Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3 x + y = 12 which is intercepted between the axes of coordinates.

#### Solution

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

At x = 0

0 + y = 12

\[\Rightarrow\] y = 12

At y = 0

3x + 0 = 12

\[\Rightarrow\] x = 4

\[\therefore A \equiv \left( 4, 0 \right) \text{and } B \equiv \left( 0, 12 \right)\]

Let

\[y = m_1 x \text { and } y = m_2 x\] be the lines that pass through the origin and trisect the line 3x + y = 12 at P and Q.

∴ AP = PQ = QB

Let us find the coordinates of P and Q.

\[P \equiv \left( \frac{2 \times 4 + 1 \times 0}{2 + 1}, \frac{2 \times 0 + 1 \times 12}{2 + 1} \right) \equiv \left( \frac{8}{3}, 4 \right)\]

\[Q \equiv \left( \frac{1 \times 4 + 2 \times 0}{2 + 1}, \frac{1 \times 0 + 2 \times 12}{2 + 1} \right) \equiv \left( \frac{4}{3}, 8 \right)\]

Clearly, P and Q lie on \[y = m_1 x \text { and } y = m_2 x\] ,respectively.

\[\therefore 4 = m_1 \times \frac{8}{3} \text { and }8 = m_2 \times \frac{4}{3}\]

\[ \Rightarrow m_1 = \frac{3}{2} \text { and} m_2 = 6\]

Hence, the required lines are

\[y = \frac{3}{2}x \Rightarrow 2y = 3x \text { and } y = 6x\]