A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1: 2x + y + 6 = 0$ and $L_2: 4x+2y-p = 0$, $p > 0$, at the points A and B, respectively. If $AB = \frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point A on the line $L_2$ is M, then $\frac{AM}{BM}$ is equal to
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Initial Analysis of the Line
The line passes through the origin (0, 0) and makes equal angles with the positive coordinate axes. This implies the line must have a slope of $1$ or $-1$. Therefore, the equation of the line is either $y = x$ or $y = -x$.
Analyzing the Given Lines
The given lines are:
- $L_1: 2x + y + 6 = 0$
- $L_2: 4x + 2y - p = 0$, where $p > 0$.
Notice that $L_2$ can be written as $2(2x + y) - p = 0$. This shows that $L_1$ and $L_2$ are parallel lines, both having a slope of $-2$.
Determining the Correct Line Equation
Let's find the intersection points A and B for both possible lines ($y=x$ and $y=-x$).
Case 1: Line is $y = x$
- Point A (intersection of $y=x$ and $L_1$): Substituting $y=x$ into $2x + y + 6 = 0$ gives $2x + x + 6 = 0 \implies 3x = -6 \implies x = -2$. So, $A = (-2, -2)$.
- Point B (intersection of $y=x$ and $L_2$): Substituting $y=x$ into $4x + 2y - p = 0$ gives $4x + 2x - p = 0 \implies 6x = p \implies x = \frac{p}{6}$. So, $B = (\frac{p}{6}, \frac{p}{6})$.
Case 2: Line is $y = -x$
- Point A (intersection of $y=-x$ and $L_1$): Substituting $y=-x$ into $2x + y + 6 = 0$ gives $2x - x + 6 = 0 \implies x = -6$. So, $A = (-6, 6)$.
- Point B (intersection of $y=-x$ and $L_2$): Substituting $y=-x$ into $4x + 2y - p = 0$ gives $4x - 2x - p = 0 \implies 2x = p \implies x = \frac{p}{2}$. So, $B = (\frac{p}{2}, -\frac{p}{2})$.
Using the Distance AB to Find p
We are given $AB = \frac{9}{\sqrt{2}}$.
Checking Case 1 ($y=x$):
- $A = (-2, -2)$, $B = (\frac{p}{6}, \frac{p}{6})$.
- $AB^2 = (\frac{p}{6} - (-2))^2 + (\frac{p}{6} - (-2))^2 = 2(\frac{p+12}{6})^2$.
- $(\frac{9}{\sqrt{2}})^2 = \frac{81}{2}$.
- So, $2(\frac{p+12}{6})^2 = \frac{81}{2} \implies (\frac{p+12}{6})^2 = \frac{81}{4} \implies \frac{p+12}{6} = \pm \frac{9}{2}$.
- $\frac{p+12}{6} = \frac{9}{2} \implies p+12 = 27 \implies p = 15$. (Since $p>0$, this is valid).
- $\frac{p+12}{6} = -\frac{9}{2} \implies p+12 = -27 \implies p = -39$. (Not valid as $p>0$).
- So, $p=15$ and the line is $y=x$. $A = (-2, -2)$ and $B = (\frac{15}{6}, \frac{15}{6}) = (\frac{5}{2}, \frac{5}{2})$.
Checking Case 2 ($y=-x$):
- $A = (-6, 6)$, $B = (\frac{p}{2}, -\frac{p}{2})$.
- $AB^2 = (\frac{p}{2} - (-6))^2 + (-\frac{p}{2} - 6)^2 = 2(\frac{p+12}{2})^2$.
- $2(\frac{p+12}{2})^2 = \frac{81}{2} \implies (\frac{p+12}{2})^2 = \frac{81}{4} \implies \frac{p+12}{2} = \pm \frac{9}{2}$.
- $\frac{p+12}{2} = \frac{9}{2} \implies p+12 = 9 \implies p = -3$. (Not valid as $p>0$).
- $\frac{p+12}{2} = -\frac{9}{2} \implies p+12 = -9 \implies p = -21$. (Not valid as $p>0$).
Conclusion: The line is $y=x$, $p=15$, $A=(-2,-2)$, $B=(\frac{5}{2}, \frac{5}{2})$, and $L_2$ is $4x+2y-15=0$.
Finding the Foot of the Perpendicular M
M is the foot of the perpendicular from A to $L_2$. $A = (-2, -2)$ and $L_2: 4x+2y-15=0$.
