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\gt 0 , at the points A and B , respectively. If A B=\frac9\sqrt2 and the foot of the perpendicular from the point A on the line L_2 is M , then \fracA MB M is equal to

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q44 with Solution

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Correct answer: B

A line making equal angles with the positive $x$ - and $y$ -axes has slope $1$ . Hence its equation is $y=x.$

So points $A$ and $B$ are the intersections of $y=x$ with $L_1$ and $L_2$ respectively.

Given $L_1: 2x+y+6=0.$ Substitute $y=x$ : $2x+x+6=0$ $3x+6=0$ $x=-2,\, y=-2.$ Thus, $A=(-2,-2).$

Given $L_2: 4x+2y-p=0.$ Substitute $y=x$ : $4x+2x-p=0$ $6x=p$ $x=\frac p6,\, y=\frac p6.$ Thus, $B=\left(\frac p6,\frac p6\right).$

Distance between $A(-2,-2)$ and $B\left(\dfrac p6,\dfrac p6\right)$ is

=\sqrt{2\left(\frac p6+2\right)^2} =\sqrt2\left|\frac p6+2\right|.$$ Since $p>0$, we have $\dfrac p6+2>0$, so $$AB=\sqrt2\left(\frac p6+2\right).$$ Given $$\sqrt2\left(\frac p6+2\right)=\frac{9}{\sqrt2}.$$ Multiply by $\sqrt2$: $$2\left(\frac p6+2\right)=9$$ $$\frac p3+4=9$$ $$\frac p3=5$$ $$p=15.$$ Therefore, $$B=\left(\frac{15}{6},\frac{15}{6}\right)=\left(\frac52,\frac52\right).$$ --- 5. **Find $AM$** $M$ is the foot of the perpendicular from $A$ to line $L_2$. So $AM$ is the perpendicular distance from $A(-2,-2)$ to $$L_2:4x+2y-15=0.$$ Distance formula gives $$AM=\frac{|4(-2)+2(-2)-15|}{\sqrt{4^2+2^2}} =\frac{|-8-4-15|}{\sqrt{20}} =\frac{27}{2\sqrt5}.$$ --- 6. **Find $BM$** Since $B$ lies on $L_2$, and $M$ is the foot of the perpendicular from $A$ to $L_2$, triangle $ABM$ is right-angled at $M$. Hence, $$AB^2=AM^2+BM^2.$$ We know $$AB=\frac{9}{\sqrt2}\implies AB^2=\frac{81}{2}.$$ Also, $$AM=\frac{27}{2\sqrt5}\implies AM^2=\frac{729}{20}.$$ So, $$BM^2=AB^2-AM^2 =\frac{81}{2}-\frac{729}{20} =\frac{810-729}{20} =\frac{81}{20}.$$ Thus, $$BM=\frac{9}{2\sqrt5}.$$ --- 7. **Compute the ratio** $$\frac{AM}{BM}=\frac{\frac{27}{2\sqrt5}}{\frac{9}{2\sqrt5}}=3.$$ So the correct option is $$\boxed{\text{B: }3}.$$ --- 8. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** Hence, the answers agree.

View written solutionFree

Correct answer: B

A line making equal angles with the positive $x$ - and $y$ -axes has slope $1$ . Hence its equation is $y=x.$

So points $A$ and $B$ are the intersections of $y=x$ with $L_1$ and $L_2$ respectively.

Given $L_1: 2x+y+6=0.$ Substitute $y=x$ : $2x+x+6=0$ $3x+6=0$ $x=-2,\, y=-2.$ Thus, $A=(-2,-2).$

Given $L_2: 4x+2y-p=0.$ Substitute $y=x$ : $4x+2x-p=0$ $6x=p$ $x=\frac p6,\, y=\frac p6.$ Thus, $B=\left(\frac p6,\frac p6\right).$

Distance between $A(-2,-2)$ and $B\left(\dfrac p6,\dfrac p6\right)$ is

=\sqrt{2\left(\frac p6+2\right)^2} =\sqrt2\left|\frac p6+2\right|.$$ Since $p>0$, we have $\dfrac p6+2>0$, so $$AB=\sqrt2\left(\frac p6+2\right).$$ Given $$\sqrt2\left(\frac p6+2\right)=\frac{9}{\sqrt2}.$$ Multiply by $\sqrt2$: $$2\left(\frac p6+2\right)=9$$ $$\frac p3+4=9$$ $$\frac p3=5$$ $$p=15.$$ Therefore, $$B=\left(\frac{15}{6},\frac{15}{6}\right)=\left(\frac52,\frac52\right).$$ --- 5. **Find $AM$** $M$ is the foot of the perpendicular from $A$ to line $L_2$. So $AM$ is the perpendicular distance from $A(-2,-2)$ to $$L_2:4x+2y-15=0.$$ Distance formula gives $$AM=\frac{|4(-2)+2(-2)-15|}{\sqrt{4^2+2^2}} =\frac{|-8-4-15|}{\sqrt{20}} =\frac{27}{2\sqrt5}.$$ --- 6. **Find $BM$** Since $B$ lies on $L_2$, and $M$ is the foot of the perpendicular from $A$ to $L_2$, triangle $ABM$ is right-angled at $M$. Hence, $$AB^2=AM^2+BM^2.$$ We know $$AB=\frac{9}{\sqrt2}\implies AB^2=\frac{81}{2}.$$ Also, $$AM=\frac{27}{2\sqrt5}\implies AM^2=\frac{729}{20}.$$ So, $$BM^2=AB^2-AM^2 =\frac{81}{2}-\frac{729}{20} =\frac{810-729}{20} =\frac{81}{20}.$$ Thus, $$BM=\frac{9}{2\sqrt5}.$$ --- 7. **Compute the ratio** $$\frac{AM}{BM}=\frac{\frac{27}{2\sqrt5}}{\frac{9}{2\sqrt5}}=3.$$ So the correct option is $$\boxed{\text{B: }3}.$$ --- 8. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** Hence, the answers agree.