Question

Find the equation of a straight line passing through the point (-1,2) and making an angle of 45 degree with the line x-y+1=0.

Answer 1

EXPLANATION.

Equation of line passing through the point

(-1,2).

Makes an angle of 45°.

Equation of line = x - y + 1 = 0.

Slope of the line = -a/b = -1/-1 = 1.

Angle between the lines.

$\sf \implies \: | \dfrac{ m_{1} - m_{2} }{1 + m_{1}m_{2} } | = \tan( \theta) \\ \\ \sf \implies \: | \frac{ m_{1} - 1 }{1 + m_{1}(1) } | = \tan(45 \degree) \\ \\ \sf \implies \: | \frac{ m_{1} - 1 }{1 + m_{1} } | = 1 \\ \\ \sf \implies \: \frac{ m_{1} - 1 }{1 + m_{1} } = 1$

$\sf \implies \: m_{1} - 1 = 1 + m_{1} \\ \\ \sf \implies \: m_{1} - m_{1} \: = 1 + 1 \\ \\ \sf \implies \:0$

$\sf \implies \: - ( \dfrac{ m_{1} - 1 }{1 + m_{1}}) = 1 \\ \\ \sf \implies \: - m_{1} + 1 = 1 + m_1 \\ \\ \sf \implies \: \: 2 m_{1} = 0 \\ \\ \sf \implies \: m_{1} = 0$

$\sf \implies \:equation \: of \: line \\ \\ \sf \implies \:(y - y_{1}) = m(x - x_{1}) \\ \\ \sf \implies \:(y - 2) = 0(x + 1) \\ \\ \sf \implies \:y = 2$