Why is vector (a,b) perpendicular to line ax+by=c?

Answer 1

If ax + by = c, then by = -ax + c

So, y = (-a/b)x + c/b. This line has slope -a/b.

The vector from (0,0) to (a,b) is parallel to any line with slope b/a.

Since -a/b and b/a are negative reciprocals, the vector from (0,0) to (a,b) is perpendicular to the line

ax + by = c.

Answer 2

Depending on your knowledge of vectors or calculus, I will try to explain!

The gradient of a function is perpendicular to the function. The function here is f(x,y) = ax + by -c = 0

The gradient is defined as Grad(f) = df/dx i + df/dy j which uses i ( unit vector in x direction) and

j (unit vector in y direction). The partial derivative df/dx = a and df/dy = b

So, the perpendicular vector is a i + b j which is (a,b).

Keep in mind that (-a,-b) is also perpendicular since f(x,y) can also be expressed as -ax - by +c = 0

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You can also prove this if you use the dot product with any vector V1 & V2. The dot product says

V1 * V2 =ABS(V1)ABS(V2)cosine(theta) where ABS means absolute valve & theta is the angle between the 2 vectors V1& V2 and * denotes the dot product. The cosine(theta) term will always be zero for 2 perpendicular vectors so V1 *V2 = 0.

From algebra, you can find the perpendicular line to y = c/b -(a/b)x which is the equation above.

Then construct any vector from the origonal line and any vector from the perpendicular line. Use the dot product with those 2 vectors and you will get 0 proving they are perpendicular.