Question

8. Find the equations of the straight lines which pass
through the origin and trisect the portion of the
st. line X/a+y/b=1
E
= 1, which is intercepted between the
a
b
axes.​

Answer 1

Answer:

$y=(\frac{2b}{a} )x$

Step-by-step explanation:

Let A(a,0) and B(0,b) are the points intercepted by the straight line  $\frac{x}{a} +\frac{y}{b}=1$ ........ (1) with the X-axis and Y-axis respectively.

Assume again  that C is the point where AB is trisected i.e. intercepted in the ratio 2:1

Hence, coordinates of C will be given by $(\frac{a*1+2*0}{1+2},\frac{0*1+b*2}{2+1})$ =$(\frac{a}{3},\frac{2b}{3})$ ....... (2)

Let us assume that the equation of the straight line passing through origin and point C is y=mx ..... (3)

Now solving (1) and (3), we get the coordinates of C.

Hence, $\frac{x}{a}+\frac{mx}{b}=1$

⇒ x(b+am)=ab

⇒x=$\frac{ab}{b+am}$

So, from equation (3), y=$\frac{mab}{b+am}$

Hence, coordinates of C are ($\frac{ab}{b+am} ,\frac{mab}{b+am}$). ...... (4)

Now, from (2) and (4), we can write,

$\frac{a}{3}=\frac{ab}{b+am}$

⇒b+am=3b

⇒m=$\frac{2b}{a}$.

Therefore, the equation of the required straight line is $y=(\frac{2b}{a} )x$ { From equation (3)} (Answer)

Now, given that the length of the AB line is 1.

So, a and b are related to each other by the equation, a²+b²=1.