Question-16789

Question Number 16789 by ajfour last updated on 26/Jun/17

Commented by ajfour last updated on 26/Jun/17

solution to Q . 16065

find locus of M such that

Area (△ MAB) = 2 Area (△ MCD) .

Answered by ajfour last updated on 26/Jun/17

Area (△ MAB) = \frac{1}{2} (ay)

Area (△ MCD) = \frac{1}{2} | \begin{matrix} x & y & 1 \\ b & c & 1 \\ d & e & 1 \end{matrix} |

2 \times Area (△ MCD) = Area (△ MAB)

| \begin{matrix} x & y & 1 \\ b & c & 1 \\ d & e & 1 \end{matrix} | = \frac{ay}{2}

so a straight line .

Commented by Tinkutara last updated on 26/Jun/17

Yes Sir the answer is a straight line but

here it is required to draw that line / or

identify it . I am giving the figure in

which it is located . There is some

theory part of the solution but it is

not clear to me . Please explain .

Commented by Tinkutara last updated on 26/Jun/17

Commented by Tinkutara last updated on 26/Jun/17

Here it is given that J K is the

required locus of points M . Solution is

given below .

Commented by Tinkutara last updated on 26/Jun/17

Commented by ajfour last updated on 26/Jun/17

pretty lengthy, cannot comprehend

right now .

Answered by mrW1 last updated on 26/Jun/17

Let AB = a and CD = b

Case 1 : AB / / CD

A_{Δ ABM} = \frac{1}{2} AB \times h_{1} = \frac{1}{2} {ah}_{1}

A_{Δ CDM} = \frac{1}{2} CD \times h_{2} = \frac{1}{2} {bh}_{2}

A_{Δ ABM} = {2 A}_{Δ CDM}

\Rightarrow {ah}_{1} = {2 bh}_{2}

\frac{h_{1}}{h_{2}} = \frac{2 b}{a}

h_{1} + h_{2} = h

\Rightarrow h_{1} = \frac{2 b}{a + 2 b} h = constant

\Rightarrow h_{2} = \frac{a}{a + 2 b} h = constant

\Rightarrow M lies on a straight line / / to AB

and CD

Commented by mrW1 last updated on 26/Jun/17

Commented by mrW1 last updated on 27/Jun/17

Case 2 :

AB and CD meets at point O, and

∠ AOD = α \neq 0

let OM = d

∠ AOM = β

h_{1} = d \sin β

h_{2} = d \sin (α - β)

A_{Δ ABM} = \frac{1}{2} {ah}_{1}

A_{Δ CDM} = \frac{1}{2} {bh}_{2}

A_{Δ ABM} = {2 A}_{Δ CDM}

\Rightarrow {ah}_{1} = {2 bh}_{2}

\Rightarrow ad \sin β = 2 bd \sin (α - β)

\Rightarrow a \sin β = 2 b \sin (α - β)

\Rightarrow \frac{a}{2 b} \sin β = \sin (α - β)

with μ = \frac{a}{2 b}

\Rightarrow μ \sin β = \sin α \cos β + \cos α \sin β

\Rightarrow (μ - \cos α) \sin β = \sin α \cos β

\Rightarrow \tan β = \frac{\sin α}{μ - \cos α}

\Rightarrow β = \tan^{- 1} (\frac{\sin α}{μ - \cos α}) = constant

\Rightarrow M lies on a straight line which

passes the point O .

Commented by mrW1 last updated on 26/Jun/17

Commented by mrW1 last updated on 28/Jun/17

This method is valid generally for

A_{Δ ABM} = {nA}_{Δ CDM}

we just need to make D^{'} E^{'} / E^{'} A^{'} = 1 / n

Commented by mrW1 last updated on 26/Jun/17

This is how to draw the line :

Make {OA}^{'} = AB = a

Make {OD}^{'} = CD = b

Make D^{'} E^{'} = \frac{1}{3} A^{'} D^{'}

i . e . A^{'} E^{'} = {2 D}^{'} E^{'}

\Rightarrow A_{Δ {OA}^{'} E^{'}} = {2 A}_{Δ {OD}^{'} E^{'}}

Connecting {OE}^{'} we get a line which

intersects with AD at E and with BC

at F . EF is the line for point M .

Commented by mrW1 last updated on 26/Jun/17

Commented by mrW1 last updated on 27/Jun/17

As we can see in my calculation above

the line with angle β for the condition

A_{Δ ABM} = {2 A}_{Δ CDM} is independent from

the position of points A and D, but

only dependent from a, b and α .

Therefore we can move the point A

and D to the point O, and the line for

locus of point M (expressed by β)

is the same .

And in this special situation the line

for locus of point M can be easily

determined through point E^{'} .

Commented by mrW1 last updated on 28/Jun/17

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