We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula above, is given by
= \(\frac { 1 }{ 2 } \)[1(6 + 5) +(-4) (-5 + 1) + (-3)(-1 - 6)]
= \(\frac { 1 }{ 2 } \)[11 + 16 + 21] = 24 square units.

We know that diagonals at a parallelogram bisect each other.
So, the coordinates of the mid-point at AC = coordinates of the mid-point at BD.
i.e., \(\left[ \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 2+3 }{ 2 } \right] \)
\(\left[ \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 5 }{ 2 } \right] \)
\(\frac { 15 }{ 2 } =\frac { 8+P }{ 2 } \)
P = 7

Let P and Q be the points of trisection at AB.
i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1:2. Therefore, the coordinates at P, by applying the section formula, are
\(\left[ \frac { 1(-7)+2(2) }{ 1+2 } ,\frac { 1(7)+2(-2) }{ 1+2 } \right] \) i.e., (-1,10)
Now, Q also divides AB internally in the ratio 2:1, so, the coordinates at Q are
\(\left[ \frac { 2(-7)+1(2) }{ 2+1 } ,\frac { 2(4)+(-2) }{ 2+1 } \right] \) i.e., (-4,2)
Therefore, the coordinates at the points at trisection of the line segment joining A and B are (-1, 0) and (-4, 2).

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

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

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 points (- 3, - 4), (7, 2) and (12, 5)
Let the points be A (- 3, - 4),8 (2, 2) and C (12, 5)
Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
Slope of AB = \(\frac{-4-2}{-3-7}=\frac{-6}{-10}=\frac{3}{5} \)
Slope of BC = \(\frac{2-5}{7-12}=\frac{-3}{-5}=\frac{3}{5} \)
Slope of AB = Slope of BC
The points A, B and C are collinear

Given points are (2, 3), (4, a) and (6, - 3)
Since the points are colinear, Area of triangle is zero
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
2(a + 3) + 4(- 3 -3) + 6(3 - a) = 0
2a + 6 - 24 + 18 - 6a = 0
-4a + 0 = 0
-4a = 0
a = 0