Use the dot product to show that the line c = by + ax is perpendicular to vector v = <a,b>

They seem like they would be parallel. doesn't c also = <a,b>?

changing c = by + ax to the form y = mx + b gives us

ax + by = c
by = -ax + c
y = (-a/b)x + c/b

so the slope of the line is -a/b
or a vector in the direction of the line is (b,-a)
(remember, slope is rise/run, so -a/b has a run of b and a rise of -a, or direction vector (b,-a))

then (b,-a).(a,b) = ab - ab = 0
so the line is perpendicular to the given vector.

To show that the line given by c = by + ax is perpendicular to the vector v = <a,b>, we need to calculate their dot product.

The dot product is calculated using the formula:

v · w = a1 * b1 + a2 * b2

Here, a1 and a2 are the components of vector v, and b1 and b2 are the components of vector w. In this case, vector v is <a,b>, and we want to find the dot product with the line c = by + ax.

To find the dot product, we need to compare the components of both vector v and the line equation. While it may seem like c equals <a,b>, we actually need to take the coefficients of the variables a and b to compare with vector v.

Comparing the components, we have:

v · c = <a,b> · <a,b>
= a*a + b*b
= a^2 + b^2

If the dot product of two vectors is zero, it indicates that the vectors are perpendicular. So, in this case, if a^2 + b^2 = 0, then the line c = by + ax is perpendicular to vector v = <a,b>.

Note that if a^2 + b^2 ≠ 0, then the line c = by + ax is not perpendicular to vector v and would be at some angle to it.