifted by one unit inwards distance of new sides from origin are $\sqrt{2} - 1, 2\sqrt{2} - 1, \frac{4}{\sqrt{34}} + 1$ respectively.

Clearly new position of AC is nearest to the origin.

Equation of AC when shifted by one unit inwareds can be chosen as x + y = c

Thus, we have $\frac{|c - 2|}{\sqrt{2}} = 1 \Rightarrow c = 2 - \sqrt{2}$

$\therefore$  Required equation is $x + y - 2 + \sqrt{2} = 0$.

(3)

Let $P(r \cos \theta , r \sin \theta )$ and $Q(-r \sin \theta , r \cos \theta )$

$\therefore O{P}^2 = O{Q}^2 = r^2$

Now, in $△POQ, \angle POQ = 90°, P{Q}^2 = 2r^2$

Now, $l = O{P}^2 + P{Q}^2 + O{Q}^2 = 4r^2$

$\because$   P and Q lies on 3x + 4y = 12

$\therefore 3(r \cos \theta ) + 4(r \sin \theta ) = 12$

$\Rightarrow 3 \cos \theta + 4 \sin \theta = \frac{12}{r}$            ... (i)

and $3(-r \sin \theta ) + 4(r \cos \theta ) = 12$

$\Rightarrow -3 \sin \theta + 4 \cos \theta = \frac{12}{r}$          ... (ii)

From (i) and (ii), we have

$25 = \frac{288}{r^2} \Rightarrow r^2 = \frac{288}{25}$

Now, $l = 4r^2 = 4 \times \frac{288}{25} = 46.08$

So, $\left[l\right] =46$.

(4)

Let coordinates of P be (h, k).

Area of $△PAB$ = 10sq. units

$\Rightarrow 10 = \frac{1}{2}$$|\begin{array}{c}h\\ -1\\ 2\end{array} \begin{array}{c}k\\ 1\\ 3\end{array} \begin{array}{c}1\\ 1\\ 1\end{array}|$

$\Rightarrow \pm 20 = h(-2) - k(-3) + 1(-5)$

$\Rightarrow -2h + 3k = 25$

($\because$  P is above the line AB, so we take only +ve value)

$\Rightarrow \frac{-2 \times 3}{5} h + \frac{3 \times 3}{5} k = \frac{3 \times 25}{5}$

$\Rightarrow \frac{-6}{5} h + \frac{9}{5} k = 15$

On comparing it with ax + by = 15, we get $a = \frac{-6}{5}$ and $b = \frac{9}{5}$

.$5a + 2b = - 6 + \frac{18}{5} = \frac{-12}{5}$

(1)

Let $D(2, \frac{-4}{3})$ be the point on BC which divides BC in ratio 2 : 1.

Let coordinates of B be $(x_1, 14 - 4x_1)$ and coordinates of C be $(x_2, \frac{3x_2 - 5}{2})$

Now, $\frac{2x_2 + x_1}{3} = 2$, $\frac{2\left({\displaystyle \frac{3x_2 - 5}{2}}\right) + 14 - 4x_1}{3} = \frac{-4}{3}$

$\Rightarrow 2x_2 + x_1 = 6, 3x_2 - 4x_1 = -4 + 5 - 14$

$\Rightarrow 2x_2 + x_1 = 6, 3x_2 - 4x_1 = -13$

On solving these two equations, we get, $x_1 = 4, x_2 = 1$

So B(4, –2) and C(1, –1) are the required points.

Equation of BC : $y+1=\frac{-1}{3}(x-1)\Rightarrow 3y+x+2=0$

(1)

Lines $L_1$ and $L_2$ trisect the line 4x + 5y = 20.

$x_1 = \frac{5 \times 1 + 0 \times 2}{1 + 2} = \frac{5}{3}$

$y_1 = \frac{0 \times 1 + 4 \times 2}{1 + 2} = \frac{8}{3}$

$(x_1, y_1) \equiv (\frac{5}{3}, \frac{8}{3})$

Similarly,

$x_2 = \frac{0 \times 1 + 5 \times 2}{1 + 2} = \frac{10}{3}$

$y_2 = \frac{4 \times 1 + 0 \times 2}{1 + 2} = \frac{4}{3}, (x_2, y_2) \equiv (\frac{10}{3}, \frac{4}{3})$

Slope of line $L_1 : m_1 = \frac{8}{3} \times \frac{3}{5} = \frac{8}{5}$

Slope of line $L_2 : m_2 = \frac{4}{3} \times \frac{3}{10} = \frac{2}{5}$

Tangent angle between the lines $L_1$ and $L_2$:

$\tan \theta =$$\left|\frac{m_1 - m_2}{1 + m_1{m}_2}\right|$$=$$\left|\frac{{\displaystyle \frac{8}{5} - \frac{2}{5}}}{1 + {\displaystyle \frac{8}{5}} \times {\displaystyle \frac{2}{5}}}\right|$$= \frac{30}{41}$

(4)

It is given that, region R lies between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0.

