If the area of the triangle whose vertices are (3,-2),(2,-3) and (p,-4) is 8 square units, then find the value of p.

Correct option is B

Given:

Vertices: (3, -2),(2, -3), and (p,−4)

Formula used:

The formula for the area of a triangle with vertices ​$$(x_1,y_1), (x_2,y_2),$$ and $$(x_3,y_3)$$

​$${Area} = \frac{1}{2} x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$$​​

Solution:

Given Vertices (3, -2),(2, -3), and (p,−4) 

Substitute the values into the formula-

​$$8=\frac{1}{2}(3(−3+4)+2(−4+2)+p(−2+3))$$

$$8=\frac{1}{2}(3(1)+2(−2)+(1))$$

8 = $$\frac{1}{2}$$ [3 - 4+ p]

8= $$\frac{1}{2}$$ (p-1)

16 = p-1

p-1 = 16

values of p -

p - 1 = 16