Asked by nichole
the lines described by y=(5a+3x) and y=-1/2x are perpendicular. what is the value of a?
Answers
Answered by
Reiny
the slopes of your lines are determined by the coefficients of the x terms of your equations.
They are 3 for the first one, and -1/2 for the second.
Since their product is not equal to -1, they cannot be perpendicular no matter what the value of a is
The 5a term has nothing to do with the slopes.
Perhaps you have the brackets in the wrong place
e.g. if it had been y = (5a+3)x then it would make sense.
set (5a+3)*(-1/2) = -1 and solve for a
They are 3 for the first one, and -1/2 for the second.
Since their product is not equal to -1, they cannot be perpendicular no matter what the value of a is
The 5a term has nothing to do with the slopes.
Perhaps you have the brackets in the wrong place
e.g. if it had been y = (5a+3)x then it would make sense.
set (5a+3)*(-1/2) = -1 and solve for a
Answered by
drwls
The slope product of two perpendicular lines must be -1.
In your case, the slope of the first line is 3, and of the second line is -1/2, so the lines cannot be perpendicular whatever a is.
Are you sure you did not mean to write
y = ax + 5 for the first line? The a needs to be part of the slope coefficient if you want to make the lines perpendicular.
In your case, the slope of the first line is 3, and of the second line is -1/2, so the lines cannot be perpendicular whatever a is.
Are you sure you did not mean to write
y = ax + 5 for the first line? The a needs to be part of the slope coefficient if you want to make the lines perpendicular.
Answered by
Reiny
Proof that great minds think alike, lol
Answered by
DrBob222
I was thinking exactly the same thing as I read the two responses.
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