Question

Write an equation of the line that it is perpendicular to y=7x-3 and passes though the origin

Answer 1

bearing in mind that perpendicular lines slopes are negative reciprocal of each other.

`\bf y=\stackrel{\stackrel{m}{\downarrow }}{7}x-3\qquad \impliedby \begin{array}c \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill`

`\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{7\implies \cfrac{7}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{7}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{7}}}`

so we're really looking for the equation of a line whose slope is -1/7 and runs through (0,0),

`\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies - \cfrac{1}{7} \\\\\\ \begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{1}{7}}(x-\stackrel{x_1}{0})\implies y=-\cfrac{1}{7}x`