Time remaining:
Find an equation of the line that satisfies the given conditions.

Algebra
Tutor: None Selected Time limit: 0 Hours

Through (−96)parallel to the line x = 1

Through (27);  perpendicular to the line y = 6

Through 

(−1−3);perpendicular to the line 2x + 7y + 9 = 0

Through 

1
2
, − 
2
3
;

  perpendicular to the line 

4x − 8y = 1


Jan 27th, 2015

1. Lint through (-9,6) and parallel to x=1 is

y-y1 = m ( x-x1).  So y-6 =1(x+9). Because its slope is

y-6 = x+9

2. Slope is 0 so y-7 = 0.

The eq.  y= 7

3.  2x+3y+9= 0 has slope. -2

aline perpendicular has inverse negative slope so its slope is +1/2

The equation is: (y+3) = 1/2 ( x+1)

y+3 = 1/2x +1/2

Jan 27th, 2015

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Jan 27th, 2015
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Jan 27th, 2015
Dec 8th, 2016
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