Let P(a, b) Q(c, d) and R(e, f ) be the vertices of a triangle.
Let S be the mid-point of PQ and T be the mid-point of PR
Therefore, S = \(\left( \frac { a+c }{ 2 } ,\frac { b+d }{ 2 } \right) \) and \(\left( \frac { a+e }{ 2 } ,\frac { b+f }{ 2 } \right) \)
Now, slope of ST = \(\frac { \frac { b+f }{ 2 } -\frac { b+d }{ 2 } }{ \frac { a+e }{ 2 } -\frac { a+c }{ 2 } } =\frac { f-d }{ e-c } \)
And slope of QR  \(=\frac { f-d }{ e-c } \)

Therefore, ST is parallel to QR. (since, their slopes are equal)
Also ST = \(\sqrt { { \left( \frac { a+e }{ 2 } -\frac { a+c }{ 2 } \right) }^{ 2 }{ +\left( \frac { a+e }{ 2 } ,\frac { b+f }{ 2 } \right) }^{ 2 } } \)
= \(\frac { 1 }{ 2 } \sqrt { { \left( e-c \right) }^{ 2 }{ +\left( f-d \right) }^{ 2 } } \)
ST = \(\frac { 1 }{ 2 } \)QR
Thus ST is parallel to QR and half of it.

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 equation of a straight line passing through the two points (x₁, y₁) and (x₂, y₂) is \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
Substituting the points we get, \(\frac { y+3 }{ -4+3 } =\frac { x-5 }{ 7-5 } \)
gives 2y + 6 = − x + 5
Therefore, x + 2y + 1 = 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 = \(\frac { 6 }{ 7 } \)

Th e vertices are A(-2, 5) , B(6, -1) and C(2, 2).
Slope of AB = \(\frac { -1-5 }{ 6+2 } =\frac { -6 }{ 8 } =\frac { -3 }{ 4 } \)
Slope of BC = \(\frac { 2+1 }{ 2-6 } =\frac { 3 }{ -4 } =\frac { -3 }{ 4 } \)
We get, Slope of AB = Slope of BC
Therefore, the points A, B, C all lie in a same straight line.
Hence 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