Given a line of a known slope, how can I find the point on a perpendicular line that is a given distance from a point on the first line?

Question

I'm trying to build a program that can take a line and turn it into a rectangle. So, the known information would be the x1,y1 and x2,y2 of that first line. What I need to be able to do is find the point that is on a line that is perpendicular to the given line and passes through one of the given points and is X units of distance away from the given point.

 

To give a concrete example, let's say the given points are:

(5,3)

(10,7)

 

So the slope is 5/4. This is simple to calculate

so we need a line with a slope of -4/5 (perpendicular) that passes through the point 5,3

And then we need to find the point that is on that line and is a distance of 2 (2 is given) from the point 5,3

 

We could then repeat the calculation for the other point (10,7), and we'd have the 4 points needed to make a rectangle.

Answer 1

Hi Mike,

 

This seems to be a interesting code. First, knowing the slope and one point in the line, you can now have the first equation by solving for b in y = m*x + b, which is the line equation: 1) y_3 = (newSlope)*(x_3) + b. (I will cal the new point as (x_3,y_3))

Next, we need to use the other known variable  to get another equation to solve the problem. There are two cases, when the slope of the first line is positive and when it is negative.  This can help you writing the code. 

We want to use the length of the new line as hypotenuse, and then we use Pythagoras'Theorem to find our second mathematical expression which will lead us to find our two variables ( new point). In this last part, one cathetus (side or leg of the imaginary right triangle that you should make in order to use the variable length) should be = (y_3) - (x_1) and the other one should be = (x_1) - (x_3) in my case the point (x_1,y_1) is the closest one to the point (x_3,y_3) and i took the outside right triangle.

Now we should know the new point , the next line will have a slope equal to -(1 /m_2). Then we can find the first equation as we did it before. Because the length of this new line is the same as the first one, again you need to imagine a right triangle out or in-side the triangle, then try to find a way to express those sides of the triangle in terms of x_4,x_3, y_4, y_3.  Again using the Pythagoras'Theorem you create our second equation. Finally, you got the the last point! I am assuming everything is in the positive plane and that you take only the positive answer from the square roots.

 

Hope it helps.

Answer 2

Actually, the slope of this line is 4/5 and the perpendicular slope is -5/4.

 

Once you have the slope of the perpendicular (call it m), find the equation for the perpendicular that goes through your starting point (call that (x₀,y₀) ).  In this case, it would be:

 

So (y - y₀) / (x - x₀) = m

 

(y - y₀) = m(x - x₀)

y - y_{0 =} mx - mx₀

y = mx - mx₀ + y₀

 

In your example, x₀ = 5 and y₀ = 3, so

 

y = -5/4 x - (-5/4) 5 + 3

y = -5/4 x + (5/4) 5 + 3

y = -5/4 x + 25/4 + 12/4

y = -5/4 x + 37/4 = (-5x + 37) / 4

 

So the equation of the perpendicular line in your example is y = (-5x + 37) / 4

 

Now you want to find a point on this line that is a specific distance from your starting point (x₀,y₀),  Call this target point (h,k).

 

Use the distance formula:

√((h-x₀)²+(k-y₀)²) = d

 

Square both sides:

(h-x₀)²+(k-y₀)² = d²

 

In your example, this would be (h-5)² + (y-3)² = 4.

 

Now you have one equation with two unknowns, h and k.  x_0, y₀ and d² are not unknowns.  They are just numbers that depend on the specific points you started with and the distance you want.

 

HOWEVER, you know that k = mh - mx₀ + y₀, because the point you are looking for satisfies the equation for the line we figured out earlier, since it is supposed to be on that line.

 

So, substitute mh - mx₀ + y₀ for k into the equation (h-x₀)²+(k-y₀)² = d2.  The result will be a bit messy, and I'm not going to work it out here.

 

You're going to end up with a quadratic equation with the one variable h, and then you have to solve that.  The result will be even messier!

 

Since it's a quadratic equation, it's going to have two solutions.  But that makes sense!  There are actually two points on the line that are 2 units away from the point (5,3), one in one direction and one in the other.

 

 

 

Answer 3

Hi Mike, slope (change in y over change in x is (7-3)/(10-5)=4/5 and not 5/4. This makes the slope of the perpendicular line -4/5. Then you can use the y=mx+b equation to find the equation of the line that passes through both (5,3) and the point on the perpendicular line and 2 units away from (5,3).

 

y=-4/5x+b, Now substitute 5 for x and 3 for y and solve for b, the y-intercept.

 

3=-4/5(5)+b => 3=-4+b => 3+4=-4+4+b => b=7

 

The equation is y=-4/5x+7 Notice how the distance 2 is not information you need to come up with the equation for the perpendicular line. Is this what you were looking for?