Question

Find the shortest distance between the (in units) lines 2y = -2x - 4 and -3y = 3x - 6.​

Answer 1

Answer:

$2y = - 2x - 4(1) \\ - 3y = 3x - 6(2) \\ from \: equcation \: (1) \\ y = \frac{ - 2 x- 4}{2} (3) \\ y = \frac{ - 1x - 2}{1} \\ putting \: (3)equation \: in \: (2) \\ - 3( - 1x - 2) = 3x - 6 \\ 3x + 6 = 3x - 6 \\ x = 12 \\ putting \: the \: value \: of \: x = 12 \: in \: equation \: (1) \\ 2y = - 2(12) - 4 \\ 2y = - 24 - 4 \\ 2y = - 28 \\ y = - 14$

Answer 2

Answer:

The shortest distance between 2y = -2x -4 and -3y = 3x - 6 is 2√2 or √8 units

Step-by-step explanation:

1) Simplying both the straight lines

$2y = - 2x - 4$

$2x + 2y = - 4$

$x + y = - 2$

... equation (1)

$- 3y = 3x - 6$

$- 3x - 3y = - 6$

$3x + 3y = 6$

$x + y = 2$

... equation (2)

So the two lines are x + y = 2 and x+ y = -2

2) Finding the nearest points

Both the lines are parallel to each other. The distance between two exactly opposite points will be the shortest. Opposite points means the points of intersection between perpendicular line to given straight lines and the lines itself.

If (-2,0) lies on the line x + y = -2 then it's opposite point will be (0,2) lying on the line x + y = 2 (Refer image)

3) Finding the Shortest distance.

Distance between (-2,0) and (0,2) is

$= \sqrt{ { (- 2 - 0)}^{2} + {(0 - 2)}^{2} }$

$= \sqrt{ { (- 2)}^{2} + {(2)}^{2} }$

$= \sqrt{4 + 4}$

$= \sqrt{8}$

$= 2 \sqrt{2}$

Therefore 2√2 units is the shortest distance between lines x + y = -2 and x + y = 2

Therefore the shortest distance between 2y = -2x -4 and -3y = 3x - 6 is 2√2 or √8 units