Cartesian Coordinate System

Question 1

composed of two axes intersecting and perpendicular to eachother. Note that when we say“perpendicular”, it means the lines form right angles or L-shaped angles. The horizontal axis (i.e., the horizontal line) that you see is called the X-axis. On the other hand, the vertical axis (i.e., the vertical line) is called the Y-axis.

Answer 1

Cartesian coordinate system

Question 2

A point in the coordinate plane is also called

Answer 2

coordinate or an ordered pair

Question 3

Example 1:Determine the abscissa and ordinate of (1, 4). Afterward, plot the point in the coordinate plane.

Answer 3

Solution:The abscissa is the x-coordinate of (1, 4) which is 1. On the other hand, the ordinate is the y-coordinate of (1, 4) which is 4. To plot (1, 4) in the coordinate plane, we count 1 unit to the right of the origin and move 4 units upward.

Question 4

Example 2:Plot (-2, -1) in the coordinate plane.

Answer 4

Solution:To plot (-2, -1) in the coordinate plane, we count 2 units on the left of the origin and move one unit downward.

Question 5

Example 3:Try to plot the following points by yourself:

A. (3, -1)

B. (1, 0)

C. (-2, 1)

D. (0, 5)

Answer 5

Answer

Question 6

If you look closely at the coordinate plane, you will notice that the entire rectangular plane is divided into four sections. These sections are called the

Answer 6

quadrants of the coordinate plane.

Question 7

The quadrants of the coordinate plane are arranged in a counter-clockwise manner. The first quadrant is the quadrant where the positive portions of the X-axis and Y-axis are located. The following list provides an overview of all four quadrants:

Answer 7

All coordinates with positive first and second components are in thefirst quadrant.All coordinates with a negative first component and a positive second component are in thesecond quadrant.All coordinates with negative first and second components are in thethird quadrant.All coordinates with a positive first component and a negative second component are in thefourth quadrant.

Question 8

Example:In which quadrant can you locate (-2, 1)?

Answer 8

Solution:(-2, 1) has a negative first component and a positive second component. Therefore, we can locate (-2, 1) in the second quadrant of the coordinate plane.

Question 9

Example:Form a straight line using the points (2, 3) and (5, 1).

Answer 9

Solution:We start by plotting the points in the coordinate plane.

Afterwards, we connect the points to form a straight line.

There you go! We have just graphed a line that passes through the points (2, 3) and (5, 1).

Question 10

the part of the line that touches the X-axis and Y-axis. For instance, look at the graph ofx + y = 10again:

Answer 10

Intercepts of a Linear Equation

Question 11

the point where the line touches the x-axis

Answer 11

x-intercept

Question 12

the point where the line touches the y-axis.

Answer 12

y-intercept

Question 13

Example:Solve for the intercepts of the line x + 2y = 8.

Answer 13

Solution:To solve for the x-intercept of the line, set y = 0

x + 2y = 8

x + 2(0) = 8Set y = 0

x = 8

Hence, the x-intercept is (8, 0).

On the other hand, to find the y-intercept, we letx = 0

x + 2y = 12

(0) + 2y = 8Set x = 0

2y = 8

2y⁄2 = 8⁄2Divide both sides by 2

y = 4

Hence, the y-intercept is (0, 4).

Question 14

Example:Graph the line 2x + 5y = 10 using its intercepts.

Answer 14

Solution:Let us start by solving for thexandy-intercepts:

As shown in the computation above, the intercepts are (5, 0) and (0, 2).

We then plot these intercepts in the coordinate plane and connect them to form a straight line:

Question 15

tells us the direction and steepness of the line.

Answer 15

Slope of a Line

Question 16

When determining the slope of a line, remember the following:

Answer 16

If the value ofmis positive (i.e.,m > 0):This means that the graph of the line is rising to the right.If the value ofmis negative (i.e.,m < 0):This means that the graph of the line is rising to the left.If the value ofmis zero (i.e.,m = 0):This means that the graph of the line is horizontal.If the value ofmis undefined (this happens when the denominator turns out to be zero):This means that the graph of the line is vertical.The larger the absolute value ofm,the steeper the line is.

