SOLUTION: Write the standard form equation of the line that is perpendicular to
6x - 7y - 1 = 0 and passes through (0, -2).
Algebra.Com
Question 1181493: Write the standard form equation of the line that is perpendicular to
6x - 7y - 1 = 0 and passes through (0, -2).
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
Write the standard form equation of the line that is perpendicular to
and passes through (, ).
first write in slope intercept form
=> a slope is
the line that is perpendicular to given line is negative reciprocal to
use slope point formula to find equation of the line that is perpendicular to given line
........substitute a slope, and coordinates of given point (, )
-> write it in standard form
.......both sides multiply by
Answer by greenestamps(13216) (Show Source): You can put this solution on YOUR website!
Especially if you are just learning basic algebra, you should understand how to solve the problem as shown by the other tutor:
determine the slope of the given line;
determine the slope of the line perpendicular to the given line; and
find the equation of the line with that slope passing through the given point.
Here is a more advanced method that is faster and easier.
Any line parallel to the given line 6x-7y-1=0 will have an equation in standard form of 6x-7y+C=0, where C can be any constant (the coefficients of x and y will be the same as in the given equation).
Any line perpendicular to the given line 6x-7y-1=0 will have an equation in standard form of 7x+6y+D=0, where again D is some constant. Note to get that form, the coefficients of x and y switch places, and one of them changes sign.
Use that form and the given point (0,-2) to determine the constant.
7(x)+6(y)+D=0
7(0)+6(-2)+D=0
-12+D=0
D=12
ANSWER: 7x+6y+12=0
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