1. ⇒ (MHT CET 2023 12th May Morning Shift )

The co-ordinates of the points on the line $2 x - y = 5$ which are the distance of 1 unit from the line $3 x + 4 y = 5$ are

A. $(\frac{30}{11}, \frac{- 5}{11}), (\frac{20}{11}, \frac{15}{11})$

B. $(\frac{- 30}{11}, \frac{5}{11}), (\frac{- 20}{11}, \frac{15}{11})$

C. $(\frac{30}{11}, \frac{5}{11}), (\frac{20}{11}, \frac{- 15}{11})$

D. $(\frac{- 30}{11}, \frac{5}{11}), (\frac{- 20}{11}, \frac{- 15}{11})$

Correct answer option is (C)

Let $(x_{1}, y_{1})$ be the required point

$∴ 2 x_{1} - y_{1} = 5$ ..... (i)

Also, $(x_{1}, y_{1})$ is at the distance of 1 unit from line $3 x + 4 y = 5$

$\begin{array}{ll} ∴ & 1 = | \frac{3 x_{1} + 4 y_{1} - 5}{\sqrt{9 + 16}} | \\ ∴ & \pm 5 = 3 x_{1} + 4 y_{1} - 5 \\ ∴ & 3 x_{1} + 4 y_{1} - 5 = 5 or 3 x_{1} + 4 y_{1} - 5 = - 5 \\ ∴ & 3 x_{1} + 4 y_{1} = 10 .... (ii) \end{array}$

or

$3 x_{1} + 4 y_{1} = 0$ .... (iii)

Solving equations (i) and (ii), we get

$x_{1} = \frac{30}{11}$ and $y_{1} = \frac{5}{11}$

Solving equation (i) and (iii), we get

$x_{1} = \frac{20}{11}$ and $y_{1} = \frac{- 15}{11}$

$∴ (\frac{30}{11}, \frac{5}{11})$ and $(\frac{20}{11}, \frac{- 15}{11})$ are the required points.

2. ⇒ (MHT CET 2021 21th September Evening Shift )

If $p$ is the length of the perpendicular from origin to the line whose intercepts on the axes are a and $b$ , then $\frac{1}{a^{2}} + \frac{1}{b^{2}} =$

A. $p^{2}$

B. $\frac{1}{2 p^{2}}$

C. $2 p^{2}$

D. $\frac{1}{p^{2}}$

Correct answer option is (D)

MHT CET 2021 21th September Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 16 English Explanation

Refer figure

Equation of given line is

$\frac{x}{a} + \frac{y}{b} = 1 i.e. b x + a y = a b$ ..... (1)

Distance of line (1) from origin is

$\begin{array}{ll} \frac{| - a b |}{\sqrt{a^{2} + b^{2}}} = p & \Rightarrow a^{2} + b^{2} = \frac{a^{2} b^{2}}{p^{2}} \\ ∴ \frac{a^{2} + b^{2}}{a^{2} b^{2}} = \frac{1}{p^{2}} & \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} \end{array}$

3. ⇒ (MHT CET 2021 21th September Morning Shift )

The product of the perpendicular distances from $(2, - 1)$ to the pair of lines $2 x^{2} - 5 x y + 2 y^{2} = 0$ is

A. $\frac{9}{\sqrt{5}}$ units

B. $\frac{1}{\sqrt{5}}$ units

C. 4 units

D. 9 units

Correct answer option is (C)

$2 x^{2} - 5 x y + 2 y^{2} = 0$

$∴ 2 x^{2} - 4 x y - x y + 2 y^{2} = 0 \Rightarrow 2 x (x - 2 y) - y (x - 2 y) = 0$

$∴ (2 x - y) (x - 2 y) = 0$ Thus lines are $2 x - y = 0$ and $x - 2 y = 0$

Distance of point $(2, - 1)$ from these two lines are respectively $| \frac{(2) (2) + (- 1) (- 1)}{\sqrt{4 + 1}} |$ and $| \frac{(1) (2) + (- 1) (- 2)}{\sqrt{1 + 4}} |$ i.e. $\frac{5}{\sqrt{5}}$ and $\frac{4}{\sqrt{5}}$

Hence required answer is $\frac{5}{\sqrt{5}} \times \frac{4}{\sqrt{5}} = 4$

⇒Straight Line