Straight Lines JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Straight Lines, free to read — no sign-in needed. The full chapter has 22 questions; sign in to attempt the remaining 17 in the exam simulator.

1. Q1JEE Advanced Adv 2024 (Paper 1)
   Let $\mathbb{R}^2$ denote $\mathbb{R} \times \mathbb{R}$. Let $S=\left\{(a, b, c): a, b, c \in \mathbb{R} \text { and } a x^2+2 b x y+c y^2\gt 0 \text { for all }(x, y) \in \mathbb{R}^2-\{(0,0)\}\right\} .$ Then which of the following statements is (are) TRUE?
   1. A.$\left(2, \frac{7}{2}, 6\right) \in S$
   2. B.If $\left(3, b, \frac{1}{12}\right) \in S$, then $|2 b| \lt 1$.
   3. C.For any given $(a, b, c) \in S$, the system of linear equations $\begin{aligned}& a x+b y=1 \\& b x+c y=-1\end{aligned}$ has a unique solution.
   4. D.For any given $(a, b, c) \in S$, the system of linear equations $\begin{aligned}& (a+1) x+b y=0 \\& b x+(c+1) y=0\end{aligned}$ has a unique solution.
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   Answer: B,C,D

   $\begin{aligned} & a x^2+2 b x y+c y^2>0 \\ & y, x \in \mathbb{R}-\{(0,0)\} \\ & \Rightarrow c\left(\frac{y}{x}\right)^2+2 b\left(\frac{y}{x}\right)+a>0 \\ & 4 b^2-4 a c<0 \\ & \Rightarrow b^22 \times 6 \end{aligned}$ $\therefore$ option A is incorrect $\begin{aligned} & \text {(B) if }\left(3, b, \frac{1}{12}\right) \in S \\ & \Rightarrow b^2<3 \cdot \frac{1}{12} \\ & \Rightarrow b^2<\frac{1}{4} \\ & \Rightarrow 4 b^2<1 \\ & \Rightarrow|2 b|<1 \text { option } B \text { is correct } \end{aligned}$ $\begin{aligned} & \text {(C) } a x+b y=1 \\ & b x+c y=-1 \\ & D=\left|\begin{array}{ll} a & b \\ b & c \end{array}\right|=a c-b^2 \neq 0 \end{aligned}$ $\therefore$ unique solution option C is correct. $\begin{aligned} & \text {(D) }(a+1) \mathrm{x}+\mathrm{by}=0 \\ & \mathrm{bx}+(\mathrm{c}+1) \mathrm{y}=0 \\ & D=\left|\begin{array}{cc} (a+1) & b \\ b & (c+1) \end{array}\right| \\ & =(a+1)(c+1)-b^2 \\ & \Rightarrow a c-b^2+a+c+1 \\ & b^20 \end{aligned}$ $\therefore$ unique solution $\therefore$ option D is correct.
2. Q2JEE Advanced Adv 2021 (Paper 1)
   Consider the lines $L_{1}$ and $L_{2}$ defined by $$L_{1}: x \sqrt{2}+y-1=0 \text { and } L_{2}: x \sqrt{2}-y+1=0$$ For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_{1}$ and the distance of $P$ from $L_{2}$ is $\lambda^{2}$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$. Let the perpendicular bisector of $R S$ meet $C$ at two distinct points $R^{\prime}$ and $S^{\prime}$. Let $D$ be the square of the distance between $R^{\prime}$ and $S^{\prime}$. The value of $D$ is
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   Answer: 77.14

