JEE Main Straight Lines and Pair of Straight Lines Practice Questions With Solutions

A variable line $\mathrm{L}$L passes through the point $(3,5)$\left(\right. 3 , 5 \left.\right) and intersects the positive coordinate axes at the points $\mathrm{A}$A and $\mathrm{B}$B. The minimum area of the triangle $\mathrm{OAB}$OAB, where $\mathrm{O}$O is the origin, is :

A ray of light coming from the point $\mathrm{P}(1,2)$P \left(\right. 1 , 2 \left.\right) gets reflected from the point $\mathrm{Q}$Q on the $x$x-axis and then passes through the point $R(4,3)$R \left(\right. 4 , 3 \left.\right). If the point $S(h,k)$S \left(\right. h , k \left.\right) is such that $PQRS$P Q R S is a parallelogram, then $h{k}^2$h k^{2} is equal to:

If the line segment joining the points $(5,2)$\left(\right. 5 , 2 \left.\right) and $(2,a)$\left(\right. 2 , a \left.\right) subtends an angle $\frac{\pi }{4}$\frac{\pi}{4} at the origin, then the absolute value of the product of all possible values of $a$a is :

The equations of two sides $\mathrm{AB}$AB and $\mathrm{AC}$AC of a triangle $\mathrm{ABC}$ABC are $4x+y=14$4 x + y = 14 and $3x-2y=5$3 x - 2 y = 5, respectively. The point $(2,-\frac{4}{3})$\left(\right. 2 , - \frac{4}{3} \left.\right) divides the third side $\mathrm{BC}$BC internally in the ratio $2:1$2 : 1, the equation of the side $\mathrm{BC}$BC is

If the locus of the point, whose distances from the point $(2,1)$\left(\right. 2 , 1 \left.\right) and $(1,3)$\left(\right. 1 , 3 \left.\right) are in the ratio $5:4$5 : 4, is $a{x}^2+b{y}^2+cxy+dx+ey+170=0$a x^{2} + b y^{2} + c x y + d x + e y + 170 = 0, then the value of $a^2+2b+3c+4d+e$a^{2} + 2 b + 3 c + 4 d + e is equal to :

Let a variable line of slope $m>0$m > 0 passing through the point $(4,-9)$\left(\right. 4 , - 9 \left.\right) intersect the coordinate axes at the points $A$A and $B$B. The minimum value of the sum of the distances of $A$A and $B$B from the origin is

Let $\mathrm{A}(-1,1)$A \left(\right. - 1 , 1 \left.\right) and $\mathrm{B}(2,3)$B \left(\right. 2 , 3 \left.\right) be two points and $\mathrm{P}$P be a variable point above the line $\mathrm{AB}$AB such that the area of $\mathrm{△}\mathrm{PAB}$\triangle PAB is 10. If the locus of $\mathrm{P}$P is $\mathrm{a}x+\mathrm{by}=15$a x + by = 15, then $5\mathrm{a}+2\mathrm{b}$5 a + 2 \textrm{ } b is :

Let two straight lines drawn from the origin $\mathrm{O}$O intersect the line $3x+4y=12$3 x + 4 y = 12 at the points $\mathrm{P}$P and $\mathrm{Q}$Q such that $\mathrm{△}\mathrm{OPQ}$\triangle OPQ is an isosceles triangle and $\mathrm{\angle }\mathrm{POQ}={90}^{\circ }$\angle POQ = 90^{\circ}. If $l={\mathrm{OP}}^2+{\mathrm{PQ}}^2+{\mathrm{QO}}^2$l = OP^{2} + PQ^{2} + QO^{2}, then the greatest integer less than or equal to $l$l is :

The vertices of a triangle are $\mathrm{A}(-1,3),\mathrm{B}(-2,2)$A \left(\right. - 1 , 3 \left.\right) , B \left(\right. - 2 , 2 \left.\right) and $\mathrm{C}(3,-1)$C \left(\right. 3 , - 1 \left.\right). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

Let $A(a,b),B(3,4)$A \left(\right. a , b \left.\right) , B \left(\right. 3 , 4 \left.\right) and $C(-6,-8)$C \left(\right. - 6 , - 8 \left.\right) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P(2a+3,7b+5)$P \left(\right. 2 a + 3 , 7 b + 5 \left.\right) from the line $2x+3y-4=0$2 x + 3 y - 4 = 0 measured parallel to the line $x-2y-1=0$x - 2 y - 1 = 0 is