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Question

A point is at a distance of $$\sqrt {10}$$ unit from the point $$(4,3)$$. Find the coordinates of point $$A$$, if its ordinate is twice its abscissa.


Solution

The coordinates of point $$A$$ are such that its ordinate is twice its abscissa.
Then, let the coordinates of point $$A=(x,2x)$$
$$\sqrt { { \left( x-4 \right)  }^{ 2 }+{ \left( 2x-3 \right)  }^{ 2 } } =\sqrt { 10 } $$
$${ \left( x-4 \right)  }^{ 2 }+{ \left( 2x-3 \right)  }^{ 2 }=10\Rightarrow { x }^{ 2 }+16-8x+4{ x }^{ 2 }+9-12x=10$$
$$\Rightarrow 5{ x }^{ 2 }-20x+15=0\Rightarrow { x }^{ 2 }-4x+3=0\Rightarrow { x }^{ 2 }-x-3x+3=0$$
$$\Rightarrow x(x-1)(-3(x-1)=0\Rightarrow (x-1)(x-3)=0$$
$$x=1,3$$
So, the coordinates of the point $$A$$ are $$(1,2)$$ and $$(3,6)$$

Mathematics

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