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Question

If the area of triangle $$ABC$$ formed by $$A(x, y), B(1, 2)$$ and $$C(2, 1)$$ is $$6$$ square units, then prove that $$x + y = 15$$.


Solution

Given: $$A(x, y)\equiv(x_1,y_1), B(1, 2)\equiv(x_2,y_2)$$ and $$C(2, 1)\equiv(x_3,y_3)$$

We know than area of triangle $$\displaystyle=\frac { 1 }{ 2 } \left[ x_{ 1 }(y_{ 2 }-y_{ 3})+x_{ 2 }((y_{ 3 }-y_{ 1 }))+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right] $$ 

(hint: using determinant formula of the area of triangle)

Then area of triangle whoes vertex is $$A(x,y) ,B(1,2)$$ and $$C(2,1)$$ is

Area of $$\displaystyle \Delta ABC=\left [ x(2-1)+1(1-y)+2(y-2) \right ]=\frac{1}{2}(x+1-y+2y-4)= \frac{1}{2}(x+y-3)$$

But given that that the area of triangle is $$6$$ sq unit 

$$\therefore \dfrac{1}{2}(x+y-3)=6$$

$$\Rightarrow x+y-3=12$$

$$\Rightarrow x+y=15$$

Mathematics

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