Question

Sum of intercepts cut off by the axes on lines x+y=a , x+y=ar , x+y=ar^2......... Where a not equals to 0 r=1|2

Answer 1

Answer:

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Answer 2

Answer:

(2√2)a

Step-by-step explanation:

The first straight line is x+y =a, ⇒$\frac{x}{a}+\frac{y}{a}=1$ ....... (1)

Now, the points of intersection of (1) with X-axis and with Y-axis are (a,0) and (0,a) respectively.

So, the length of intercept cut off by the axes will be=$\sqrt{a^{2}+a^{2}  }=a\sqrt{2}$ ........ (2)

Again, the second straight line is x+y =ar, ⇒$\frac{x}{ar}+\frac{y}{ar}=1$ ....... (3)

Now, the points of intersection of (3) with X-axis and with Y-axis are (ar,0) and (0,ar) respectively.

So, the length of intercept cut off by the axes will be=$\sqrt{(ar)^{2}+(ar)^{2}  }=ar\sqrt{2}$ ........ (4)

Similarly, the third straight line is x+y =ar², ⇒$\frac{x}{ar^{2} }+\frac{y}{ar^{2} }=1$ ....... (5)

Now, the points of intersection of (5) with X-axis and with Y-axis are (ar²,0) and (0,ar²) respectively.

So, the length of intercept cut off by the axes will be=$\sqrt{(ar^{2} )^{2}+(ar^{2} )^{2}  }=ar^{2} \sqrt{2}$ ........ (6)

And so on.

So, the sum =a√2+ar√2+ar²√2+ar³√2+...... up to ∞ {From equations (2), (4),(6), ........}

                    =a√2[1+r+r²+r³+ ...... up to ∞]

                    =a√2[$\frac{1}{1-r}$]

                    =$\frac{a\sqrt{2} }{1-r}$

Now, given that, a≠0 and r= $\frac{1}{2}$

Therefore, the sum = $\frac{a\sqrt{2} }{1-\frac{1}{2} }$

                                =(2√2)a (Answer)