Write the equation of a line that passws through the point (1,7) and is perpendicular to y=1/2x-5

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Write the equation of a line that passws through the point (1,7) and is perpendicular to y=1/2x-5

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation:

[tex]y=-2x+5[/tex]

Step-by-step explanation:

For one line to be perpendicular to another, their slopes must multiply to -1. So, we really want the line with slope [tex]$-1/(1/2)$ = -2$[/tex] that passes through the point [tex](1, 7)[/tex].

Recall the point-slope form of a line. If you have a point [tex]$(x_{1}, y_{1})$[/tex] and a slope [tex]m[/tex], the line through that point with the given slope is [tex]$y-y_{1}=m(x-x_{1})$[/tex]. Plugging in our given point and slope gives our line: [tex]y-7=-2(x-1)[/tex](point-slope form).

We can convert this line to slope-intercept form, the most common form by expanding the right side and adding [tex]7[/tex] to get [tex]y[/tex] by itself:

[tex]y-7=-2x+2[/tex]

[tex]y=-2x+5[/tex]

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)