A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines and and , p > 0, at the points A and B, respectively. If and the foot of the perpendicular from the point A on the line is M, then is eq

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Solution

Q.
A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_{1} : 2 x + y + 6 = 0$ and $L_{2} : 4 x + 2 y - p = 0$ , p > 0, at the points A and B, respectively. If $A B = \frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point A on the line $L_{2}$ is M, then $\frac{A M}{B M}$ is equal to [2025]

1 4

2 3

3 5

4 2

Ans.
(2)

We have, $L_{1} : 2 x + y + 6 = 0$ and $L_{2} : 4 x + 2 y - p = 0$

Line y = x passes through origin and makes equal angles with the positive coordinates axes.

Slope of y = x is $m_{1} = 1$ , Slope of $L_{2}$ is $m_{2} = - 2$

So, $\tan θ =$ $| \frac{1 + 2}{1 - 2} |$ $= 3$

Now, in $△$ ABM

$\Rightarrow \tan θ = \frac{A M}{B M} = 3$ .

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