Solving Scalar Product Problem: Find Perpendicular Distance

AI Thread Summary
The discussion centers on solving the scalar product problem to find the perpendicular distance from a point A(x_1, y_1) to the line y = mx + c. The parametric vector equation of the line is given as r = cj + λ(i + mj). To derive the perpendicular distance, the formula |(mx_1 - y_1 + c)/√(m^2 + 1)| is established. Additionally, when the line is expressed in the form ax + by + d = 0, the perpendicular distance is given by |(ax_1 - by_1 + d)/√(a^2 + b^2)|. The discussion emphasizes using vectors and the dot product to approach the problem.
pavadrin
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Hey,
Today I was given a problem to solve in class and was told to complete it for homework. This problem is as follows:

The line y=mx + c has a gradient m and cuts the y-axis at (0,c). Thus we can write the parametric vector equation of the line as:

r = cj +\lambda (i + mj)

Using this fact show that that the perpendicular distance from point A(x_1 , y_1) to y = mx + c is:

\mid(\frac{mx_{1} - y_{1} + c}{\sqrt{m^2 + 1}})\mid

If y = mx + c is instead written as ax + by + d = 0 show that the perpendicular distance of point A(x_1 , y_1) to as ax + by + d = 0 is given by:

\mid(\frac{ax_{1} - by_{1} + d}{\sqrt{a^2 + b^2}})\mid

This diagram which I drew to help me may help:

PF.JPG


______________________​

I have tried solving this problem by using vectors:

PF4.GIF


and I know that the dot product of ( x_1 , y_1 ) and y = mx + c is equal to zero but from there onwards I am not sure on how to approach this problem. All help is appreciated,
thanks, Pavadrin
 
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First, you need to find a vector perpendicular to the line y = mx + c. Call this vector u. Let v be the vector that points to A. Then for some n, nu + v = r. The magnitude of nu is the perpendicular distance from the point A to the line y = mx + c.
 
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