# What is k if the angle between the lines 3x + 2y + 6 = 0 and kx - y +2 = 0 is 45 degrees?

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We'll know that the angle between the lines is pi/4 = 45 degrees.

We'll apply the formula:

tan 45 = |(m1 - m2)/(1+m1*m2)| (1)

m1 and m2 are the slopes of the given lines.

We'll put the given lines in the standard form, to determine the slopes.

The standard form of the equation of the line is:

y = mx + n

We'll put d1 into the standard form.

3x + 2y + 6 = 0

For this reason, we'll isolate 2y to the left side.

2y = -3x - 6

We'll divide by 2:

y = -3x/2 - 3

The slope of d1 is m1 = -3/2.

We'll put d2 into the standard form.

kx - y +2 = 0

For this reason, we'll isolate -y to the left side.

-y = -kx - 2

We'll multiply by -1:

y = kx + 2

The slope of d2 is m2 = k.

Now, we'll substitute m1 and m2 in (1):

tan 45 = |(-3/2 - k)/(1 - 3k/2)|

But tan 45 = 1

|(-3/2 - k)/(1 - 3k/2)| = 1

(-3/2 - k) = 1 - 3k/2

We'll move the terms that contain k to the left side:

-k + 3k/2 = 3/2 + 1

-2k + 3k = 3 + 2

**k = 5**

**The equation of the line, that makes an angle of 45 degrees to the line 3x + 2y + 6 = 0, is: 5x - y +2 = 0.**

To determine k the angle between the lines 3x + 2y + 6 = 0 and kx - y +2 = 0 is 45 degree.

We write both lines in the slope and y axis intercept form;

3x+2y+6= 0 implies 2y = -3x-6 .

Or y = (-3/2)x -3.....(1)

kx-y +2 = 0.

Ory = kx+2 ............(2).

Therefore the slope m1 ad m2 of the lines are m1 = -3 and m2 = k.

Since the angle between the lines is 45 degree,

tan 45 = (m1- m2)/(1+ m1m2) = (-3/2 - k)/{ 1+(-3)k}

1 =( -1.5-k)/(1-1.5*k)

1-1.5k = -1.5 -k

-1.5k+k = -1.5-1 = -2.5

= -0.5k = -2.5

k = 5.

Therefore for k = 5, the lines: 3x+2y+6 = 0 and kx-y+ 2 = 0 they make 45 degree inbetween the lines.