Question

Given that the equation of line one (L1) is 2x-y=3 and equation of line two (L2) is kx+2y=3 where k is a constant.

(a) Find the value of k such that L1 and L2 are parallel.

(b) Find the value of k such that L1 and L2 are perpendicular.


This question is related to Linear functions, Answers: (a) k = -4, (b) k=1. Someone pls tell me how to get there though

Answer 1

see explanation

Explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

(a)

given the equation of L1 is

2x - y = 3 ( subtract 2x from both sides )

- y = - 2x + 3 ( multiply through by - 1 )

y = 2x - 3 ← in slope- intercept form

with slope m = 2

• Parallel lines have equal slopes

then slope of L2 is m = 2

given the equation of L2 is

kx + 2y = 3 ( subtract kx from both sides )

2y = - kx + 3 ( divide through by 2 )

y = -
`(k)/(2)` +
`(3)/(2)` ← in slope- intercept form

with slope m = -
`(k)/(2)`

since slope of L1 and L2 are equal, both 2 , then

-
`(k)/(2)` = 2 ( multiply both sides by 2 to clear the fraction )

- k = 4 ( multiply both sides by - 1 )

k = - 4

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(b)

given the slope of a line is m then the slope of a line perpendicular to it is

`m_(perpendicular)` = -
`(1)/(m)` = -
`(1)/(2)`

Then

for L2 to be perpendicular to L1

-
`(k)/(2)` = -
`(1)/(2)` ( multiply both sides by - 1 )

`(k)/(2)` =
`(1)/(2)` , thus

k = 1