# The area of a triangle is 5. Two of its vertices are $(2, 1)$ and $(3, -2)$. The third vertex lies on $y = x + 3$. Find the third vertex.

Given:

The area of a triangle is 5. Two of its vertices are $(2, 1)$ and $(3, -2)$. The third vertex lies on $y = x + 3$.

To do:

We have to find the third vertex.

Solution:

Let $A(2, 1), B(3, -2)$ and C(x, y)$be the vertices of$\triangle ABC$. We know that, Area of a triangle with vertices$(x_1,y_1), (x_2,y_2), (x_3,y_3)$is given by, Area of$\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$Therefore, Area of triangle $ABC=\frac{1}{2}[2(-2-y)+3(y-1)+x(1+2)]$ $5=\frac{1}{2}[-4-2y+3y-3+3x]$ $5(2)=(y-7+3x)$ $10+7=3x+y$ $3x+y=17$ The third vertex lies on$y = x + 3$. This implies, $y=x+3$ $\Rightarrow 3x+(x+3)=17$ $\Rightarrow 4x=17-3$ $\Rightarrow 4x=14$ $\Rightarrow x=\frac{14}{4}$ $\Rightarrow x=\frac{7}{2}$ $\Rightarrow y=\frac{7}{2}+3$ $\Rightarrow y=\frac{7+2(3)}{2}$ $\Rightarrow y=\frac{13}{2}$ The third vertex is$(\frac{7}{2}, \frac{13}{2})\$.

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Updated on: 10-Oct-2022

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