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Equation of a Plane : Vector Form

L1: 2x+y=50 a...

Question

$L_{1} : 2 x + y = 50$ and $L_{2} : y = m x + 1$ are two lines.
Square of the distance between the points of intersection of $L_{1}$ and $L_{2}$ for the greatest and the least integer values of $m$ is

A

5

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B

20

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C

25

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D

30

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Solution

The correct option is B $20$
The point of intersection of the given lines is $(\frac{49}{2 + m}, \frac{2 + 50 m}{2 + m})$ .
For the x coordinate $\frac{49}{2 + m}$ to be an integer 49 should be a multiple of $2 + m .$
Hence $m$ can be $47, 5, - 1$ etc.

The greatest value of $m$ is $47.$

Similarly the least value occurs when $2 + m = - 49 \Rightarrow m = - 51$ .
For greatest value of m:
For the point of intersection of lines $2 x + y = 50$ and $47 x - y + 1 = 0$
$2 x + y = 50 \Rightarrow x = \frac{50 - y}{2}$
$\Rightarrow 47 (\frac{50 - y}{2}) - y + 1 = 0$ i.e $(x, y) = (1, 48)$
For the least value of m:
For the point of intersection of lines $2 x + y = 50$ and $51 x + y - 1 = 0$
$2 x + y = 50 \Rightarrow x = \frac{50 - y}{2}$
$\Rightarrow 51 (\frac{50 - y}{2}) + y - 1 = 0$ i.e $(x, y) = (- 1, 52)$
The square of the distance between these two points is $(1 + 1)^{2} + (48 - 52)^{2} = 20$

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