Use your graph to find the coordinates of C the point of intersection between lines a and b. Hence find the area of Triangle ABC (extra information - line a passes through point (1,2) with gradient 1/2 and line b passes through (-4,7) with gradient -2

Let's break down the problem step by step:

### Step 1: Find the Equations of Lines a and b

#### Line a:
- Point on the line: \((1, 2)\)
- Gradient (slope): \(\frac{1}{2}\)

Using the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Substitute the given point \((1, 2)\) and the gradient \(m = \frac{1}{2}\):
\[
y - 2 = \frac{1}{2}(x - 1)
\]
Simplify the equation:
\[
y - 2 = \frac{1}{2}x - \frac{1}{2}
\]
\[
y = \frac{1}{2}x + \frac{3}{2}
\]
So, the equation of line \(a\) is:
\[
y = \frac{1}{2}x + \frac{3}{2}
\]

#### Line b:
- Point on the line: \((-4, 7)\)
- Gradient (slope): \(-2\)

Using the point-slope form again:
\[
y - y_1 = m(x - x_1)
\]
Substitute the given point \((-4, 7)\) and the gradient \(m = -2\):
\[
y - 7 = -2(x + 4)
\]
Simplify the equation:
\[
y - 7 = -2x - 8
\]
\[
y = -2x - 1
\]
So, the equation of line \(b\) is:
\[
y = -2x - 1
\]

### Step 2: Find the Intersection Point C
To find the coordinates of \(C\), the point of intersection between lines \(a\) and \(b\), set the equations equal to each other:

\[
\frac{1}{2}x + \frac{3}{2} = -2x - 1
\]

Multiply through by 2 to eliminate the fraction:
\[
x + 3 = -4x - 2
\]

Add \(4x\) to both sides:
\[
5x + 3 = -2
\]

Subtract 3 from both sides:
\[
5x = -5
\]

Divide by 5:
\[
x = -1
\]

Now, substitute \(x = -1\) back into the equation of either line to find \(y\). Let's use line \(a\)'s equation:

\[
y = \frac{1}{2}(-1) + \frac{3}{2} = -\frac{1}{2} + \frac{3}{2} = \frac{2}{2} = 1
\]

So, the coordinates of point \(C\) are:
\[
C(-1, 1)
\]

### Step 3: Find the Area of Triangle ABC
We now have the coordinates of points \(A(1, 2)\), \(B(-4, 7)\), and \(C(-1, 1)\).

The area of triangle \(ABC\) can be found using the determinant formula:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substitute the coordinates of points \(A\), \(B\), and \(C\):

\[
\text{Area} = \frac{1}{2} \left| 1(7 - 1) + (-4)(1 - 2) + (-1)(2 - 7) \right|
\]

Simplify inside the absolute value:

\[
\text{Area} = \frac{1}{2} \left| 1(6) + (-4)(-1) + (-1)(-5) \right|
\]
\[
\text{Area} = \frac{1}{2} \left| 6 + 4 + 5 \right| = \frac{1}{2} \times 15 = 7.5
\]

### Final Answer
The area of triangle \(ABC\) is \(7.5\) square units.

Would you like more details or have any questions?

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Here are some related questions to explore:
1. How would you find the intersection point if one of the lines were vertical?
2. Can the area of a triangle be negative using the determinant method?
3. What is the geometric interpretation of the slopes of these lines?
4. How would the area change if the coordinates of one point were moved along a line?
5. How would you find the area if the points were given as parametric equations?

**Tip:** Always double-check your sign calculations when working with determinants or any formula involving multiple terms.

Learn how to find the intersection point of two lines using their equations and calculate the area of Triangle ABC with given vertices (1, 2), (-4, 7), and (-1, 1). This problem covers coordinate geometry, linear equations, and the determinant formula for triangle area. Suitable for students in Grades 9-12.

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