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CLP-3 Multivariable Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

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Section 1.3 Equations of Lines in 2d

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A line in two dimensions can be specified by giving one point (x0,y0) on the line and one vector d=⟨dx,dy⟩ whose direction is parallel to the line.
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If (x,y) is any point on the line then the vector ⟨x−x0,y−y0⟩, whose tail is at (x0,y0) and whose head is at (x,y), must be parallel to d and hence must be a scalar multiple of d. So
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These are called the parametric equations of the line, because they contain a free parameter, namely t. As t varies from −∞ to ∞, the point (x0+tdx,y0+tdy) traverses the entire line.
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It is easy to eliminate the parameter t from the equations. Just multiply x−x0=tdx by dy, multiply y−y0=tdy by dx and subtract to give
(x−x0)dy−(y−y0)dx=0
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In the event that dx and dy are both nonzero, we can rewrite this as
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This is called the symmetric equation for the line.
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A second way to specify a line in two dimensions is to give one point (x0,y0) on the line and one vector n=⟨nx,ny⟩ whose direction is perpendicular to that of the line.
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If (x,y) is any point on the line then the vector ⟨x−x0,y−y0⟩, whose tail is at (x0,y0) and whose head is at (x,y), must be perpendicular to n so that
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Observe that the coefficients nx,ny of x and y in the equation of the line are the components of a vector ⟨nx,ny⟩ perpendicular to the line. This enables us to read off a vector perpendicular to any given line directly from the equation of the line. Such a vector is called a normal vector for the line.

Example 1.3.4.

Consider, for example, the line y=3x+7. To rewrite this equation in the form
nxx+nyy=nxx0+nyy0
we have to move terms around so that x and y are on one side of the equation and 7 is on the other side: 3x−y=−7. Then nx is the coefficient of x, namely 3, and ny is the coefficient of y, namely −1. One normal vector for y=3x+7 is ⟨3,−1⟩.
Of course, if ⟨3,−1⟩ is perpendicular to y=3x+7, so is −5⟨3,−1⟩=⟨−15,5⟩. In fact, if we first multiply the equation 3x−y=−7 by −5 to get −15x+5y=35 and then set nx and ny to the coefficients of x and y respectively, we get n=⟨−15,5⟩.

Example 1.3.5.

In this example, we find the point on the line y=6−3x (call the line L) that is closest to the point (7,5).
We’ll start by sketching the line. To do so, we guess two points on L and then draw the line that passes through the two points.
  • If (x,y) is on L and x=0, then y=6. So (0,6) is on L.
  • If (x,y) is on L and y=0, then x=2. So (2,0) is on L.
Denote by P the point on L that is closest to (7,5). It is characterized by the property that the line from (7,5) to P is perpendicular to L. This is the case just because if Q is any other point on L, then, by Pythagoras, the distance from (7,5) to Q is larger than the distance from (7,5) to P. See the figure on the right above.
Let’s use N to denote the line which passes through (7,5) and which is perpendicular to L.
Since L has the equation 3x+y=6, one vector perpendicular to L, and hence parallel to N, is ⟨3,1⟩. So if (x,y) is any point on N, the vector ⟨x−7,y−5⟩ must be of the form t⟨3,1⟩. So the parametric equations of N are
⟨x−7,y−5⟩=t⟨3,1⟩orx=7+3t, y=5+t
Now let (x,y) be the coordinates of P. Since P is on N, we have x=7+3t, y=5+t for some t. Since P is also on L, we also have 3x+y=6. So
3(7+3t)+(5+t)=6⟺10t+26=6⟺t=−2⟹x=7+3×(−2)=1, y=5+(−2)=3
and P is (1,3).
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Exercises Exercises

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Exercise Group.

Exercises — Stage 1
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1.
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A line in R2 has direction d and passes through point c.
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Which of the following gives its parametric equation: ⟨x,y⟩=c+td, or ⟨x,y⟩=c−td?
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2.
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A line in R2 has direction d and passes through point c.
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Which of the following gives its parametric equation: ⟨x,y⟩=c+td, or ⟨x,y⟩=−c+td?
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3.
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Two points determine a line. Verify that the equations
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⟨x−1,y−9⟩=t⟨8,4⟩
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and
⟨x−9,y−13⟩=t⟨1,12⟩
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describe the same line by finding two different points that lie on both lines.
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4.
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A line in R2 has parametric equations
x−3=9ty−5=7t
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There are many different ways to write the parametric equations of this line. If we rewrite the equations as
x−x0=dxty−y0=dyt
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what are all possible values of ⟨x0,y0⟩ and ⟨dx,dy⟩?
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Exercise Group.

Exercises — Stage 2
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5.
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Find the vector parametric, scalar parametric and symmetric equations for the line containing the given point and with the given direction.
  1. point (1,2), direction ⟨3,2⟩
  2. point (5,4), direction ⟨2,−1⟩
  3. point (−1,3), direction ⟨−1,2⟩
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6.
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Find the vector parametric, scalar parametric and symmetric equations for the line containing the given point and with the given normal.
  1. point (1,2), normal ⟨3,2⟩
  2. point (5,4), normal ⟨2,−1⟩
  3. point (−1,3), normal ⟨−1,2⟩
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7.
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Use a projection to find the distance from the point (−2,3) to the line 3x−4y=−4.
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8.
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Let a, b and c be the vertices of a triangle. By definition, a median of a triangle is a straight line that passes through a vertex of the triangle and through the midpoint of the opposite side.
  1. Find the parametric equations of the three medians.
  2. Do the three medians meet at a common point? If so, which point?
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9.
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Let C be the circle of radius 1 centred at (2,1). Find an equation for the line tangent to C at the point (52,1+32).
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