Question

The equation of the lines joining the origin to the points of trisection of the portion of line 3x+y=12 intercepted between the axes are​

Answer 1

Giνєи Qυєѕтiσи —

The equation of the lines joining the origin to the points of trisection of the portion of line 3x + y = 12 intercepted between the axes are ?

Cσиcєρт —

Straight Lines - equation of family of lines pássing through the point of intersection of two lines .

Sσℓυтiσи –

Let the line 3x + y = 12 intersect the x - axis and the y - axis at A and B ,

✪ At x = 0

➙ 3x + y = 12

➙ 3 × 0 + y = 12

➙ 0 + y = 12

➙ y = 12

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✪ At y = 0

➙ 3x + y = 12

➙ 3 x + 0 = 12

➙ 3x = 12

➙ x = 12/3

➙ x = 4

Therefore , A = ( 4, 0 ) and B = ( 0, 12 )

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➻ Let , $\sf{y=m_{1}x}$ and $\sf{y=m_{2}x}$ be the lines that páss through the origin and trisect the line 3x + y = 12 at P and Q .

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Therefore , AP = PQ = QB

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❍ Let is find the co-ordinates of P and Q.

$\sf \longrightarrow P \: = \left ( \dfrac{2 \times 4 + 1 \times 0}{ 2 + 1} ,\dfrac{2 \times 0 + 1 \times 12}{2 + 1} \right)$

$\sf \longrightarrow P = \left ( \dfrac{8}{3} ,\dfrac{ 12}{3} \right)$

$\sf \longrightarrow p = \left ( \dfrac{8}{3} , 4 \right)$

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$\sf \longrightarrow Q\: = \left ( \dfrac{1 \times 4 + 2 \times 0}{2 + 1} ,\dfrac{1 \times 0 + 2 \times 12}{2 + 1} \right)$

$\sf \longrightarrow \:Q = \left ( \dfrac{4}{3} , \dfrac{24}{3} \right)$

$\sf \longrightarrow q = \left ( \dfrac{4}{3} , 8 \right)$

➻ Clearly , P and Q lie on $\sf{y=m_{1}x}$ and $\sf{y=m_{2}x}$

✪ Therefore :

$\sf{\longrightarrow 4 = m_{1}\times \dfrac{8}{3}}$

$\sf{\longrightarrow m_{1} = \dfrac{3}{2}}$

and,

$\sf{\longrightarrow 8 = m_{2}\times \dfrac{4}{3}}$

$\sf{\longrightarrow m_{2} = 6}$

Hence , the required lines are :

${\pmb{\sf{\longrightarrow y = \dfrac{3}{2}x}}}$

${\pmb{\sf{\longrightarrow y = 6x}}}$