Rectangular Cartesian Coordinate System (Points) JEE Main Mathematics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: b

Explanation: (15, –10) $\begin{matrix}L e t   p o i n t   \left(x_{1} , y_{1}\right)   b e   o n   t h e   l i n e   3 x + 4 y = 5 \\ \therefore 3 x_{1} + 4 y_{1} = 5 . . . \left(i\right) \\ A l s o ,   \left(x_{1} - 1\right)^{2} + \left(y_{1} - 2\right)^{2} = \left(x_{1} - 3\right)^{2} + \left(y_{1} - 4\right)^{2} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} - 2 x_{1} - 4 y_{1} + 5 \\ = x_{1}^{2} + y_{1}^{2} - 6 x_{1} - 8 y_{1} + 25 \\ \Rightarrow 4 x_{1} + 4 y_{1} = 20 . . . \left(i i\right) \\ O n   s o l v i n g   E q s . \left(i\right)   a n d   \left(i i\right) ,   w e   g e t \\ x_{1} = 15 , y_{1} = - 10\end{matrix}$

Correct Answer: b

Explanation: y + 3x = 20xy $\begin{matrix}L e t t h e e q u a t i o n o f t h e l i n e b e \frac{x}{a} + \frac{y}{b} = 1 \\ S i n c e , i t p a s s e s t h r o u g h \left(\frac{1}{5} , \frac{1}{5}\right) \\ T h e r e f o r e , \frac{1}{5 a} + \frac{1}{5 b} = 1 \\ a + b = 5 a b \\ S i n c e , t h e p o i n t P \left(x , y\right) d i v i d e s A B j o i n i n g A \left(a , 0\right) a n d B \left(0 , b\right) \\ i n t e r n a l l y i n r a t i o 3 : 1 \\ T h e r e f o r e , x = \frac{a}{4} , y = \frac{3 b}{4} \\ a = 4 x a n d b = \frac{4 y}{3} \\ O n p u t t i n g t h e v a l u e o f a a n d b i n e q u a t i o n \left(i\right) , w e g e t \\ 4 x + \frac{4 y}{3} = 5 \left(4 x\right) \left(\frac{4 y}{3}\right) \\ 3 x + y = 20 x y\end{matrix}$

Correct Answer: b

Explanation: – 4 Given that the mid-point of diagonals lies on line _y_ = 2_x_ + _c_, Here mid point is (3, 2) Hence, _c_ = – 4

Correct Answer: 6

Explanation: For $P R = R Q$ to be minimum it should be the path of light $\therefore \angle P R A = \angle Q R M$ From similar $\Delta P A R$ and $\Delta Q M R$ $\frac{A R}{R M} = \frac{P A}{Q M}$ $\begin{matrix}\Rightarrow \frac{h + 3}{0 - h} = \frac{4}{1} \\ \Rightarrow h = - \frac{3}{5} \\ \Rightarrow 10 \left|h\right| = 10 \left|- \frac{3}{5}\right| = 6\end{matrix}$

Correct Answer: a

Explanation: $\begin{matrix}L e n g t h o f t h e c h o r d \\ = \sqrt{\left[\right. 4 c o s \left(\right. + 60^{} \left.\right) 4 c o s \left]\right.^{2} + \left[\right. 4 s i n \left(\right. + 60^{} \left.\right) 4 s i n \left]\right.^{2}} \\ = 4 \sqrt{c o s^{2} \left(\right. + 60^{} \left.\right) + c o s^{2} + s i n^{2} \left(\right. + 60^{} \left.\right) + s i n^{2} 2 c o s cos \left(\right. + 60^{} \left.\right) 2 s i n sin \left(\right. + 60^{} \left.\right)} \\ = 4 \sqrt{1 + 1 2 c o s 60^{}} \\ = 4\end{matrix}$