Gb) Graph the relation : $$ x-4 y=4 $$ and if the straight line representing the relation intersects the $$ X $$-axis at the point $$ A $$ and the $$ y $$-axis at the point $$ B $$, find the area of the triangle $$ O A B $$ where $$ O $$ is the origin point.

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To graph the relation $$ x - 4y = 4 $$, we can first find the intercepts of the line.

### Finding the Intercepts

1. **X-intercept**: Set $$ y = 0 $$ in the equation.
   $$
   x - 4(0) = 4 \implies x = 4
   $$
   So, the X-intercept is the point $$ A(4, 0) $$.

2. **Y-intercept**: Set $$ x = 0 $$ in the equation.
   $$
   0 - 4y = 4 \implies -4y = 4 \implies y = -1
   $$
   So, the Y-intercept is the point $$ B(0, -1) $$.

### Points of Interest
- The origin $$ O $$ is at the point $$ (0, 0) $$.
- The points $$ A $$ and $$ B $$ are $$ A(4, 0) $$ and $$ B(0, -1) $$.

### Area of Triangle $$ OAB $$

The area $$ A $$ of triangle $$ OAB $$ can be calculated using the formula:
$$
A = \frac{1}{2} \times \text{base} \times \text{height}
$$
In this case, we can consider the base as the distance along the X-axis from $$ O $$ to $$ A $$ and the height as the distance along the Y-axis from $$ O $$ to $$ B $$.

- The base $$ OA $$ is the distance from $$ O(0, 0) $$ to $$ A(4, 0) $$, which is $$ 4 $$.
- The height $$ OB $$ is the distance from $$ O(0, 0) $$ to $$ B(0, -1) $$, which is $$ 1 $$.

Thus, the area of triangle $$ OAB $$ is:
$$
A = \frac{1}{2} \times 4 \times 1 = 2
$$

### Conclusion
The area of triangle $$ OAB $$ is $$ \boxed{2} $$.
The area of triangle $$ OAB $$ is 2.

Gb) Graph the relation : and if the straight line representing the relation
intersects the -axis at the point and the -axis at the point , find the area of the
triangle where is the origin point.

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Gb) Graph the relation : and if the straight line representing the relation intersects the -axis at the point and the -axis at the point , find the area of the triangle where is the origin point.