   From the first question The equation of the locus is $\left|2{x}^{2}-(y-1{)}^{2}\right|=27$ The line is $y=2x+1$ or $y-1=2x$ By substituting the value of $y$ in the equation of the curve $C$, we get $\left|2{x}^{2}-(y-1{)}^{2}\right|=27$ $\Rightarrow \left|2{x}^{2}-(2x{)}^{2}\right|=27$ $\Rightarrow 2{x}^{2}=27$ $\Rightarrow x=\pm 3\sqrt{\dfrac{3}{2}}$ $\Rightarrow {x}_{1},{x}_{2}=\pm 3\sqrt{\dfrac{3}{2}}$ Let $M$ be the mid-point of ${R}^{'}\&{S}^{'}$ So, the $x$ coordinate of $T$ is $\dfrac{{x}_{1}+{x}_{2}}{2}=0$ It lies on $y=2x+1$ So, the coordinates of $T$ are $\left(0,1\right)$ Slope of the line perpendicular to $y=2x+1$ is $\dfrac{-1}{2}$ So, the equation of perpendicular bisector is $y-1=\dfrac{-1}{2}\left(x-0\right)$ Or, $x+2y=2$ Let coordinate of ${R}^{'}\&{S}^{'}$ are $\left({p}_{1},{q}_{1}\right)\&\left({p}_{2},{q}_{2}\right)$, we get $D={\left({p}_{2}-{p}_{1}\right)}^{2}+{\left({q}_{2}-{q}_{1}\right)}^{2}$ Since both the points satisfy the equation of the line, we get $D={\left[2\left({q}_{2}-{q}_{1}\right)\right]}^{2}+{\left({q}_{2}-{q}_{1}\right)}^{2}$ $D=5{\left({q}_{2}-{q}_{1}\right)}^{2}$ Solving, $x+2y=2$ with $2{x}^{2}-(y-1{)}^{2}=27$, we get $2{\left(2-2y\right)}^{2}-(y-1{)}^{2}=27$ $\Rightarrow 7(y-1{)}^{2}=27$ $\Rightarrow y-1=\pm 3\sqrt{\dfrac{3}{7}}$ $\Rightarrow y=1\pm 3\sqrt{\dfrac{3}{7}}$ $\Rightarrow {q}_{1},{q}_{2}=1\pm 3\sqrt{\dfrac{3}{7}}$ So, ${\left({q}_{2}-{q}_{1}\right)}^{2}={\left(6\sqrt{\dfrac{3}{7}}\right)}^{2}$ Hence, $D=5{\left({q}_{2}-{q}_{1}\right)}^{2}=5\times {\left(6\sqrt{\dfrac{3}{7}}\right)}^{2}=77.14$
3. Q3JEE Advanced Adv 2021 (Paper 1)
   Consider the lines $L_{1}$ and $L_{2}$ defined by $$L_{1}: x \sqrt{2}+y-1=0 \text { and } L_{2}: x \sqrt{2}-y+1=0$$ For a fixed constant $\lambda$, let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_{1}$ and the distance of $P$ from $L_{2}$ is $\lambda^{2}$. The line $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$. Let the perpendicular bisector of $R S$ meet $C$ at two distinct points $R^{\prime}$ and $S^{\prime}$. Let $D$ be the square of the distance between $R^{\prime}$ and $S^{\prime}$. The value of ${\lambda }^{2}$ is
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   Answer: 9

