Straight Line – JEE Advanced Previous Year Questions with Solutions

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Q. Let $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ be the points on the plane with position vectors $-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}, 4 \hat{\mathrm{i}}, 3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}$ and $-3 \hat{\mathrm{j}}$ and $-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}$ respectively. The quadrilateral PQRS must be a

(A) parallelogram, which is neither a rhombus nor a rectangle

(B) square

(C) rectangle, but not a square

(D) rhombus, but not a square

[JEE 2010, 3]

Sol. (A)

$\Rightarrow$ PQRS is a parallelogram but neither a rhombus nor a rectangle.


Q. A straight line L through the point $(3,-2)$ is inclined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersect the x-axis, then the equation of $L$ is

(A) $y+\sqrt{3} x+2-3 \sqrt{3}=0$

(B) $\mathrm{y}-\sqrt{3} \mathrm{x}+2+3 \sqrt{3}=0$

(C) $\sqrt{3} y-x+3+2 \sqrt{3}=0$

(D) $\sqrt{3} y+x-3+2 \sqrt{3}=0$

[JEE 2011, 3 (–1)]

Sol. (B)


Q. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines $a x+b y+c=0$ and $b x+a y+c=0$ is less than $2 \sqrt{2} .$ Then

(A) a + b – c > 0

(B) a – b + c < 0

(C) a – b + c > 0

(D) a + b – c < 0

[JEE-Advanced 2013, 2]

Sol. (A or C or A,C)

Point of intersection of both lines is $\left(-\frac{c}{(a+b)},-\frac{c}{(a+b)}\right)$

Distance between $\left(-\frac{c}{(a+b)},-\frac{c}{(a+b)}\right) \&(1,1)$ is

Distance $=\sqrt{\frac{(a+b+c)^{2}}{(a+b)^{2}} \times 2}<2 \sqrt{2}$

$a+b+c<2(a+b)$

$a+b-c>0$

According to given condition option (C) also correct.


Q. For a point $P$ in the plane, let $d_{1}(P)$ and $d_{2}(P)$ be the distances of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_{1}(P)+d_{2}(P) \leq 4,$ is

[JEE(Advanced)-2014, 3]

Sol. 6


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