Question

Find the equation of the straight lines which pass through the origin and trisect the intercept of line 3x+4y=12 between the axes.

Answer 1

Answer:  2y = 3x  and y = 6x

Step-by-step explanation:

Given  line 3x +4y = 12

Its point of intersection on graph

x = 0 ,4

y = 12, 0

The point passes through the line are A(4,0) and B(0, 12)

Let P and Q be the point on of intersection trisecting AB.

Here, P divides AB in the ratio 1:2

Coordinates of P =( $\frac{1*0 +2*4}{1+2}$  , $\frac{1 * 12+2*0}{1+2}$ )= $\frac{8}{3}$ , 4

Given Point O is origin whose coordinates are (0,0)

The equation of line OP

 y - 0 = $\frac{4-0}{8/3-0}$ (x - 0)

y =3/2 x

2y = 3x ..........(i)

The point Q will divide AB in the ratio 2:1 , so its coordinates are

= ( $\frac{2*0 +1*4}{2+1}$ , $\frac{2*12+1*0}{2+1}$ ) =  ($\frac{4}{3}$ ,8)

The equation of OQ

y - 0 = $\frac{8-0}{4/3-0}$ (x -0)

y = 6x ......(ii)

Equation of straight lines are 2y =3x , y=6x

Answer 2

Answer:

the Equations are

• 3x-8y=0
• 3x-2y=0