solve x²p+y²q=z² ..find this answer
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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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