Question

The area of the triangle whose vertices are $ (-2, a) $, $ (2, -6) $, and $ (5, 4) $ is 35 square units. Then the value of $ a $ is:

• A. 4
• B. \( \frac{53}{3} \)
• C. \( -\frac{23}{3} \)
• D. \( \frac{128}{3} \)

Hint:

When using the area formula for a triangle with given vertices, make sure to simplify the expression carefully and solve for the unknown variable.

Answer 1

The Correct Option is A

To find the value of \( a \), we use the formula for the area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \): \[ \text{Area of triangle} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substitute the given vertices into the formula. The vertices are \( (-2, a) \), \( (2, -6) \), and \( (5, 4) \). So, \( x_1 = -2 \), \( y_1 = a \), \( x_2 = 2 \), \( y_2 = -6 \), \( x_3 = 5 \), and \( y_3 = 4 \). Now, substitute these into the area formula: \[ \text{Area} = \frac{1}{2} \left| (-2)[(-6) - 4] + 2[(4) - a] + 5[a - (-6)] \right| \] Simplifying the expression: \[ \text{Area} = \frac{1}{2} \left| (-2)(-10) + 2(4 - a) + 5(a + 6) \right| \] \[ \text{Area} = \frac{1}{2} \left| 20 + 8 - 2a + 5a + 30 \right| \] \[ \text{Area} = \frac{1}{2} \left| 58 + 3a \right| \] Given that the area is 35 square units, set the equation equal to 35: \[ \frac{1}{2} \left| 58 + 3a \right| = 35 \] \[ \left| 58 + 3a \right| = 70 \] Now, solve for \( a \): 1. \( 58 + 3a = 70 \) \[ 3a = 12 \quad \Rightarrow \quad a = 4 \] 2. \( 58 + 3a = -70 \) \[ 3a = -128 \quad \Rightarrow \quad a = -\frac{128}{3} \] Since the value of \( a \) must be a reasonable integer and corresponds to the first equation, we conclude that: \[ a = 4 \] Thus, the correct answer is (A) 4.