Given: #&\overleftrightarrow{\sf PQ} \perp \overleftrightarrow{\sf P'Q'}#&

Prove: #&\left(m_{\overleftrightarrow{\sf PQ}}\right)\left(m_{\overleftrightarrow{\sf P'Q'}}\right) = -1#&

✓ 1. #&m_{\overleftrightarrow{\sf PQ}} = {\large \frac{y_2-y_1}{x_2-x_1}}= {\large \frac{\clubsuit}{c-a}}#&    #&\clubsuit = #& d − b b − d a − c

✓ 2. #&m_{\overleftrightarrow{\sf P'Q'}} = {\large \frac{y_2-y_1}{x_2-x_1}}= {\large \frac{c-a}{\heartsuit}}#&    #&\heartsuit = #& −d − (−b) b − d c − a

✓ 3. #&m_{\overleftrightarrow{\sf P'Q'}} = {\large \frac{c-a}{-d - (-b)}} = {\large \frac{c-a}{\spadesuit}}#&    #&\spadesuit = #& d − b −d + b −d − b

✓ 4. #&\left(m_{\overleftrightarrow{\sf PQ}}\right)\left(m_{\overleftrightarrow{\sf P'Q'}}\right) = \left({\large \frac{d-b}{c-a}}\right) \left({\large \frac{c-a}{-d+b}}\right) =-1#&