MA005: Calculus I

If we move from a point $P = (x_1,y_1)$ to a point $Q = (x_2,y_2)$ in the plane, then we will have two increments or changes to consider. The increment in the $x$ or horizontal direction is $x_2 – x_1$ which is denoted by $∆x = x_2 – x_1$. The increment in the $y$ or vertical direction is $∆y = y_2 – y_1$. These increments are shown in Fig. 5 . $∆x$ does not represent $∆$ times $x$, it represents the difference in the $x$ coordinates: $∆x = x_2 – x_1$.

The distance between the points $P = (x_1,y_1)$ and $Q = (x_2,y_2)$ is simply an application of the Pythagorean formula for right triangles, and

dist(P,Q) = $\sqrt{(∆x)^2 + (∆y)^2} = \sqrt{ (x_2–x_1)^2 + (y_2–y_1)^2}$

The midpoint $M$ of the line segment joining $P$ and $Q$ is $M = (\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2})$

Example 2: Find an equation describing the points $P = (x,y)$ which are equidistant from $Q = (2,3)$ and $R = (5,–1)$. (Fig. 6) 

Solution: The points $P=(x,y)$ must satisfy $dist(P,Q) = dist(P,R)$ so

$\sqrt{(x–2)^2+(y–3)^2} =\sqrt {(x–5)^2+(y–(–1))^2}$

By squaring each side we get $(x–2)^2+(y–3)^2 = (x–5)^2+(y+1)^2$

Then $x^2 – 4x + 4 + y^2 – 6y + 9 = x^2 – 10x + 25 + y^2 + 2y + 1$

so $–4x – 6y + 13 = –10x + 2y + 26$ and $y = .75x – 1.625$, a straight line. Every point on the line $y = .75x – 1.625$ is equally distant from $Q$ and $R$.

Practice 2: Find an equation describing all points $P = (x,y)$ equidistant from $Q = (1,–4)$ and $R = (0,–3)$.

A circle with radius $r$ and center at the point $C = (a,b)$ consists of all points $P = (x,y)$ which are at a distance of $r$ from the center $C$: the points $P$ which satisfy $dist(P,C) = r$.

Example 3: Find the equation of a circle with radius $r = 4$ and center $C = (5,–3)$. (Fig. 7)

Solution: A circle is the set of points $P=(x,y)$ which are at a fixed distance $r$ from the center point $C$, so this circle will be the set of points $P=(x,y)$ which are at a distance of 4 units from the point $C = (5,–3)$. P will be on this circle if $dist(P,C) = 4$.
Using the distance formula and simplifying,

$\sqrt{(x–5)^2 + (y+3)^2} = 4$ so $(x–5)^2 + (y+3)^2=16$ or

$x^2 – 10x + 25 + y^2 + 6y + 9 = 16$.

Practice 3: Find the equation of a circle with radius $r = 5$ and center $C = (–2,6)$.