Hello!

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...

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Hello!

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

`5m^2+24m-5=0.`

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.`