\(3 x-5 y+7=0\)
\(\text { Slope } m_{1}=\frac{-\text { coefficient of } x}{\text { coefficient of } y}\)
\(=\frac{-3 x}{-5} \Rightarrow \frac{3}{5}\)
\(15 x+9 y+4=0\)
\(\text { Slope } m_{2}=\frac{-\text { coefficient of } x}{\text { coefficient of } y}\)
\(=\frac{-15}{9}\)
\(\therefore \mathrm{m}_{1} \times \mathrm{m}_{2}\)
\(\frac{3}{5} \times \frac{-15}{9}=-1\)
Hence proved

\(\text { point }=(-4,3) ; \mathrm{m}=\frac{-7}{5}\)
\(y-y_{1} =m\left(x-x_{1}\right) \)
\(y-3 =\frac{-7}{5}(x+4) \)
\(5(y-3) =-7(x+4) \)
\(5 y-15 =-7 x-28 \)
\(7 x+5 y-15+28 =0 \)
\(7 x+5 y+13 =0 \)

x = a is parallel to Y-axis and \(\mathrm{by}+\mathrm{c}=0 \Rightarrow \mathrm{y}=-\frac{c}{b} \text { is parallel to } \mathrm{X} \text {-axis }\)
\(\therefore \text { The angle between the two lines is } 90^{\circ} \text {. }\)

\(\text { Slope of the line } 4 x+3 y+5=0 \text { is }-\frac{4}{3}=m_{1}\)
Slope of the line joining (a, 2a) and(- 2, 3) is
\(\frac{2 a-3}{a+2}=m_{2}\)
Since, the lines are perpendicular, m₁ x m₂ = -1
\(\Rightarrow -\frac{4}{3} \times \frac{2 a-3}{a+2}=-1 \)
\(\Rightarrow 8 \mathrm{a}-12=3 \mathrm{a}+6 \)
\(\Rightarrow \mathrm{a}=\frac{18}{5} \)

The two lines are x(2 + 6k) + y(3 - k) + 4 + 12k = 0 and 7x + 5y- 4 = 0
\(\text { Slope of the first line }=m_{1}=-\frac{(2+6 k)}{3-k}\)
\(\text { Slope of the second line }=m_{2}=-\frac{7}{5}\)
\(\text { The lines are perpendicular, } \therefore \mathrm{m}_{1} \times \mathrm{m}_{2}=-1\)
\(-\frac{(2+6 k)}{3-k} x-\frac{7}{5} =-1 \)
\(14+42 \mathrm{k} =-15+5 \mathrm{k} \)
\(\mathrm{k} =-\frac{29}{37} \)

Equation of straight line is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\Rightarrow \frac{x}{4}+\frac{y}{-\frac{3}{2}}=1\)
\(\Rightarrow \frac{x}{4}-\frac{2 y}{3}=1\)
\(\Rightarrow 3 x-8 y-12=0\)

Given that the line makes an angle 30^o. with y-axis. i.e., the line makes 60^o with the positive direction of  X-axis.
Slope of the line = m = tan 60^o = \(\sqrt{3}\)
Equation of the line passing through the point (x₁, y₁,) and having slope 'm' is y - y₁, = m (x - x₁)
\(\Rightarrow y-2 =\sqrt{3}(x-1) \)
\(\Rightarrow y-2 =\sqrt{3} x-\sqrt{3} \)
\(\Rightarrow \sqrt{3} x-y+2-\sqrt{3}=0 \)

Equation of the straight line passing through the points (x₁, y₁) and (x₂, y₂) is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
\(\Rightarrow \frac{y-b}{a-b-b}=\frac{x-a}{a+b-a}\)
\(\Rightarrow \frac{y-b}{a-2 b}=\frac{x-a}{b}\)
\(\Rightarrow b(y-b)=(x-a)(a-2 b) \)
\(\Rightarrow b y-b^{2}=a x-2 b x-a^{2}+2 a b \)
\(\Rightarrow(a-2 b) x-b y-a^{2}+2 a b+b^{2}=0 \)

\(\text { Slope of } \mathrm{AB}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
\(=\frac{3-6}{2-5}=\frac{-3}{-3}=1\)
\(\text { Slope of } \mathrm{BC}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
\(=\frac{6+2}{5-0}=\frac{8}{5}\)
Slope of AB \(\ne\)Slope of BC
\(\therefore\) The points are not collinear.

