If the point of intersection of the lines 2ax | vedclass

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Question

(B) Let the coordinates of the point of intersection be $(\alpha, -\alpha)$ because it lies in the $4^{th}$ quadrant and is equidistant from the axes.

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(B) Let the coordinates of the point of intersection be $(\alpha, -\alpha)$ because it lies in the $4^{th}$ quadrant and is equidistant from the axes.Since $(\alpha, -\alpha)$ lies on $2ax + 4ay + c = 0$, we have:$2a(\alpha) + 4a(-\alpha) + c = 0$$-2a\alpha + c = 0 \Rightarrow \alpha = \frac{c}{2a} \quad (i)$Since $(\alpha, -\alpha)$ lies on $7bx + 3by - d = 0$, we have:$7b(\alpha) + 3b(-\alpha) - d = 0$$4b\alpha - d = 0 \Rightarrow \alpha = \frac{d}{4b} \quad (ii)$Equating $(i)$ and $(ii)$:$\frac{c}{2a} = \frac{d}{4b}$$4bc = 2ad$$\frac{ad}{bc} = \frac{4}{2} = \frac{2}{1}$Therefore, $ad : bc = 2 : 1$.

If the point of intersection of the lines $2ax + 4ay + c = 0$ and $7bx + 3by - d = 0$ lies in the $4^{th}$ quadrant and is equidistant from the two axes, where $a, b, c,$ and $d$ are non-zero numbers, then $ad : bc$ is equal to

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