Step-by-step explanation:
Given equation of line is:
[tex] 2x + y = 3\\\\ \therefore y = - 2x + 3\\\\\[/tex]
Equating it with[tex] y = m_1x+c [/tex] we find:
[tex] m_1=-2m [/tex]
Let the slope of required line be [tex] m_2 [/tex]
[tex] \because [/tex] Required line is parallel to the given line.
[tex] \therefore\: m_2=\frac{-1}{m_1}=\frac{-1}{-2}=\frac{1}{2}[/tex]
[tex] \because [/tex] Required line passes through (5, 4) and has a slope [tex] m_2=\frac{1}{2}m [/tex]
[tex] \therefore [/tex] Equation of line in slope point form is given as:
[tex] y - y_1 = m_2(x-x_1) \\\\ \therefore y- 4=\frac{1}{2}(x-5)\\\\ \therefore y- 4=\frac{1}{2}x-\frac{5}{2}\\\\ \therefore y=\frac{1}{2}x+4-\frac{5}{2}\\\\ \therefore y=\frac{1}{2}x+\frac{2\times 4-5}{2}\\\\ \therefore y=\frac{1}{2}x+\frac{3}{2}\\\\
\therefore 2y= x +3\\\\
\therefore x -2y + 3=0 \\[/tex]
Hence, x -2y + 3=0 is the required equation of line.