When the lines are parallel, their slopes are equal Slope of the required line = 2.

Slope of the line joining two points (x₁, y₁) and (x₂, y₂) is
\(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{0-4}{2-3}=\frac{-4}{-1}=4=m\)

(i) Angle of inclination \(\theta\) = 45^o
Slope m = tan\(\theta\)
m = tan 45^o = 1
(ii) When \(\theta\) = 60^o,
slope m = tan 60^o = \(\sqrt{3}\)

\(\text { Area of Quadrilateral }=\frac{1}{2}\left[\left(\mathrm{x}_{1}-\mathrm{x}_{3}\right)\left(\mathrm{y}_{2}-\mathrm{y}_{4}\right)-\right.\left.\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right] \text { sq. units }\)
\(=\frac{1}{2}[(-1-5)(4+1)-(-1-5)(-1-4)\)
\(=\frac{1}{2}[-30-30] \)
\(=-\frac{60}{2}=-30 \)
Area = 30 sq. units
[\(\because\) area cannot be negative.]

For collinear points, Area of triangle is zero.
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
\(7(1+k)+5(-k+2)+3(-2-1)=0\)
\(7+7 \mathrm{k}-5 \mathrm{k}+10-9=0\)
\(2 \mathrm{k}=-8\)
\( \mathrm{k}=-4\)

\(\text { Area of triangle }=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+\right.\left.\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \text { sq. units }\)
\(=\frac{1}{2}[0(4-3)+3(3-0)+0(0-4)] \)
\(=\frac{1}{2}(9)=\frac{9}{2} \text { sq. units } \)

\(\text { Area of triangle }=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+\right.\left.\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \text { sq. units }\)
\(=\frac{1}{2}[1(3-5)+2(5-1)+4(1-3)] \)
\(=\frac{1}{2}[-2+8-8] \)
\(=\frac{-2}{2}=-1 \)
\(=1 \mathrm{sq} . \text { unit }\) \([\because \text{area cannot be negative}].\)

By joining B to D, you will get too triangles ABD and BCD.
Now, the area of ΔABD
=\(\frac { 1 }{ 2 } \)[ -5(-5 - 5) + (-4)(5 -7) + 4(7 + 5)]
=\(\frac { 1 }{ 2 } \)(50 + 8 + 48)
=\(\frac { 106 }{ 2 } \) = 53 square units.
Also, the area of ΔBCD
=\(\frac { 1 }{ 2 } \) = [-4(-6 - 5) - 1(5 + 5) + 4(-5 + 6)]
=\(\frac { 1 }{ 2 } \)(44 - 10 + 4)
= 19 square units.
So, the area of quadrilateral ABCD
= 53 + 19 = 72 square units.

Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given paints. One way at showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonals should be equal.
Now,
AB = \(\sqrt { (1-4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
BC = \(\\ \sqrt { (4+1)^{ 2 }+(2+1)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
CD = \(\sqrt { (-1+4)^{ 2 }+(-1-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
DA =\(\sqrt { (1+4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
AC = \(\sqrt { (1+1)^{ 2 }+(7+1)^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 } \)
BD = \(\\ \sqrt { (4+4)^{ 2 }+(2-4)^{ 2 } } =\sqrt { 64+4 } =\sqrt { 68 } \)
Since, AB = BC = CD = DA and AC = BD, all the four sides at the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

Let P(x, y) be equidistant from the points A (7, 1) and B (3, 5).
We are given that AP = BP. So, AP² = BP²
(x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
x²- 14x + 49 + y - 2y + 1 = x²- 6x + 9 + y -10y + 25
x - y = 2
Which is the required relation.