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Find the lines slope and a point on the line y+4=-1/2(x-5)

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y + 4 = -\cfrac{1}{2}(x-5)\implies y-(\stackrel{y_1}{-4})= \stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{5}) \impliedby \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\\\\ ~\hfill m=-\cfrac{1}{2}\hspace{5em}(5~~,~~-4)~\hfill

User R Quijano
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4 votes

Explanation:

y+4 = -1/2(x-5)

y = (-x+5)/2 -4

y= (-x+5-8)/2

y= (-x-3)/2

when x=1

y=(-1-3)/2

y= -2

point on graph = (1,-2)

when x =3

y=(-3-3)/2

y= -3

point = (3,-3)

slope = (-2-(-3))/(1-3)

=1/-2

slope = 1/-2 , a point = (1,-2)

User Aliasbody
by
8.2k points

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