# Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3x + y=12 which is intercepted between the axes of the coordinates. Ans-y=6x...

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

Ans-y=6x ; 2y=3x

how????

*print*Print*list*Cite

The line `l1 = 3x+y = 12` is graphed above. What we need is to find the two points on l1 which trisect the portion of the straight line between the axes of the coordinates.

So the coordinates of the axes are the x and y intercepts which are shown by open and bold red dots respectively.

`3x+y = 12`

x interception is given when y = 0

`3x+0 = 12`

`x = 4`

So the coordinates of the open dot is (4,0).

Similarly y intercept is given when x = 0.

`3xx0+y = 12`

`y = 12`

So the coordinates of the bold dot is (0,12).

Let us say;

Bold dot as point A and open dot as point B.

`A = (0,12)`

`B = (4,0)`

Length of `AB = sqrt((0-4)^2+(12-0)^2) = 4sqrt10` .

So if we take trisect points as C and D where C is close to A and D is close to B;

`AC = BD = (4/3)sqrt10`

Let us say x coordinate of C is a. Since C is on l1;

`3a+y = 12`

`y = (12-3a)`

`C = (a,(12-3a)`

Length of `AC = sqrt((0-a)^2+(12-(12-3a)^2) = sqrt(10a^2) = asqrt10`

We know that `AC = 4/3sqrt10`

`asqrt10 = 4/3sqrt10`

`a = 4/3`

`C = (4/3,8)`

Let us say x coordinate of D is b.

Doing same for BD will give you the following.

`B = (b,(12-3b)`

`AD = sqrt((0-b)^2+(12-3b)^2) = bsqrt10`

But `AD = 2xx(4/3)sqrt10`

`bsqrt10 = 2xx(4/3)sqrt10`

`b = 8/3`

`D = (8/3,4)`

Origin `(O) = (0,0)`

Now we need to get the equation of lines between OC and OD.

The lines will satisfy the equation of y = mx since they are going through origin.

equation of OC;

`y = (8-0)/(4/3-0)x = 6x`

`y = 6x`

Equation of OD;

`y = (4-0)/(8/3-0) = 3/2x`

`2y = 3x`

*So the answer is*

*y = 6x*

*2y = 3x*

This is graph of the line 3x+y=12

Let this line meets x- axis at A(4,0)

and y-axis at B(0,12).

Let C and D are points on the line 3x+y=12 such that AC=CD=DB

Then point C divides AB in ratio or 1:2

i.e AC/CB=1/2

By section formula ,coordinate of C will be

`((1xx0+2xx4)/(1+2),(1xx12+2xx0)/(1+2))`

`=(8/3,4)`

The point D divides AB in ratio of 2:1 i.e AD/DB=2/1

By section formula , coordinate of D is

`((2xx0+1xx4)/(2+1),(2xx12+1xx0)/(2+1))`

`=(4/3,8)`

The slope of line posses through O(0,0) and C(8/3,4)

=4/(8/3)

=3/2

Thus equation of line OC

`y-4=(3/2)(x-8/3)`

`y-4=(3/2)x-4`

`y=(3/2)x`

`2y=3x` (i)

Slope of line posses through O(0,0) and D(4/3,8)

=8/(4/3)

=6

Thus equation of line OD

`y=8=6(x-4/3)`

`y-8=6x-8`

`y=6x` (ii)

Thus (i) and (ii) are required staright lines.