Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then

Solution :

Since, (\(\alpha\), -\(\alpha\)) lie on 4ax+2ay+c=0 and 5bx+2by+d=0.

\(\therefore\) 4a\(\alpha\) + 2a\(\alpha\) + c = 0 \(\implies\) \(\alpha\) = \(-c\over 2a\)  …..(i)

Also, 5b\(\alpha\) – 2b\(\alpha\) + d = 0 \(\implies\) \(\alpha\) = \(-d\over 3b\)    …..(i)

from equation (i) and (ii),

\(-c\over 2a\) = \(-d\over 3b\)

3bc = 2ad


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