23 Coordinates and Distance

Let $O$ be a fixed point in the plan, called the origin.

Two perpendicular lines passing through $O$ are called the coordinate axes and labelled $x$ -axis, and $y$ -axis.

The axes divide the plan into 4 parts, called 1st, 2nd, 3rd, and 4th quadrants.

A point $P$ of the plan is represented by the ordered couple $(a,b)$ of real numbers $a$ and $b$ , called coordinates of $P$ . $a$ is the $x$ -coordinate, $b$ is the $y$ -coordinate.

$|a|$ is the distance of $P$ to the $y$ -axis, $|b|$ is the distance of $P$ to the $x$ -axis.

The distance between two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is given by: $\|P_1 P_2\|= \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2}$

The coordinates of the the midpoint of a line segment joining $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is given by:
$\displaystyle{\Big( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\Big)}$

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Exercise 1

Exercise 2

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The coordinates of the midpoint of the segment that joins the two points $A= ({4}, {-1})$ and $B= ({-2}, {-3})$ are

$\displaystyle{\Big( \frac{4+(-2)}{2}, \frac{ -1 +(-3)}{2} \Big) = \Big( 1, -2 \Big) }.$

Exercise 3

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The length of the sides of the triangle are:

$AB= \sqrt{ (1-(-1))^2 + (3-(2))^2} = \sqrt{4 + 1} = \sqrt{5}$

$AC= \sqrt{ (1-5)^2 + (3-(-5))^2} = \sqrt{16 + 64} = \sqrt{80}$

$CB= \sqrt{ (5-(-1))^2 + (-5-(2))^2} = \sqrt{36 + 49} = \sqrt{85}$ .

We have $CB^2 = AB^2 + AC^2 =85$ . Thus, by the Pythagorean theorem, the triangle is a right triangle at $A$ .

Exercise 4

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The distance between the point $A ({a}, {b})$ and the origin is given by

$\sqrt{ ( {a} -0 )^2 + ( {b}-0 )^2 } = \sqrt{ a^2 + b^2 }$ .

The distance between the point $B({b}, {a})$ and the origin is given by

$\sqrt{ ( {b} -0 )^2 + ( {a}-0 )^2 } = \sqrt{ b^2 + a^2}$ .

Thus the points $A({a}, {b})$ and $B({b}, {a})$ are at the same distance from the origin.

Exercise 5

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The distance between the point $S =(-2,3)$ and the $x$ -axis is

$\sqrt{ ( {-2} - ({-2}) )^2 + ( {3} - ({0}) )^2 } = \sqrt{ {0} + {9} } =\sqrt{ {9} } =3.$

The distance between the point $S =(-2,3)$ and the $y$ -axis is

$\sqrt{ ( {-2} - ({0}) )^2 + ( {3} - ({3}) )^2 } = \sqrt{ {4} + {0} } =\sqrt{ {4} } =2.$

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Guide to Precalculus Review Copyright © 2025 by Samia CHALLAL is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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