Question

write the equation that goes through the point (4,-7) and is perpendicular to the line y+6=-2/5(x-1) all i need is slope intercept form.

Answer 1

To find the equation of the line that goes through the point (4, -7) and is perpendicular to the line y + 6 = -2/5 (x - 1), we can use the fact that two lines are perpendicular if their slopes are negative reciprocals of each other.

The slope of the line y + 6 = -2/5 (x - 1) is -2/5, so the slope of the line that is perpendicular to it must be -5/2. We can use this slope and the point (4, -7) to write the equation of the line in slope-intercept form.

To do this, we can use the point-slope formula, which is: y - y1 = m(x - x1), where (x1, y1) is the point through which the line passes, and m is the slope of the line. In our case, the point is (4, -7) and the slope is -5/2, so the equation of the line is: y - (-7) = (-5/2)(x - 4).

We can simplify this equation to get: y + 7 = -5/2 x + 10. Finally, we can rearrange the terms to get the equation in slope-intercept form, which is: y = -5/2 x - 25/2.

Therefore, the equation of the line that goes through the point (4, -7) and is perpendicular to the line y + 6 = -2/5 (x - 1) is y = -5/2 x - 25/2 in slope-intercept form.

Answer 2

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

`y+6=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{5}}(x-1)\qquad \impliedby \begin{array} \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{-2}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{-2}\implies \cfrac{5}{2}}}`

so we're really looking for the equation of a line whose slope is 5/2 and that it passes through (4 , -7)

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{-7})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{2} \\\\\\ \begin{array} \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}{(-7)}=\stackrel{m}{ \cfrac{5}{2}}(x-\stackrel{x_1}{4}) \implies {\large \begin{array}{llll} y +7= \cfrac{5}{2} (x -4) \end{array}}`