Given intercepts are  -5, \(\frac 34\)
\(a=-5, b=\frac{3}{4}\)
Equation of the line in the intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{-5}+\frac{y}{\left(\frac{3}{4}\right)}=1 \)
\(\frac{x}{-5}+\frac{4 y}{3}=1 \)
3x - 20y = -15
3x - 20y + 15 = 0

After how many seconds, the song will be downloaded completely.
Remaining megabytes is 0
Y = 0.1 x + 1
0 = -0.1 x + 1
0.1 x = 1
\(x=\frac{1}{0.1}=10 \text { seconds. }\)

Use the slope and y intercept to write an equation 
The equation is y =  \(\frac { 9 }{ 5 } x\) + 32

(a) Slope of BC = \(\frac { 1+2 }{ 5-6 } =\frac { 3 }{ -1 } =3\)
(b) Slope of AD = \(\frac { 1+2 }{ 2-1 } =\frac { 3 }{ 1 } =3\)

Given vertices are A (- 5, - 4), B (1, 6) and C (7, - 4)
Area of triangle \(=\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\right. \left.\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \text { sq. units } \)
Area of triangle ABC \(=\frac{1}{2}[-5(6+4)+ 1(-4+4)+7(-4-6)] \)
\(=\frac{1}{2}[-50+0-70]=\frac{-120}{2}=-60\)
[Area cannot be negative].
Area = 60 sq. units.
Given that one bucket of paint can be applied for 6 sq. feet
No. of buckets \(=\frac{60}{6}=10\)

Given vertices of a rectangle are P (-1, -1),
Q(- 1, 4),R (5, 4) and S (5, -1)
Mid point of PQ = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)
\(=A\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right)=A\left(-1, \frac{3}{2}\right)\)
Mid point of QR = \(B\left(\frac{-1+5}{2}, \frac{4+4}{2}\right)=B(2,4) \)
Mid point of RS = \(C\left(\frac{5+5}{2}, \frac{4-1}{2}\right)=C\left(5, \frac{3}{2}\right) \)
Mid point of SP = \(D\left(\frac{-1+5}{2}, \frac{-1-1}{2}\right)=D(2,-1) \)
\(\mathrm{AB} =\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\(=\sqrt{(2+1)^{2}+(4-3 / 2)^{2}}=\sqrt{9+\frac{25}{4}}=\frac{\sqrt{61}}{2} \)
\(\mathrm{BC}=\sqrt{(5-2)^{2}+(3 / 2-4)^{2}}=\frac{\sqrt{61}}{2} \)
\(\mathrm{CD}=\sqrt{(2-5)^{2}+(-1-3 / 2)^{2}}=\frac{\sqrt{61}}{2} \)
\(\mathrm{DA}=\sqrt{(2+1)^{2}+(-1-3 / 2)^{2}}=\frac{\sqrt{61}}{2} \)
From triangle ABC,
\( A B^{2}+B C^{2}=\frac{61}{4}+\frac{61}{4}=\frac{122}{4}=\frac{61}{2} \neq 36 \)
\(\text { i.e., } \mathrm{AB}^{2}+\mathrm{BC}^{2} \neq \mathrm{AC} \)
The points A, B, C and D Cannot be the vertices of a square or Rectangle.
Hence, ABCD is a Rhombus.

Let us solve 8x + 3y = 18 and 4x + 5y = 9
8x + 3y = 18         ....(1)
4x + 5y = 9         ....(2)  

Substitute in (1)
8x + 3(0) = 18

The point of intersection of (1) and (2) is \(\left( \frac { 9 }{ 4 } ,0 \right) \)
Mid point of the line segment joining (5, -4) and (-7,6) is
\(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{5-7}{2}, \frac{-4+6}{2}\right)=(-1,1)\)
Equation of the straight line joining \(\left( \frac { 9 }{ 4 } ,0 \right) \) and (-1, 1) is
\(\frac{y-y_{1}}{y_{2}-y_{1}} =\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-0}{1-0} =\frac{x-\frac{9}{4}}{-1-\frac{9}{4}} \)
\(\frac{y}{1} =\frac{4 x-9}{-13} \)
-13y = 4x - 9
4x + 13y - 9 = 0

Given equation is y = - 0.1 x + 1 where 'x' is time (in seconds) and 'y' is percentage of megabytes remaining.
Graph of y = -0.1x + 1 
\(y=-\frac{x}{10}+1\)
10y = -x + 10
Points to be plotted

(i) East Avenue is the straight line joining C(0, 2) and B(7, 2). Thus the equation of East Avenue is obtained by using two-point form which is
\(\frac { y-2 }{ 2-2 } =\frac { x-0 }{ 7-0 } \)
\(\frac { y-2 }{ 0 } =\frac { x }{ 7 } \) gives y = 2
(ii) Since the point D lie vertically above C(0, 2). The x coordinate of D is 0. Since any point on North Street has x coordinate value 0.
Therefore, the equation of North Street is x = 0
(iii) To find equation of Cross Road.
Center of circular garden M is at (7, 7), A is (3, 10)
We first find slope of MA, which we call m₁
Thus m₁ = \(\frac { 10-7 }{ 3-7 } =\frac { -3 }{ 4 } \)
Since the Cross Road is perpendicular to MA, if m₂ is the slope of the Cross Road then,
m₁m₂ = -1 gives \(\frac { { -3 } }{ 4 } \) m₂ = -1 so m₂ = \(\frac { 4 }{ 3 } \)
Now, the cross road has slope \(\frac { 4 }{ 3 } \) and it passes through the point A(3, 10).
The equation of the Cross Road is y - 10 = \(\frac { 4 }{ 3 } \) (x - 3)
3y - 30 = 4x - 12
Hence, 4x - 3y + 18 = 0