question_answer

The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (\[3,\text{ }2\]). If the third vertex is \[\left( \frac{7}{2},y \right)\], find the value of y.

Answer:

Given,

\[A(2,1),B(3,-2)\] and \[C\left( \frac{7}{2},y \right)\]

Now, Area \[(\Delta ABC)=\frac{1}{2}|{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})|\]

\[5=\frac{1}{2}|2(-2-y)+3(y-1)+\frac{7}{2}(1+2)|\]

\[\Rightarrow \]   \[10=|-4-2y+3y-3+\frac{7}{2}+7|\]

\[\Rightarrow \]   \[10=\left| y+\frac{7}{2} \right|\]

\[\Rightarrow \]   \[10=y+\frac{7}{2}\]     or         \[-10=\left( y+\frac{7}{2} \right)\]

\[\Rightarrow \]   \[y=\frac{13}{2}\]                     or         \[y=\frac{-27}{2}\]