set 3 - chapter 1 3 - chapter 1 - 2 slides.pdfis the intersection point of a horizontal line, known...

Analytic Geometry

الھندسـة التحليليـة

1 - 1 Cartesian Coordinate System نظام اإلحداثيات الديكارتي• The Cartesian coordinate system, or the rectangular coordinate system, is a

geometrical system that is used to determine the locations of points in a plane.

• Points are located with respect to a reference point called the origin which is the intersection point of a horizontal line, known as x-axis, and a verticalline called y-axis.

• The x and y axes divide the Cartesian plane into four regions called quadrants.

• Each point in the plane is defined by an ordered pair (x, y) of real numbers called the coordinates of the point.

• An example of ordered pairs or coordinates is the point P below:

2SET 3 - Chapter 1 GFP - Sohar University


1 - 2 The Distance Formula قانون إيجاد المسافة

• The distance d between two points A(x1, y1) and B(x2, y2) can be found from the distance formula:


Example 1: Find the distance between the points C(3, ‒ 4) and D( ‒ 13, ‒ 11).

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Solution:

SET 3 - Chapter 1 GFP - Sohar University

1 - 3 The Midpoint Formula قانون إيجاد نقطة المنتصف

• The coordinates of the midpoint of a line segment joining the two points A(x1, y1) and B(x2, y2) are found by averaging the coordinates of the endpoints.

• The midpoint formula is:


Example 2: F is the midpoint between points C(3, ‒ 4) and D( ‒ 13, ‒ 11). Find its coordinates.

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Solution:


1 - 4 The Slope of a Line ميل الخط المستقيم

• The slope of a line is a measurement of its steepness and direction.

• Slope of a line m is calculated from the following formula which is called the slope formula:


• Depending on the direction of the line, its slope could be positive, negative, zero or undefined and as shown below.


Example 3: Find the slope of the line that passes through points P(‒ 4, 8)and R(9, ‒ 7).

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Solution:


Example 4: Find the slope of the lines a, b, c and d shown in the figure below.

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Solution:


1 - 5 Parallel Lines الخطوط المتوازية


1 - 6 Perpendicular Lines الخطوط المتعامدة


Example 5: Lines m and n are parallel. If the slope of line m is ‒ 0.48, what is the slope of line n?

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Solution:

Since the two lines are parallel, then they have the same slope.

So, the slope of line n = ‒ 0.48

Example 6: Line c is perpendicular to line d and the slope of line c is 0.5.Find the slope of line d.

Solution: Since lines c and d are perpendicular, then their slopes are opposite reciprocals of one another.Therefore, the slope of line d =


1 - 7 x-Intercept and y-Intercept المقطع السيني و المقطع الصادي

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• For a non-horizontal line, x-intercept is the x-coordinate of the point where the line intersects x-axis.

• In the same way, for a non-vertical line, y-intercept is the y-coordinate of the point where it intersects y-axis.

a


1 - 8 Equations of Lines معادلة الخط المستقيم

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• The equation of a line is a mathematical sentence that describes the relationship between the x-coordinate and the y-coordinate of all its points.

• The equation of a line is of the first degree and is therefore called a linear equation.

• Straight line equation may be written in any of the following three forms:

where m is the slope, and b is the y-intercept.

where m is the slope.


Example 7: Draw the graph of the line whose equation is 2x ‒ 3y = 6 usingtwo randomly selected points.

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Solution: Let x = 1 , then:

Thus, (1, ‒1.33) is the first point.

Let y = 1 , then:

Thus, (4.5, 1) is the second point.


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The graph of 2x ‒ 3y = 6 is as shown in the figure below.


Example 8: Draw the graph of the line whose equation is 2x ‒ 3y = 6 usingthe x-intercept and the y-intercept.

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Solution: x-intercept y = 0

So, (3, 0) is the first point.

Therefore, (0, ‒ 2) is the first point.

y-intercept x = 0


Example 9: Draw the graph of : (a) x = 4 (b) y = ‒ 2

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Solution:

(a) The graph of x = 4 is a vertical linewith x-coordinate = 4 for all its points

(b) The graph of y = ‒ 2 is a horizontal linewith y-coordinate = ‒ 2 for all its points


Example 10: Find the equation of the line that passes though the points (4, –5)and (–11, 3). Write the equation in point-slope form, standardform and slope-intercept form.

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Solution:

Find the slope first:

Write the equation in point-slope form:


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To write the equation in the standard form, multiply both sides of theequation by 15:

Rearrange the equation 8x + 15y = ‒ 43 to write it in the slope-intercept form:

Divide both sides by 15:


Example 11: Determine whether the lines 6x + 4y = ‒ 9 and 8x ‒ 12y = ‒ 7are parallel or perpendicular or neither.

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Solution: Find the slopes of the two lines and compare them:

m1 = m2 =

So, the two lines are perpendicular since their slopes are opposite reciprocalsof one another.


Example 12: Which of the points A(2, 1.6) and B(1, –2.2) lie on the graph ofthe line 3x + 5y = 14?

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Solution: If a point lies on the graph of a line then it satisfies its equation.

Point A(2, 1.6) lies on the graph of 3x + 5y = 14

Check point A(2, 1.6):

Check point B(1, –2.2) :

Point B(1, –2.2) doesn’t lie on the graph of 3x + 5y = 14


1 - 9 Equations of Circles معادلة الدائرة

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• The radius r of a circle with a centre at thepoint (h, k) can be found using the distanceformula between the centre and any pointon the circle (x, y) and as follows:

(x – h)2 + (y – k)2 = r2

• The standard form of the equation ofa circle of radius r with centre at thepoint (h, k) is:


Example 13: The point (3, 4) lies on a circle whose centre is at (‒ 1, 2), as shownin the Figure below. Write the standard form of the equation of thiscircle.

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Solution: The radius r of the circle is the distance between (‒ 1, 2) and (3, 4):

Using (h, k) = (‒ 1, 2) and r ,The equation of the circle is:

Standard Form


1 - 10 Symmetry of Equations تناظر المعادالت

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• A graph has symmetry with respect to the y-axis if whenever (x, y) is on the graph, so is the point (‒ x, y).

• A graph has symmetry with respect to the origin if whenever (x, y) is on the graph, so is the point (‒ x, ‒ y).

• A graph has symmetry with respect to the x-axis if whenever (x, y) is on the graph, so is the point (x, ‒ y).


Example 14: Test y = x 2 + 2 for symmetry with respect to the x-axis,the y-axis, and the origin. .

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Solution: x-AxisWe replace y with ‒ y:

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

Multiplying both sides by ‒ 1:

y-AxisWe replace x with ‒ x:

Simplifying gives:

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis.


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OriginWe replace x with ‒ x and y with ‒ y:

Simplifying gives:

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.


Example 15: Test x 2 + y 4 = 5 for symmetry with respect to the x-axis,the y-axis, and the origin. .

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Solution: x-AxisWe replace y with ‒ y:

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis.

y-AxisWe replace x with ‒ x:

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis.


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OriginWe replace x with ‒ x and y with ‒ y:

The resulting equation is equivalent tothe original equation, so the graphis symmetric with respect to the origin.


set 3 - chapter 1 3 - chapter 1 - 2 slides.pdfis the intersection point of a horizontal line, known...

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