The point $(a^2, a + 1)$ and (0, 0) lie in the region R.

$\Rightarrow (a^2, a + 1)$ and (0, 0) are on same side of both the line.

For Line 3x – y + 1 = 0, O(0, 0) is on the right side of the line.

So, point $(a^2, a + 1)$ will also be on right side of the line.

$\Rightarrow 3a^2 - a - 1 + 1 > 0$

$\Rightarrow a(3a - 1) > 0$

$\Rightarrow a \in (-\infty , 0) \cup (\frac{1}{3}, \infty )$          ... (i)

For line x + 2y – 5 = 0, O(0, 0) is on the left side of the line.

So, point $(a^2, a + 1)$ will also be on left side of the line.

$\Rightarrow a^2 + 2a + 2 - 5 < 0$

$\Rightarrow (a + 3)(a - 1) < 0$

$\Rightarrow a \in (-3, 1)$         ... (ii)

From the intersection of (i) and (ii), we get

$a \in (-3, 0) \cup (\frac{1}{3}, 1)$

(2)

We have, y = x          ... (i)

and    2x – y = 2          ... (ii)

Solving (i) and (ii), we get x = 2 and y = 2

As, $△ABD \widetilde{=} △CBD$

$\therefore \frac{AB}{BC} = \frac{AD}{DC} = \frac{1}{2}$

$D (\frac{\lambda + 8}{3}, \frac{2\lambda - 2 + 12}{3}) \mathrm{lies} \mathrm{on} y = x$

$\therefore \frac{\lambda + 8}{3} = \frac{2\lambda + 10}{3} \Rightarrow \lambda = -2. So, C\equiv (-2, -6)$

The image of the point A will lies on the line BC.

Let A' = (6, 4)

Now, $\frac{\beta - 4}{\alpha - 6} = \frac{10}{8} \Rightarrow \frac{\alpha - 4}{\alpha - 6} = \frac{5}{4} (\because \alpha = \beta )$

$\Rightarrow 4\alpha - 16 = 5\alpha - 30 \Rightarrow 14 = \alpha = \beta$

$\therefore \alpha + 2\beta = 14 + 2 \times 14 = 42$

(1)

Slope of a line passing through (2, 3) and parallel to line $\sqrt{3}x - y + 1 = 0$ is same as that of line $\sqrt{3}x - y + 1 = 0$

So, slope of line = $\sqrt{3}$

$\Rightarrow \tan \theta = \sqrt{3} \Rightarrow \sin \theta = \frac{\sqrt{3}}{2}, \cos \theta = \frac{1}{2}$

So, equation of line passing through (2, 3) and having slope $\sqrt{3}$ in normal form is,

$\frac{x - 2}{1/2} = \frac{y - 3}{\sqrt{3}/2} = r \Rightarrow x =\frac{r}{2} + 2, y = \frac{\sqrt{3}}{2} r + 3$

This line passes through 2x – 3y + 28 = 0

$\therefore 2(\frac{r}{2} + 2) - 3(\frac{\sqrt{3}}{2} r + 3) + 28 = 0 \Rightarrow \left(\frac{2 - 3\sqrt{3}}{2}\right) r = -23$

$\Rightarrow r = \frac{-23 \times 2}{2 - 3\sqrt{3}} \times \frac{2 + 3\sqrt{3}}{2 + 3\sqrt{3}} \Rightarrow r = 2(2 + 3\sqrt{3}) = 4 + 6\sqrt{3}$

(4)

Point of intersection of lines 3x + 2y = 14 and 5x – y = 6 is A(2, 4)

Point of intersection of lines 4x + 3y = 8 and 6x + y = 5 is $B = (\frac{1}{2}, 2)$

So, equation of line AB is, $y - 4 = \frac{2 - 4}{{\displaystyle \frac{1}{2}} - 2} (x - 2)$

$\Rightarrow y - 4 = \frac{4}{{\displaystyle 3}} (x - 2) \Rightarrow 4x - 3y + 4 = 0$

$\therefore$  Distance of point P(5, –2) from 4x – 3y + 4 is given as

$d =$$\left|\frac{4 \times 5 - 3 \times (-2) + 4}{\sqrt{16 + 9}}\right|$$\Rightarrow d = 6 \mathrm{units}$

(2)

Let $l$ be the line passing through point A(9, 0) making angle 30° with x-axis.

Line is rotated clockwise by 15° then $l\text{'}$ is the new position of line where it make 15° angle with x-axis.

So, equation of line passing through (9, 0) and making angle 15° with x-axis is (y – 0) = tan 15° (x – 9)

$\Rightarrow$   y = tan (45° – 30°)(x – 9) = $\left(\frac{\tan 45° – \tan 30°}{1 + \tan 45° · \tan 30°}\right) (x – 9)$

$\Rightarrow y = (2 – \sqrt{3}) (x – 9)$

$\Rightarrow \frac{y}{2 – \sqrt{3}} = x – 9 \Rightarrow \frac{y}{\sqrt{3} – 2} + x = 9$