Sample problems and solutions in finding the distance between a point and the line?

Answer 1

The distance between a point and a line is the length of the perpendicular line drawn from the point to the line.

The easiest way is with vectors. Call P1 and P2 two points on the line, and P3 the third point.

Call the intersection of the line and the perpendicular from the point to the line P.

Now if the dot product of two vectors is zero, they are perpendicular. So

we look at the dot product of (P2-P1)dot(P3-P)=0

We can solve for P and find the intersection. Then find the length of that segment. Here is an example using the line 2x+3y=7 and the point (9,5) First, lets find a vector representation of the line. If x=0, y=7/3 and if y=0, x=7/2. So the vector (7/2,-7/3)t represents the line. Now we want a line through the point (9,5) perpendicular to the line 2x+3y=7 So let's call the point of intersection (x,y). Then the vector (9-x,5-x)t must be perpendicular to the first vector. We use the dot product. (9-x,5-y) dot (7/2,-7/3)=0 so evaluate the dot product and we have 7/3y-7/2x+119/6=0 Now solve the system of equations 7/3y-7/2x+119/6=0 2x+3y=7 the solution is (5,-1) To finish the problem, find the distance between (5,-1) and (9,5) using the distance formula. This distance is 2(square root 13)

If you would rather not use vectors, that's fine too.

Let the original line be y=mx+b and the point you want to find the distance to (x1,y1)

You can use the formula that this distance is

(y1-mx1-b)/square root of (m^2+1). You really need the absolute value of the expression in the numerator. The advantage of the first method is it can be used in any dimension. So it can find the distance from a point to a plane in 3d or even higher dimension. Here is a worked example. Y You can model other problems and do them the same way.

What is the distance from the point (9,5) to the line 2x+3y=7

Let A be the point (9,5) and P be the point in which the normal meets the given line.

So AP is normal to 2x+3y=7 and

we write AP as AP=(2k,3k) for some scalar k.

Now OP=OA+AP where O is the origin.

The coordinate of of P are (9+2k,5+3k) and sine P lies on the line we have

2(9+2k)+3(5+3k)=7. Now solve for k and you find k=-2

So AP =(-4,-6) and the length of AP using the distance formula is square root of (16+36)=2Square root of 13. Since OP=OA+Ap we find P is the point (5,-1)

Solution 2

The point C(0,7/3) lies on the line AC=(-9,-8/3) Let theta be the angle between AC and the normal vector N=(2,3)

AP=|ACx Cos theta)=AC|AC dot N/ACx|N|=18+8/Square root of 13=2square root of 13.

Solution 3: Here is one last approach that I would say is geared toward Alg1 or Alg 2 students. It does not use vectors or formulas, just some basic Algebra concept. the line is 2x+3y=7, we know that a perpendicular line must have a slope that is the negative reciprocal of the slope of the line2x+3y=7, So let's write that line in slope intercept form. 3y=-2x+7 and y=-2/3x+7/3 So the slope of the perpendicular line is 3/2. Now we have a point and a slope so we have a line using point slope form. The slope is 3/2 and the point is (9,5) (y-5)=3/2(x-9) which we can write as y=5+3/2x-27/2 or y=3/2x-17/2 Now we find the point where this line intersects the original line. We do this because then we need only find the distance between two points which we can do with the distance formula. y=3/2x-17/2

y=-2/3x+7/3 So solve them easily I will rewrite them both 2y=3x-34/2

3y=-2x+7 Solve these two simultaneous equations, find the point then just find the distance from that point to the original one. When we solve them, we find the intersection, which is the point we are looking for is (5,-1). The distance from that point to (9,5) is the square root of [(9-5)2 +(5+1)2 ]=2square root(13) as above.