Question

The point of intersection of the lines x/a + y/b = 1 and x/b + y/a = 1 lies on the line
a) x - y = 0
b) (x+y)(a+b) = 2ab
c) (lx + my) ( a+ b) = (l + m) ab
d) all of these
NOTE : THE ANSWER IS D ALL OF THESE . PLZ!! PROVE IT HOW

Answer 1

Mark this answer as brainliest answer

Answer 2

Answer:

(d) all of these.

Step-by-step explanation:

$\frac{x}{a}+\frac{y}{b} =1\\\\\frac{x}{b} +\frac{y}{a} =1\\\\\frac{x}{ab} + \frac{y}{b^{2} } = \frac{1}{b} ....(multiplied by \frac{1}{b} )...(i)\\\\\frac{x}{ab} + \frac{y}{a^{2} } = \frac{1}{a} ....(multiplied by \frac{1}{a} )....(ii)\\\\(i)-(ii)\\\\ \frac{y}{b^{2} } - \frac{y}{a^{2} } = \frac{1}{b}- \frac{1}{a} \\\\or , y = \frac{ab}{a+b}\\ \\x = \frac{ab}{a+b}$

The point of intersection of the lines x/a + y/b = 1 and x/b + y/a = 1 lies on the line

a) x - y = 0

b) (x+y)(a+b) = 2ab

c) (lx + my) ( a+ b) = (l + m) ab

d) all of these

This is our question .

1) $x-y=0$............ option (a) is satisfied

2)

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

option (b) is satisfied

3)

$(lx + my) ( a+ b) \\\\= \frac{ab}{a+b}(l+m)(a+b)\\\\= (l+m)ab$

option (c) is satisfied.

So , the answer is (d).

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