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A.x = y

B.2x = y

C.4x = y

D.x = 4y

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Let \[M(h,k)\] be the midpoint of line AB. Given A, B move on the ‘x’ and ‘y’ axis respectively. Then it is obvious that the coordinates of A and B are \[A(2h,0)\] and \[B(0,2k)\] .

Since point \[P(t,2t)\] is first trisection it divides A and B into \[1:2\] ratio. See in the below diagram for understanding point of view.

We have \[A({x_1},{y_1}) = A(2h,0)\] , \[B({x_2},{y_2}) = B(0,2k)\] and which divides in the ratio \[1:2\] . Using the section formula we find the coordinates of the point P.

Using the formula \[\left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}} \right)\]

\[ \Rightarrow = P\left( {\dfrac{{\left( {1 \times 0} \right) + \left( {2 \times 2h} \right)}}{{1 + 2}},\dfrac{{\left( {1 \times 2k} \right) + \left( {2 \times 0} \right)}}{{1 + 2}}} \right)\]

\[ \Rightarrow = P\left( {\dfrac{{0 + 4h}}{3},\dfrac{{2k + 0}}{3}} \right)\]

\[ \Rightarrow = P\left( {\dfrac{{4h}}{3},\dfrac{{2k}}{3}} \right)\]

But we already have point p coordinates as \[P(t,2t)\] .

Comparing ‘x’ and ‘y’ coordinates in both Points P we have

The ‘x’ coordinate \[\dfrac{{4h}}{3} = t\]

Multiply by 3 on both sides.

\[ \Rightarrow 4h = 3t{\text{ - - - - - - - (1)}}\]

The ‘y’ coordinate \[\dfrac{{2k}}{3} = 2t\]

Cancelling 2 on both sides,

\[ \Rightarrow \dfrac{k}{3} = t\]

Multiply by 3 on both sides,

\[ \Rightarrow k = 3t{\text{ - - - - - - - (2)}}\]

Now substituting equation (2) in equation (1) we have \[ \Rightarrow 4h = k\] .

But we have options in ‘x’ and ‘y’ variables, so we have

\[ \Rightarrow 4x = y\] (Because \[(h,y) = (x,y)\] )