Given area of triangle is 5
Vertices of triangle are (2, 1), (3, - 2) and (x, y)
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=5\)
2(-2 - y) + 3(y - 1) + x(1 + 2) = 10
-4 - 2y + 3y - 3 + 3x = 10
3x + y = 17
Given that Y = x + 3, Substituting in (1)
3x + x + 3 = 17
4x = 4
x = 7/2
and y = 7/2 + 3
= 13/2
Third vertex is (7/2, 13/2)

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.

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.

Plot the points in a rough diagram and take them in counter-clockwise order.
Let the vertices be


The area of Δ ABC is 
= \(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
= \(\frac{1}{2}\) { (6 + 30 + 25) - (25 - 10 18) }
= \(\frac{1}{2}\) { 61 + 3 }
= \(\frac{1}{2}\) (64) = 32 sq. units

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

The equation \(y=\frac { 4 }{ 3 } x-7\) can be written as 4x - 3y -21 = 0
Equation of a straight line perpendicular to 4x - 3y - 21 = 0 is 3x + 4y + k = 0
Since it is passes through the point (7,  – 1),
21 - 4 + k = 0 we get, k = -17
Therefore, equation of the required straight line is 3x + 4y - 11 = 0

Given 6x + 8y + 7 = 0
slope m \(=\frac { -coefficient\quad of\quad x }{ coefficient\quad of\quad y } =\frac { 6 }{ 8 } =-\frac { 3 }{ 4 } \)
Therefore, the slope of the straight line is = - \(\frac { 3 }{ 4 } \)

Let the x intercept be ‘a’ and y intercept be ‘– a’.
The equation of the line in intercept form is \(\frac { x }{ a } +\frac { y }{ b } =1\)
gives \(\frac { x }{ a } +\frac { y }{ -a } =1\) (Here b = – a)
Therefore, x − y = a ...(1)
Since (1) passes through (5, 7)
Therefore, 5 - 7 = a gives a = − 2
Thus the required equation of the straight line is x − y = − 2 ; or x − y + 2 = 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 = \(\frac { 6 }{ 7 } \)

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.

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