Question

The area of a triangle whose vertices are (-2,-2), (-1,-3) and (p, 0) is 3 sq. units what is the value of p?
(a). -2
(b). 2
(c). 3
(d). -3

Answer 1

Hint: In this question you can use the formula for the area of the triangle is equal to $\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$ if the vertices of the triangle are given and then find the value of p.

Complete step-by-step answer:

Area of triangle ABC= $\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]..........(1)$
It is given that, the area of the triangle is 3 square units and vertices of the triangle are $\left( {{x}_{1}},{{y}_{1}} \right)=(-2,-2),\left( {{x}_{2}},{{y}_{2}} \right)=(-1,-3)$ and$\left( {{x}_{3}},{{y}_{3}} \right)=(p,0)$.
The equation (1) becomes
 $3=\dfrac{1}{2}\left[ -2\left( -3+0 \right)-1\left( 0+2 \right)+p\left( -2+3 \right) \right].$
 $3=\dfrac{1}{2}\left[ -2\left( -3 \right)-1\left( 2 \right)+p\left( 1 \right) \right].$
$3=\dfrac{1}{2}\left[ 6-2+p \right].$
$3=\dfrac{1}{2}\left[ 4+p \right].$
Multiplying both sides by 2, we get
$6=4+p$
$p=2$
Hence the value of the p is 2 cm.
Therefore, the correct option for the given question is option (b).

Note: Alternatively, the vertices of the triangle are given then you can definitely use distance formula to find the length of all the sides of the triangle which can enable him to use Heron's formula to find the area of the triangle.