Explain how you got the answer please

Mathematics

2 answers:

LuckyWell [14K]2 years ago

7 0

Pick any 2 data points from the list, say (6,13) and (5,11)
compute m, slope using the slope fornula m= y2-y1/x2-x1 = 13-11/6-5=2/1=2. pick any point and write your equation in slope intercept form
y-11=2(x-5)
y-11=2x-10
y=2x+1. answer B

marta [7]2 years ago

4 0

(-1,-1) and (2,5)
slope = (5+1)/(2+1) = 6/3 = 2

using (-1,-1)
y = mx + b
b = y - mx
b = -1 - (2)(-1)
b = -1 + 2
b = 1

equation slope intercept form (y = mx + b where m = 2 and b = 1)
so
y = 2x + 1
or
f(x) = 2x + 1

answer is B. second choice

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Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).

Svetradugi [14.3K]

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

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5$

well then therefore

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

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}$