Diketahui garis g=ax+by+c dan L=px+qy+r.jika g tegak lurus L,maka pernyataan yang benar adalah...

Garis G
ax + by + c = 0
by = -ax - c
y = [tex] \frac{-a}{b}x - \frac{c}{b} [/tex]
[tex] m_G = \frac{-a}{b} [/tex]

Garis L
px + qy + r = 0
qy = -px - r
y = [tex] \frac{-p}{q}[/tex]x - [tex]\frac{r}{q} \\ m_L = \frac{-p}{q} [/tex]

Karena kedua garis tegak lurus, maka
[tex] m_G . m_L = -1 \\ \frac{-a}{b} . \frac{-p}{q} = -1 \\ \frac{ap}{bq} = -1 [/tex]
ap = -bq
ap + bq = 0