**Distance Formula:**

The distance formula is derived by using the Pythagorean theorem. Notice in the figure below that the distance between the points (*x, y*) and (*a, b*) is the hypotenuse of a right triangle. The difference *y â€“ b* is the measure of the height of the triangle, and the difference *x â€“ a* is the length of base of the triangle. Applying the Pythagorean theorem yields

*d* ^{2} = (*x - a*)^{2} + (*y - b*)^{2}

Taking the square root of both sides this equation yields

**Example:**

In the figure to the right, the circle is centered at the origin and passes through point P. Which of the following points does it also pass through?

(A) (3, 3)

(B) (-2âˆš 2,-1)

(C) (2, 6)

(D) (-âˆš 3, âˆš 3)

(E) (â€“3, 4)

Since the circle is centered at the origin and passes through the point (0, â€“3), the radius of the circle is 3. Now, if any other point is on the circle, the distance from that point to the center of the circle (the radius) must also be 3. Look at choice (B). Using the distance formula to calculate the distance between (-2âˆš 2,-1) and (0, 0) (the origin) yields

$d=\sqrt{{\left(-2\sqrt{2}-0\right)}^{2}+{\left(-1-0\right)}^{2}}=\sqrt{{\left(-2\sqrt{2}\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{8+1}=\sqrt{9}=3$

Hence, (-2âˆš 2,-1) is on the circle, and the answer is (B).

**Midpoint Formula:**

The midpoint M between points (x,y) and (a,b) is given by

M = $\left(\frac{x+a}{2},\frac{y+b}{2}\right)$

In other words, to find the midpoint, simply average the corresponding coordinates of the two points.

**Example:**

In the figure to the right, polygon PQRO is a square and T is the midpoint of side QR. What are the coordinates of T ?

(A) (1, 1)

(B) (1, 2)

(C) (1.5, 1.5)

(D) (2, 1)

(E) (2, 3)

Since point R is on the x-axis, its y-coordinate is 0. Further, since PQRO is a square and the x-coordinate of Q is 2, the x-coordinate of R is also 2. Since T is the midpoint of side QR, the midpoint formula yields

T = $\left(\frac{2+2}{2},\frac{2+0}{2}\right)$ = $\left(\frac{4}{2},\frac{2}{2}\right)$ = (2,1)

The answer is (D).

**Slope Formula:**

The slope of a line measures the inclination of the line. By definition, it is the ratio of the vertical change to the horizontal change (see figure below). The vertical change is called the **rise**, and the horizontal change is called the **run**. Thus, **the slope is the rise over the run.**

Forming the rise over the run in the above figure yields

m = $\frac{y-b}{x-a}$

**Example:**

In the figure below, what is the slope of line passing through the two points?

(A) 1/4

(B) 1

(C) 1/2

(D) 3/2

(E) 2

The slope formula yields m = $\frac{4-2}{5-1}$ = $\frac{2}{4}$= $\frac{1}{2}$. The answer is (C).

**Slope-Intercept Form:**

Multiplying both sides of the equation *m* = $\frac{y-b}{x-a}$ by *xâ€“a* yields

*y â€“ b = m(x â€“ a)*

Now, if the line passes through the *y*-axis at (0, *b*), then the equation becomes

*y â€“ b = m*( *x â€“ 0*)

or

*y â€“ b = mx*

or

*y = mx + b*

This is called the slope-intercept form of the equation of a line, where *m* is the slope and *b* is the *y*-intercept. This form is convenient because it displays the two most important bits of information about a line: its slope and its *y*-intercept.

**Example:**

**Column A**

*AO*

**ColumnB**

*BO*

The equation of the line above is

*y*= $\frac{9}{10}$

*x + k*

Since

*y*= $\frac{9}{10}$

*x + k*is in slope-intercept form, we know the slope of the line is $\frac{9}{10}$. Now, the ratio of

*BO*to

*AO*is the slope of the line (rise over run). Hence, $\frac{BO}{AO}$= $\frac{9}{10}$. Multiplying both sides of this equation by

*AO*yields

*BO*= $\frac{9}{10}$

*AO*. In other words,

*BO*is $\frac{9}{10}$the length of

*AO*. Hence,

*AO*is longer. The answer is (A).

**Intercepts:**

The

*x*-intercept is the point where the line crosses the

*x*-axis. It is found by setting

*y*= 0 and solving the resulting equation. The

*y*-intercept is the point where the line crosses the

*y*-axis. It is found by setting

*x*= 0 and solving the resulting equation.

**Example:**

Graph the equation

*x*â€“ 2

*y*= 4.

**Solution:**

To find the x-intercept, set y = 0. This yields

*x*- 2 â€¢ 0 = 4, or

*x*= 4. So the x-intercept is (4, 0). To find the

*y*-intercept, set

*x*= 0. This yields 0 â€“ 2

*y*= 4, or

*y*= â€“2. So the

*y*-intercept is (0, â€“2). Plotting these two points and connecting them with a straight line yields

**Areas and Perimeters:**

Often, you will be given a geometric figure drawn on a coordinate system and will be asked to find its area or perimeter. In these problems, use the properties of the coordinate system to deduce the dimensions of the figure and then calculate the area or perimeter. For complicated figures, you may need to divide the figure into simpler forms, such as squares and triangles. A couple examples will illustrate:

**Example:**

What is the area of the quadrilateral in the coordinate system to the right?

(A) 2

(B) 4

(C) 6

(D) 8

(E) 11

If the quadrilateral is divided horizontally through the line

*y*= 2, two congruent triangles are formed. As the figure to the right shows, the top triangle has height 2 and base 4. Hence, its area is

*A*= $\frac{1}{2}$

*bh*= $\frac{1}{2}$ â€¢ 4 â€¢ 2 = 4

The area of the bottom triangle is the same, so the area of the quadrilateral is 4 + 4 = 8. The answer is (D).

**Example:**

What is the perimeter of Triangle

*ABC*in the figure below?

(A) 5+ âˆš 5+ âˆš 34

(B) 10 + âˆš 34

(C) 5+ âˆš 5+ âˆš 28

(D) 2 âˆš 5+ âˆš 34

(E) âˆš 5+ âˆš 28

Point

*A*has coordinates (0, 4), point

*B*has coordinates (3, 0), and point

*C*has coordinates (5, 1). Using the distance formula to calculate the distances between points

*A*and

*B, A*and

*C*, and

*B*and

*C*yields

$\stackrel{\xc2\xaf}{AB}=\sqrt{{\left(0-3\right)}^{2}+{\left(4-0\right)}^{2}}=\sqrt{9+16}=\sqrt{25}=5$

$\stackrel{\xc2\xaf}{AC}=\sqrt{{\left(0-5\right)}^{2}+{\left(4-1\right)}^{2}}=\sqrt{25+9}=\sqrt{34}$

$\stackrel{\xc2\xaf}{BC}=\sqrt{{\left(5-3\right)}^{2}+{\left(1-0\right)}^{2}}=\sqrt{4+1}=\sqrt{5}$

Adding these lengths gives the perimeter of Triangle

*ABC*:

$\stackrel{\xc2\xaf}{AB}+\stackrel{\xc2\xaf}{AC}+\stackrel{\xc2\xaf}{BC}=5+\sqrt{34}+\sqrt{5}$

The answer is (A).