The slope of $L_2$ is $m_{L_2} = -2$. The slope of the line AM (perpendicular to $L_2$) is $m_{AM} = -\frac{1}{m_{L_2}} = -\frac{1}{-2} = \frac{1}{2}$.
The equation of the line AM passing through $A=(-2, -2)$ with slope $\frac{1}{2}$ is:
$y - (-2) = \frac{1}{2}(x - (-2))$
$y + 2 = \frac{1}{2}(x + 2)$
$2y + 4 = x + 2 \implies x - 2y - 2 = 0$.
Point M is the intersection of $L_2$ ($4x+2y-15=0$) and the line AM ($x-2y-2=0$).
Adding the two equations:
$(4x+2y-15) + (x-2y-2) = 0 \implies 5x - 17 = 0 \implies x = \frac{17}{5}$.
Substitute $x = \frac{17}{5}$ into $x - 2y - 2 = 0$:
$\frac{17}{5} - 2y - 2 = 0 \implies \frac{17}{5} - \frac{10}{5} = 2y \implies \frac{7}{5} = 2y \implies y = \frac{7}{10}$.
Thus, $M = (\frac{17}{5}, \frac{7}{10})$.
Calculating the Ratio AM/BM
We can calculate the lengths AM and BM.
- Length AM: This is the perpendicular distance from A to $L_2$. $AM = \frac{|4(-2) + 2(-2) - 15|}{\sqrt{4^2 + 2^2}} = \frac{|-8 - 4 - 15|}{\sqrt{16 + 4}} = \frac{|-27|}{\sqrt{20}} = \frac{27}{\sqrt{20}} = \frac{27}{2\sqrt{5}}$.
- Length BM: This is the distance between $B=(\frac{5}{2}, \frac{5}{2})$ and $M=(\frac{17}{5}, \frac{7}{10})$. $BM^2 = (\frac{17}{5} - \frac{5}{2})^2 + (\frac{7}{10} - \frac{5}{2})^2$ $BM^2 = (\frac{34 - 25}{10})^2 + (\frac{7 - 25}{10})^2$ $BM^2 = (\frac{9}{10})^2 + (\frac{-18}{10})^2 = \frac{81}{100} + \frac{324}{100} = \frac{405}{100} = \frac{81 \times 5}{100}$. $BM = \sqrt{\frac{81 \times 5}{100}} = \frac{9\sqrt{5}}{10}$.
Calculating the Ratio:
$\frac{AM}{BM} = \frac{\frac{27}{2\sqrt{5}}}{\frac{9\sqrt{5}}{10}} = \frac{27}{2\sqrt{5}} \times \frac{10}{9\sqrt{5}} = \frac{27 \times 10}{2 \times 9 \times 5} = \frac{270}{90} = 3$.
Alternative Geometric Method
Let L be the line $y=x$. Let $\alpha$ be the angle between line L and line $L_2$. The slope of L is $m_L = 1$ and the slope of $L_2$ is $m_{L_2} = -2$.
$\tan \alpha = |\frac{m_L - m_{L_2}}{1 + m_L m_{L_2}}| = |\frac{1 - (-2)}{1 + (1)(-2)}| = |\frac{3}{1 - 2}| = |-3| = 3$.
Consider the right-angled triangle formed by A, M, and the projection of A onto $L_2$ (which is M), and the point B on $L_2$. M and B lie on $L_2$. AM is perpendicular to $L_2$. The angle between the line segment AB (on line L) and the line segment BM (on line $L_2$) is $\alpha$.
In the context of the right triangle involving AM and BM, consider the projection. Point M is the foot of the perpendicular from A to $L_2$. Point B is on $L_2$. The segment BM lies on $L_2$. AM is perpendicular to $L_2$.
The distance $BM$ can be seen as the projection of the line segment AB onto the line $L_2$. Thus, $BM = AB \cos \alpha$.
From $\tan \alpha = 3$, we can form a right triangle with opposite side 3, adjacent side 1, and hypotenuse $\sqrt{3^2+1^2} = \sqrt{10}$. So, $\cos \alpha = \frac{1}{\sqrt{10}}$.
$BM = AB \cos \alpha = \frac{9}{\sqrt{2}} \times \frac{1}{\sqrt{10}} = \frac{9}{\sqrt{20}} = \frac{9}{2\sqrt{5}}$.
The length AM is the perpendicular distance from A to $L_2$, which we calculated as $AM = \frac{27}{2\sqrt{5}}$.
The ratio $\frac{AM}{BM} = \frac{\frac{27}{2\sqrt{5}}}{\frac{9}{2\sqrt{5}}} = \frac{27}{9} = 3$.
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