Given straight line is 2x - 3y + 8 = 0
Slope m = \(\frac { -2 }{ -3 } =\frac { 2 }{ 3 } \)
Since product of slope is −1 for perpendicular lines, slope of any line perpendicular to 2x - 3y + 8 = 0 is \(\frac { -1 }{ \frac { 2 }{ 3 } } =\frac { -3 }{ 2 } \)

Show of the straight lines x - 2y + 3 = 0 is
m₁ = \(\frac{-1}{-2}=\frac{1}{2}\)
Slope of the straight line 6x + 3y + 8 = 0 is
m₂ = \(\frac{-6}{3}\)=-2
Now, m₁ x m₂ = \(\frac12\) x (-2) = -1
Hence, the two straight lines are perpendicular

Slope of the straight line 2x + 3y - 8 = 0 is
m₁ = \(\frac { -coefficient\quad of\quad x }{ cofficient\quad of\quad y } \)
m₂ = \(\frac{-2}{3}\)
Slope of the straight line 4x + 6y + 18 = 0 is
m₂ = \(\frac { -4 }{ 6 } =\frac { -2 }{ 3 } \)
Here, m₁ = m₂
That is, slopes are equal. Hence, the two straight lines are parallel.

The slope of line p is m₁ = \(\frac { 4+2 }{ 12-3 } =\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
The slope of line p is m₂ = \(\frac { 2+2 }{ 12-6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Thus, slope of line p = slope of line q.
Therefore, the line p is parallel to the line q.

Th e slope of line r is m₁  \(=\frac { 8-2 }{ 5+2 } =\frac { 6 }{ 7 } \)
The slope of line θ is m₂  \(=\frac { 0-7 }{ -2+8 } =\frac { -7 }{ 6 } \)
The product of slopes \(=\frac { 6 }{ 7 } \times \frac { -7 }{ 6 } =-1\)
That is, m₁m₂ = -1

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).

The coordinate of the point of contact P of the tangent line AB with the circle is point of intersection of line.
x + y − 5 = 0 and x − y = 0
solving, we get x = \(\frac { 5 }{ 2 } \) and y = \(\frac { 5 }{ 2 } \)
Therefore, the coordinate of the P\(\left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) \)

The slope of AB and CD are equal so AB, CD are parallel.
Similarly the lines AD and BC are not parallel, since their slopes are not equal. So, we can deduce that the quadrilateral ABCD is a trapezium.