JEE MAIN Straight Line Previous Year Questions (PYQs) – Page 12 of 13

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$, respectively, is:

Consider the lines $x(3\lambda+1)+y(7\lambda+2)=17\lambda+5$, $\lambda$ being a parameter, all passing through a point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$, then the value of $d^2$ is:

Lines are drawn parallel to the line $4x-3y+2=0$, at a distance $\dfrac{3}{5}$ from the origin. Then which one of the following points lies on any of these lines?

A straight line through origin $O$ meets the lines $3y = 10 - 4x$ and $8x + 6y + 5 = 0$ at points $A$ and $B$ respectively. Then $O$ divides the segment $AB$ in the ratio :

A ray of light is incident along a line which meets another line $7x - y + 1 = 0$ at the point $(0,1)$. The ray is then reflected from this point along the line $y + 2x = 1$. Then the equation of the line of incidence of the ray of light is:

Let $\vec a=2\hat{\imath}+\hat{\jmath}+\hat{k}$, and $\vec b,\vec c$ be two nonzero vectors such that $\left\lvert \vec a+\vec b+\vec c \right\rvert=\left\lvert \vec a+\vec b-\vec c \right\rvert$ and $\vec b\cdot\vec c=0$. Consider the statements: (A) $\left\lvert \vec a+\lambda\vec c \right\rvert \ge \lvert \vec a\rvert \text{ for all } \lambda\in\mathbb{R}$. (B) $\vec a$ and $\vec c$ are always parallel. Then,

both (A) and (B) are correct  

only (A) is correct  

 neither (A) nor (B) is correct

Let $P$ and $Q$ be the points on the line $\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha,\beta,\gamma)$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is:

Let the shortest distance between the lines $\dfrac{x-3}{3}=\dfrac{y-\alpha}{-1}=\dfrac{z-3}{1}$ and $\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-\beta}{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha+\beta$ is

A point P moves on the line $2x - 3y + 4 = 0.$ If $Q(1, 4)$ and $R(3, -2)$ are fixed points, then the locus of the centroid of $\triangle PQR$ is a line :