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

Similar Questions

Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1)

If the line 2x + y = k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3:2, then k is equal to

The x-coordinate of the incenter of the triangle that has the coordinates of mid-point of its sides as (0,1), (1,1) and (1,0) is

If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\)

If PS is the median of the triangle, with vertices of P(2,2), Q(6,-1) and R(7,3), then equation of the line passing through (1,-1) and parallel to PS is

Leave a Comment

Your email address will not be published. Required fields are marked *