Question

PLSSSS HELP ITS DUE TODAY!!!!!

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

`y-2=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{5}}(x-3)\qquad \impliedby \qquad \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} \\\\[-0.35em] ~\dotfill`

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

so ANY line that is perpendicular to that equation above, will have a slope of -5, so any of these are all perpendicular to it

`\begin{array}{llll} \stackrel{ Jules }{y=-5x+1} \\\\\\ \stackrel{ Lauren }{y=-5x+7} \\\\\\ y=-5x+999999999 \\\\\\ y=-5x-93789 \end{array}\hspace{5em} \begin{array}ll \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}`

Answer 2

Both Jules' and Lauren's equations are correct because they have slopes that are the negative reciprocal of the slope of the given line, making them perpendicular to the given line.

Explanation:

Let's reevaluate the equations based on the corrected given line equation:

`\sf y - 2 = (1)/(5)(x - 3)`

The given line equation is in point-slope form:
`\sf \boxed{\sf y - y_1 = m(x - x_1)}`, where m is the slope.

Given line equation:
`\sf y - 2 = (1)/(5)(x - 3)`

While comparing, we get

`\textsf{The\underline{ slope (m) }of the given line is }(1)/(5)`

For a line to be perpendicular to the given line, its slope must be the negative reciprocal of the slope of the given line.

The negative reciprocal of
`\sf (1)/(5)` is
`\sf -5.`

Now let's check the slopes of the equations provided by Jules and Lauren:

1. Jules' equation:
`\sf y = -5x + 1`

The slope of Jules' equation is -5, which matches the negative reciprocal of the slope of the given line.

2. Lauren's equation:
`\sf y = -5x + 7`

The slope of Lauren's equation is also -5, which again matches the negative reciprocal of the slope of the given line.

Both Jules' and Lauren's equations have a slope of -5, which is the negative reciprocal of the slope of the given line
`(1)/(5)`.

Therefore, both equations are correct and satisfy the condition of being perpendicular to the given line
`\sf y - 2 = (1)/(5)(x - 3)`