Question

Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:

1. perpendicular to the line 2x – y + 7 = 0
2. parallel to it.

Answer 1

Writing the given equation in the form of y = mx + c

(k − 2)x + (k + 3)y − 5 = 0    ...(i)

`=>` (k + 3)y = −(k − 2)x + 5

`=> y = ((k - 2))/(k + 3)x + 5/(k + 3)`

`=> y = (k - 2)/(k + 3) xx x + 5/(k + 3)`

Slope of the line = `-(k - 2)/(k + 3)`

Again equation of given line be

2x − y + 7 = 0   ...(ii)

`=>` −y = −2x − 7

`=>` y = 2x + 7

Slope of the line = 2

i. When the given line (i) perpendicular to the line (ii)

`-(k - 2)/(k + 3) xx 2 = -1`   ...(Product of slopes = –1)

`=>` −(k − 2)2 = −1(k + 3)

`=>` −2k + 4 = −k − 3

`=>` 2k − 4 = k + 3

`=>` 2k − k = 3 + 4

`=>` k = 7

ii. When, the given line is parallel to (ii)

`-(k - 2)/(k + 3) = 2`

`=>` −(k − 2) = 2k + 6  ...(∵ Slopes are equal)

`=>` −k + 2 = 2k + 6

`=>` 2k + k = 2 − 6

`=>` 3k = −4

`=> k = (-4)/3`