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The lines $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$ are perpendicular to each other, if
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Verified Answer
The correct answer is:
$a_1 b_1+a_2 b_2=0$
$\ \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=0$
The equations of the lines are
$a_1 x+b_1 y+c_1=0$
Therefore, the slope $m_1=\left(-\frac{a_1}{b_1}\right)$
and $a_2 x+b_2 y+c_2=0$
Therefore, the slope $m_2=\left(-\frac{a_2}{b_2}\right)$
If the lines are perpendicular to each other, then
$\begin{aligned}
& \mathrm{m}_1 \times \mathrm{m}_2=-1 \\
& \Rightarrow\left(-\frac{\mathrm{a}_1}{\mathrm{~b}_1}\right) \times\left(-\frac{\mathrm{a}_2}{\mathrm{~b}_2}\right)=-1 \\
& \Rightarrow \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=0
\end{aligned}$
The equations of the lines are
$a_1 x+b_1 y+c_1=0$
Therefore, the slope $m_1=\left(-\frac{a_1}{b_1}\right)$
and $a_2 x+b_2 y+c_2=0$
Therefore, the slope $m_2=\left(-\frac{a_2}{b_2}\right)$
If the lines are perpendicular to each other, then
$\begin{aligned}
& \mathrm{m}_1 \times \mathrm{m}_2=-1 \\
& \Rightarrow\left(-\frac{\mathrm{a}_1}{\mathrm{~b}_1}\right) \times\left(-\frac{\mathrm{a}_2}{\mathrm{~b}_2}\right)=-1 \\
& \Rightarrow \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=0
\end{aligned}$
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