(i) If D is (0,k) then D is a point on the Cross Road.
Therefore, substituting x = 0, y = k in the equation of Cross Road,
we get, 0 - 3k + 18 = 0
Value of k = 6
Therefore, D is (0, 6)
(ii) To find E, let E be (q, 2)
Put y = 2 in the equation of the Cross Road,
we get, 4q - 6 + 18 = 0
4q = - 12 gives q = - 3
Therefore, The point E is (-3, 2)
Thus the Cross Road meets the North Street at D(0, 6) and
East Avenue at E (-3, 2)

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

If a and b are the intercepts then a + b = 7 or b = 7 − a
By intercept form \(\frac { x }{ a } +\frac { y }{ b } =1\)...(1)
We have \(\frac { x }{ a } +\frac { y }{ 7-a } =1\)
As this line pass through the point (-3,8), we have
\(\frac { -3 }{ a } +\frac { 8 }{ 7-a } =1\) gives –3(7–a) + 8a = a(7 – a)
−21 + 3a + 8a = 7a − a²
So, a² + 4a − 21 = 0
Solving this equation (a − 3) (a + 7) = 0
a = 3 or a = −7
Since a is positive, we have a = 3 and b = 7 – a = 7 – 3 = 4.
Hence \(\frac { x }{ 3 } +\frac { y }{ 4 } =1\)
Therefore, 4x + 3y − 12 = 0 is the required equation.

From the figure,
slope = \(\frac { change\quad in\quad y\quad coordinate }{ change\quad is\quad x\quad coordinate } \)
=\(\frac { 68-32 }{ 20-0 } =\frac { 36 }{ 20 } =\frac { 9 }{ 5 } \)= 1.8
The line crosses the Y axis at (0, 32)
So the slope is \(\frac { 9 }{ 5 } \) and y intercept is 32.

The parking lot is a quadrilateral whose vertices are at A(2, 2), B(5, 5), C(4, 9) and D(1, 7).
Therefore, Area of parking lot
= \(\frac{1}{2}\) {(10 + 45 + 28 + 2) - (10 + 20 + 9 + 14)}
= \(\frac{1}{2}\) {85 - 53}

= \(\frac{1}{2}\) (32) = 16.units.
So, area of parking lot = 16 sq feet
Construction rate per square feet = Rs. 1300
Therefore, total cost for constructing the parking lot = 16 x 1300 = Rs. 20800

Before determining the area of quadrilateral, plot the vertices in a graph.
Let the vertices be A(8, 6), B(5, 11), C(-5, 12) and D(-4, 3).
Therefore, area of the quadrilateral ABCD
=\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
=\(\frac{1}{2}\) { (80 + 60  - 15 - 24) - (30 - 55 - 48 + 24)}
=\(\frac{1}{2}\) {109 + 49 }
=\(\frac{1}{2}\) { 158 } = 79 sq. units

The vertices are A(1, 2), B(k, -2) and C(7, 4)
Area of triangle ABC is 22 sq. units
\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) } = 22
\(\frac{1}{2}\) { (2 + 4k + 14) - (2k - 14 - 4) } = 22
2k + 34 = 44 gives 2k = 10 so k = 5.

The points are P(-1, 5, 3), Q(6, -2) , R(-3, 4)
Area of Δ PQR = \(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
= \(\frac{1}{2}\) { (3 + 24 - 9) - (18 + 6 - 6) }
= \(\frac{1}{2}\) { 18 - 18 } = 0
Therefore, the given points are collinear.

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.

Given, θ = 45⁰, y intercept, c = 11
Slope m = tan θ  = tan 45⁰ = 1
Therefore, equation of a straight line is of the form y = mx + c
Hence we get, y = x + 11 gives x − y + 11 = 0

Let A(6,10) , B(14,12) be the points denoting the terrace of the buildings.
The equation of the rod is the equation of the straight line passing through A(6,10) and B(14,12)
\(\frac { { y-y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { { x-x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \) gives \(\frac { y-10 }{ 12-10 } =\frac { x-6 }{ 14-6 } \)
\(\frac { y-10 }{ 2 } =\frac { x-6 }{ 8 } \)
Therefore, x − 4y + 34 = 0
Hence, equation of the rod is x − 4y + 34 = 0

Let the given points be A(1, 4) , B(2, 5) and C(4, 7).
Slope of line BC = \(\frac { 7-5 }{ 4-2 } =\frac { 2 }{ 2 } =1\)
Let m be the slope of the required line.
Since the required line is perpendicular to BC,
m x 1 = −1
m = −1
The required line also pass through the point A(1, 4).
The equation of the required straight line is y − y₁ = m(x - x₁)
y - 4 = − 1(x  − 1)
y - 4 = − x + 1
we get, x + y − 5 = 0

Equation of the given straight line is 8x − 7y + 6 = 0
7y = 8x + 6 (bring it to the form y = mx + c)
\(y=\frac { 8 }{ 7 } x+\frac { 6 }{ 7 } \).... (1)
Comparing (1) with y = mx + c
Slope m = \(\frac { 8 }{ 7 } \) and y intercept c = \(\frac { 6 }{ 7 } \)