Let PQ be the line y = 3x + 1 slope of PQ = 3 The angle θ between PQ and QR is
tan Q = \(\big|\frac{m − 3}{1 + 3m}\big|\)
Slope of line PR: 2y = x + 3 is \(\frac{1}{2}\) and slope of QR is m.
∴ The angle θ between PR and QR is
⇒ ∴ m2 − m − 6 = 6m2 − m – 1
m°2 = − 1
\(\frac{m-3}{1+3m}\) = − \(\frac{2m-1}{m+2}\)
and m2 – m – 6 = – 6m2 + m + 1
7m2 – 2m – 7 = 0
∴ m = \(\frac{1\pm 5\sqrt{2}}{7}\)