Question

show that the lines x-mz-a=0=y-nz-b and x-m'z'-a'=0=y-n'z'-b' intersect , if (a-a')(n-n')=(b-b')(m-m')

Answer 1

hiii........
here is your answer. .......


Given lines are
x-mz-a=0=y-nz-b................(1)
and x-m'z'-a'=0=y-n'z'-b'..................(2)

putting z=0 in (1) we get x=a and y=b
so (a, b,0) is a point on (1).
again putting z=0 in (2) we get x=a' and y=b"
so (a', b',0) is a point on (2).
Let <L1, M1, N1> and <L2, M2, N2> be the d.dc.of the lined (1) anf (2).
ThenL1-mN1=0, M1-nN1=0
=》L1/m=N1/l=M1/n

hence <m, n, l> are the d.cs. of the line (1)
Ahain L2-m'N2=0 and M2-n'N2=0
=》L2/m'=N2/l=M2/n'

hence <m', n', l> are d.cs. of (2).
Equations of the straight lines in symmetrical form are
(x-a)/m=(y-b)/n=z/1=r1 (say).............. (4)
Any point on (3) is (mr1+a, nr1+b, r1)
Any point on (4) is (m'r2+a', n'r2+b', r2)
if they are intersrcting then

mr1+a=m'r2+a" .............. (5)
nr1+b=n'r2+b .............. (6)
r1=r2 ................ (7)
By (7) we get from (5)
or, (m-m') r1=a'-a
or, (a'-a)/(m-m')
Again r2=(a'-a)/(m-m')
putting these in (6) we get

n [(a'-a)/(m-m')]+b=n'[(a'-a)-(m-m')]+b'
or, n (a'-a)+b (m-m')=n'(a'-a)+b (m-m')
=》(a-a')(n-n')=(b-b')(m-m')


(proved)

I think it will help you then you mark me brainlist please please please. ....★★★★★