   Let the point $P$ is $\left(h,k\right)$ Distance of $P$ from ${L}_{1}=\left|\dfrac{h\sqrt{2}+k-1}{\sqrt{(\sqrt{2}{)}^{2}+{1}^{2}}}\right|$ $=\left|\dfrac{h\sqrt{2}+k-1}{\sqrt{3}}\right|$ Distance of $P$ from ${L}_{2}=\left|\dfrac{h\sqrt{2}-k+1}{\sqrt{(\sqrt{2}{)}^{2}+{1}^{2}}}\right|$ $=\left|\dfrac{h\sqrt{2}-k+1}{\sqrt{3}}\right|$ $∴$ The equation of the locus of $P$ is $\left|\dfrac{h\sqrt{2}+k-1}{\sqrt{3}}\right|\times \left|\dfrac{h\sqrt{2}-k+1}{\sqrt{3}}\right|={\lambda }^{2}$ $\left|\left(\dfrac{h\sqrt{2}+k-1}{\sqrt{3}}\right)\left(\dfrac{h\sqrt{2}-k+1}{\sqrt{3}}\right)\right|={\lambda }^{2}$ $\Rightarrow \left|2{h}^{2}-(k-1{)}^{2}\right|=3{\lambda }^{2}$ Hence, the equation of the locus is $\left|2{x}^{2}-(y-1{)}^{2}\right|=3{\lambda }^{2}$ The line is $y=2x+1$ or $y-1=2x$ By substituting the value of $y$ in the equation of the curve $C$, we get $\left|2{x}^{2}-(y-1{)}^{2}\right|=3{\lambda }^{2}$ $\Rightarrow \left|2{x}^{2}-(2x{)}^{2}\right|=3{\lambda }^{2}$ $\Rightarrow 2{x}^{2}=3{\lambda }^{2}$ $\Rightarrow x=\pm \sqrt{\dfrac{3}{2}}\lambda$ $\Rightarrow {x}_{2}-{x}_{1}=|\sqrt{6}\lambda |$ Also, $y-1=2x$ $\Rightarrow {y}_{2}-1=2{x}_{2}$ and ${y}_{1}-1=2{x}_{1}$ $\Rightarrow {y}_{2}-{y}_{1}=2\left({x}_{2}-{x}_{1}\right)$ $\Rightarrow {y}_{2}-{y}_{1}=|2\sqrt{6}\lambda |$ Given $RS=\sqrt{270}$ $\Rightarrow \sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}=\sqrt{270}$ $\Rightarrow (\sqrt{6}\lambda {)}^{2}+(2\sqrt{6}\lambda {)}^{2}=270$ $30{\lambda }^{2}=270$ ${\lambda }^{2}=9$
4. Q4JEE Advanced Adv 2013 (Paper 2)
   Paragraph: Let $S=S_{1} \cap S_{2} \cap S_{3}$, where $S_{1}=\{z \in \mathbb{C}:|z|\lt 4\}, \quad S_{2}=\left\{z \in \mathbb{C}: \operatorname{Im}\left[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right]\gt 0\right\}$ and $S_{3}=\{z \in \mathbb{C}: \operatorname{Re} z\gt 0\}$. Question: $\min _{z \in S}|1-3 i-\mathrm{z}|=$
   1. A.$\dfrac{2-\sqrt{3}}{2}$
   2. B.$\dfrac{2+\sqrt{3}}{2}$
   3. C.$\dfrac{3-\sqrt{3}}{2}$
   4. D.$\dfrac{3+\sqrt{3}}{2}$
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   Answer: (C)

   $\text{min}_{\text{z}\in \text{s}}\left|1-3\text{i}-\text{z}\right|$ perpendicular length of point (1, -3) from line $\sqrt{3}\text{x}+\text{y}=0$ $\left|\dfrac{\sqrt{3}-3}{\sqrt{3+1}}\right|=\left|\dfrac{\sqrt{3}-3}{2}\right|=\dfrac{3-\sqrt{3}}{2}$
5. Q5JEE Advanced Adv 2013 (Paper 1)
   For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than $2\sqrt{2}$.Then
   1. A.$a+b-c\gt 0$
   2. B.$a-b+c\lt 0$
   3. C.$a-b+c\gt 0$
   4. D.$a+b-c\lt 0$
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   Answer: (A)

   $\left(\text{a}-\text{b}\right)\text{x}+\left(\text{b}-\text{a}\right)\text{y}=0$ $\Rightarrow \text{x}=\text{y}$ $\Rightarrow \text{Point of intersection}\left(\dfrac{-\text{c}}{\text{a}+\text{b}}\text{,}\dfrac{-\text{c}}{\text{a}+\text{b}}\right)$ Now using distance formula between two points $d=\sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({y}_{1}-{y}_{2}\right)}^{2}}$ $\Rightarrow \sqrt{{\left(1+\dfrac{\text{c}}{\text{a}+\text{b}}\right)}^{2}+{\left(1+\dfrac{\text{c}}{\text{a}+\text{b}}\right)}^{2}}\lt 2\sqrt{2}$ $\Rightarrow \sqrt{2}\left(\left|\dfrac{\text{a}+\text{b}+\text{c}}{\text{a}+\text{b}}\right|\right)\lt 2\sqrt{2}$ $\Rightarrow \left(\dfrac{\text{a}+\text{b}+\text{c}}{\text{a}+\text{b}}\right)\lt 2$ $\text{⇒a}+\text{b}+\text{c}\lt 2\left(\text{a}+\text{b}\right)$ $\Rightarrow \text{a}+\text{b}-\text{c}\gt 0$