Question

b) Find the equations to the straight lines which go through the point (-1, -1) and trisect the portion of the straight line x + y = 6 which is intercepted between the axes of coordinates.​

Answer 1

Step-by-step explanation:

3x+y=12

x=0⇒y=12

y=0⇒x=4

So the line is interecepted between A(4,0) ans B(0,12)

Let the points P and Q trisect the portion of the line AB

Now Q divides AB in 2:1

⇒Q(

2+1

2(0)+1(4)

,

2+1

2(12)+1(0)

)

Q(

3

4

,

3

24

)

So the equation of line joining origin and Q is

y−0=

⎝

⎜

⎜

⎛

3

4

−0

3

24

−0

⎠

⎟

⎟

⎞

(x−0)

y=

4

24

x

y=6x

Now P divides AQ in 1:1 , so P is the mid point of AQ

⇒P

⎝

⎜

⎜

⎛

2

4+

3

4

,

2

0+

3

24

⎠

⎟

⎟

⎞

P(

3

8

,

3

12

)

So, the equation of line joining origin to P is

y−0=

⎝

⎜

⎜

⎛

3

8

−0

3

12

−0

⎠

⎟

⎟

⎞

(x−0)

y=

8

12

x

2y=3x

So, the desired equations are y=6x and 2y=3x.