Find the equation of a line that passes through point (0,2) and makes 45 degrees to the line 2x+3y+6=0.

Expert Answers

An illustration of the letter 'A' in a speech bubbles


As you probably know, a normal (orthogonal) vector for a line with the equation `ax+by+c=0`  is `lta, bgt.` So for the given line it is `lt2, 3gt.`  Denote the equation in question as `y=mx+b,` which is the same as `mx-y+b=0,` its orthogonal vector is `ltm, -1gt.`

For the angle between the lines to be 45 degrees, the angle between normal vectors must be `+-45` degrees. Take the dot product of these two vectors: from the one hand, it is `2m-3,` from the other it is

`|<m, -1>| * |<2, 3>|*cos(45)=sqrt(m^2+1)*sqrt(2^2+3^2)*sqrt(2)/2=+-(2m-3).`


Square this equation and obtain

`(m^2+1)*13/2=(2m-3)^2=4m^2-12m+9,` or

`13m^2+13=8m^2-24m+18,` or


The roots of this quadratic equation is

`(-12+-sqrt(144+25))/5=(-12+-13)/5,` i.e. `m_1=-5` and `m_2=1/5.`


Now we have to find the corresponding b's. The point (0, 2) satisfies the equation `y=mx+b,` so `b=y_0-mx_0=2-m*0=2.` Thus `b_1=b_2=2.`


This way, the answer is: `y=-5x+2` and `y=1/5 x+2.`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial