Find the eq of the straight lines which go through the origin and trisect the portion of the straight - Maths - Straight Lines

Question

Find the eq of the straight lines which go through the origin
and trisect the portion of the straight line 3x+y=12 which is
intetercepted between the axes of the coordinates

Ankush Jain · Accepted Answer

Given line is: 3x + y = 12 Its point of intersection on the graph can be find out as: x 0 4 y 12 0 So, the point through which the given line passes are let A(4, 0) and B(0, 12) Let P and Q be the point of intersection trisecting AB Here, P divides AB in the ratio 1 : 2 So, the co-ordinates of P $= (\frac{1 \times 0 + 2 \times 4}{1 + 2}, \frac{1 \times 12 + 2 \times 0}{1 + 2}) = (\frac{8}{3}, 4)$ Given, point O is origin whose co-ordinates must be (0, 0) Therefore, equation of line OP is: $y - 0 = \frac{4 - 0}{\frac{8}{3} - 0} \times (x - 0) \Rightarrow y = \frac{3}{2} x \Rightarrow 2 y = 3 x (1)$ Similarly, the point Q will divide AB in the ratio 2 : 1, so its coordinates are co-ordinates of Q $= (\frac{2 \times 0 + 1 \times 4}{2 + 1}, \frac{2 \times 12 + 1 \times 0}{2 + 1}) = (\frac{4}{3}, 8)$ Again, the equation of OQ is: $y - 0 = \frac{8 - 0}{\frac{4}{3} - 0} \times (x - 0) \Rightarrow y = 2 \times 3 \times x \Rightarrow y = 6 x (2)$ Hence, equation (1) and (2) are the required equation of the line

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