Let a,b,c and d be non-zero numbers.

Question

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:

Accepted Answer

Since the point of intersection lies on fourth quadrant and equidistant from the two axes,

i.e., let the point be (k, -k) and this point satisfies the two equations of the given lines.

$$\therefore$$ 4ak - 2ak + c = 0 ......... (1)

and 5bk - 2bk + d = 0 ..... (2)

From (1) we get, $$k = {{ - c} \over {2a}}$$

Putting the value of k in (2) we get,

$$5b\left( { - {c \over {2a}}} \right) - 2b\left( { - {c \over {2a}}} \right) + d = 0$$

or, $$ - {{5bc} \over {2a}} + {{2bc} \over {2a}} + d = 0$$ or, $$ - {{3bc} \over {2a}} + d = 0$$

or, $- 3bc + 2ad = 0$ or, $3bc - 2ad = 0$

Solutions · 1