1.

Solve x²p+y²q=z² ..find this answer​

Answer»

<P>Answer:

We have, x=py+q⇒y=

p

x−q

⋯(i)

And z=ry+s⇒y=

R

z−s

…(ii)

⇒

p

x−q

=

1

y

=

r

z−s

[ using Eqs. ( i ) and (ii)]⋯(III)

Similarly,

p

′

x−q

′

=

1

y

=

r

′

z−s

′

…(iv)

From Eqs. (iii) and (iv),a

1

=p,b

1

=1,c

1

=r and a

2

=p

′

;b

2

=1,c

2

=r

′

if these GIVEN lines are perpendicular to each other, then

a

1

a

2

+b

1

b

2

+c

1

c

2

=0

⇒pp

′

+1+rr

′

=0

Which is the required condition.


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