If the vertices of a triangle are `(1,k),(4,-3) and (-9,7)` and its area is 15 eq. units, then the value(s) of k is

Options

A. `-3, 21//13`
B. `-3, -21//13`
C. `3, 21//13`
D. `3, -21//13`

Answer

To find the value(s) of \( k \) for the triangle with vertices \( (1, k) \), \( (4, -3) \), and \( (-9, 7) \) and an area of 15 square units, we can use the formula for the area of a triangle given its vertices:

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

### Step 1: Substitute the vertices into the area formula

Let the vertices be:
- \( (x_1, y_1) = (1, k) \)
- \( (x_2, y_2) = (4, -3) \)
- \( (x_3, y_3) = (-9, 7) \)

Substituting these values into the area formula gives:

\[ \text{Area} = \frac{1}{2} \left| 1(-3 - 7) + 4(7 - k) + (-9)(k + 3) \right| \]

### Step 2: Simplify the expression

Calculating each term inside the absolute value:

\[ = \frac{1}{2} \left| 1(-10) + 4(7 - k) - 9(k + 3) \right| \]

Expanding this:

\[ = \frac{1}{2} \left| -10 + 28 - 4k - 9k - 27 \right| \]

Combining like terms:

\[ = \frac{1}{2} \left| -10 + 28 - 27 - 13k \right| \]
\[ = \frac{1}{2} \left| -9 - 13k \right| \]

### Step 3: Set the area equal to 15 and solve for \( k \)

Setting the area equal to 15 gives:

\[ \frac{1}{2} \left| -9 - 13k \right| = 15 \]

Multiplying both sides by 2:

\[ \left| -9 - 13k \right| = 30 \]

### Step 4: Solve the absolute value equation

This leads to two cases:

**Case 1:**
\[ -9 - 13k = 30 \]
Solving for \( k \):
\[ -13k = 30 + 9 \]
\[ -13k = 39 \]
\[ k = -3 \]

**Case 2:**
\[ -9 - 13k = -30 \]
Solving for \( k \):
\[ -13k = -30 + 9 \]
\[ -13k = -21 \]
\[ k = \frac{21}{13} \]

### Final Answer

The values of \( k \) are:

\[ k = -3 \quad \text{and} \quad k = \frac{21}{13} \]