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determine the equation of all lines that are perpendicular to 2x+3y-6=0 and that are a distance of 4 units from the origin.thanks..ive tried so many ways but i dont get it when i graph it using my graphing calculator..thanks..

2006-08-30 08:06:44 · 2 answers · asked by jeiyne_03 1 in Science & Mathematics Mathematics

2 answers

2x+ 3y -6 = 0

The line perpendicular to this will have equation

3x - 2y + c = 0

Now, for distance from origin = 4, there are 2 lines possible with this... one for + value of c and one for negative value of c

so, distance from origin is IcI = 4 x sqrt (3^3 + 2^2) = 4 x sqrt(13)

put c = + and - 4sqrt(13) in the equation and you get the answer.

2006-09-02 03:24:55 · answer #1 · answered by DG 3 · 0 0

The equation you are looking for is 3x -2y + b =0

To find b you impute the condition: distance to origin 4

4 = abs(b)/sqrt(9+4) and you will find two results

b= 4sqrt(13) and b = -4sqrt(13)

Now you have the two lines that satisfy the conditions.


d

2006-08-30 09:09:46 · answer #2 · answered by vahucel 6 · 0 1

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