Question

Line k has a slope of 1/3. Line j is perpendicular to line k and passes through the point (1,4). Create an equation for line j.

Answer 1

The equation for line j, which is perpendicular to line k with a slope of 1/3 and passes through the point (1,4), is y = -3x + 7.

Step-by-step explanation:

To find the equation of line j, which is perpendicular to line k, we first need to determine the slope of line j. Since line k has a slope of 1/3, line j, being perpendicular to it, will have a slope that is the negative reciprocal of 1/3. This means the slope of line j will be -3 (the negative reciprocal of 1/3).

Now, to create the equation of a line, we can use the point-slope form, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Since line j passes through the point (1,4), we have:

y - 4 = -3(x - 1)

Expanding this equation, we get:

y - 4 = -3x + 3

Adding 4 to both sides to solve for y, we obtain the slope-intercept form of the equation:

y = -3x + 7

This is the equation for line j, which is perpendicular to line k and passes through the point (1,4).

Answer 2

keeping in mind that perpendicular lines have negative reciprocal slopes, and that line "k" has a slope of 1/3, then

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{1}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{1}\implies -3}}`

so for line "j" we're really looking for the equation of a line whose slope is -3 and that it passes through (1 , 4)

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\hspace{10em} \stackrel{slope}{m} ~=~ - 3 \\\\\\ \begin{array}c \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}{4}=\stackrel{m}{- 3}(x-\stackrel{x_1}{1}) \\\\\\ y-4=-3x+3\implies {\Large \begin{array}{llll} y=-3x+7 \end{array}}`