12.
[tex] d:3x-4y+12=0[/tex]
[tex] d:4y=3x+12[/tex]
[tex] d:y=\frac{3}{4} *x + 3[/tex]
A(-2,3)
a)
[tex]1) m_1=? \ d || d_1 \\
d: y= \frac{3}{4} *x + 3 \\
d_1: y=mx+n \\
d || d_1 ==> m_1=\frac{3}{4} \\
2) m_2=? \ d\perp d_2 \\
d: y= \frac{3}{4} *x + 3 \\
d_2: y=mx+n \\
d\perp d_2==> m_2*\frac{3}{4} = -1 \\
m_2= \frac{-4}{3}[/tex]
b)
[tex]1)d: y= \frac{3}{4} *x + 3 \\
d_1: y=mx+n \\
d || d_1 ==> m_1=\frac{3}{4} \\
d_1:y=\frac{3}{4}*3+n \\
A(-2,3)\in d_1 \rightarrow 3=\frac{3}{4} *(-2)+n \\
3=\frac{-3}{2} +n \\
n= 3+\frac{3}{2} \\
n=\frac{9}{2} \\
d_1:y= \frac{3}{4}*x +\frac{9}{2} \\
d_1: 3x-4y+18=0[/tex]
[tex]2)d: y= \frac{3}{4} *x + 3 \\
d_2: y=mx+n \\
d\perp d_2==> m_2*\frac{3}{4} = -1 \\
m_2= \frac{-4}{3} \\
d_2:y=\frac{-4}{3}*x+n \\
A(-2,3)\in d_2 \rightarrow 3=\frac{-4}{3} *(-2)+n \\
3= \frac{8}{3} +n \\
n= 3-\frac{8}{3} \\
n= \frac{1}{3} \\
d_2:y= \frac{-4}{3}*x+\frac{1}{3} \\
d_2:4x+3y-1=0[/tex]