Question 17

Example 1:What is the slope of the line containing the points (1, 2) and (3, 5)?

Answer 17

Solution:

We havey1= 2, y2= 5, x1= 1, and x2= 3.

Using the slope formula:

Therefore, the slope is 3/2.

Note that 3/2 is positive. This means that the graph of the line will be rising to the right. You may try to graph the line using the given points and verify that the line is indeed rising to the right.

Question 18

Example 2:Compute for the slope of the line containing the points (-1, 5) and (0, 1).

Answer 18

Solution:

We havey1= 5,x1= -1, y2= 1,andx2= 0

Using the slope formula:

Therefore, the slope is -4/1 or -4.

Since we have a negative slope, this implies that the graph of the line will be rising to the left.

Question 19

Example 3:Compute for the slope of the line containing the points (5, 1) and (5, 5).

Answer 19

Solution:

Using the slope formula:

Since the slope we have computed is undefined, this means that the graph of the line will be a vertical line.

Note:If the given points of the line have the same abscissa or first coordinate, the graph of that line is a vertical line and its slope is undefined. On the other hand, if the given points have the same ordinate or second coordinate, the graph of that line is a horizontal line and its slope is 0.

Question 20

Example 4:Identify the slope of the line that contains the points (3, 4) and (7, 4).

Answer 20

Solution:Notice that the given points of the line have the same ordinate or second coordinate (both are 4). This implies that the slope of the line that contains this points is equal to 0.

To verify, let’s use the slope formula:

Here, we have x1= 3, x2= 7, y1= 4, and y2= 4.

Substituting these values in the slope formula:

Indeed, the slope of the line in the given problem is 0.

Question 21

Example 1:Determine the slope of the linear equation
2x + y = 10.

Answer 21

Solution:

The first step is to isolateyfrom other quantities. This means thatyshould be the only variable on the left-hand side of the equation. This is possible by transposing2xto the right-hand side of the equation:

2x + y = 10

y = -2x + 10Transposing 2x to the right-hand side

By havingyas the only quantity remaining on the left-hand side, we can successfully write the equation in its slope-intercept form. For the given equation,y = -2x + 10is its slope-intercept form.

To find the slope, we take the coefficient of thexterm in the slope-intercept form of the given equation. Thexterm iny = -2x + 10is-2xand the numerical coefficient is–2.

Therefore the slope is -2 orm = -2.

Since the slope is negative, it is expected that if we graph2x + y = 10in the coordinate plane, the graph will be a line rising to the left.

Question 22

Example 2:What is the slope of the linear equation
3x – y = 11?

Answer 22

Solution:

We start by isolatingyfrom other quantities by transposing3xto the right-hand side of the equation:

We have now written the given linear equation in its slope-intercept form. Now, we can take the numerical coefficient of thexterm and it would be the slope of our line.

Since thexterm ofy = 3x – 11is3xand its numerical coefficient is 3, then our slope must be 3.

Thus, the slope of the given linear equation is 3.

Question 23

If two linear equations have the same slope, then the graph of these lines will show

Answer 23

Parallel Lines

Question 24

Suppose we have two linear equations,x + 3y = 6andx + 3y = 12.

Answer 24

First, let us transform both of these equations into their slope-intercept form so we can determine their respective slopes:

x + 3y = 6

3y = -x + 6Transposing x to the right-hand side

y = -⅓x + 6Dividing both sides by 3

Slope is -⅓

x + 3y = 12

3y = -x + 12Transposing x to the right-hand side

y = -⅓x + 12Dividing both sides by 3

Slope is -⅓

Based on our computations above, the linear equationsx + 3y = 6andx + 3y = 12have the same or equal slopes (which is -⅓).

Since these linear equations have the same slope, then their graphs are parallel lines or lines that will not intersect.

Indeed, the graphs of the linear equationsx + 3y = 6andx + 3y = 12are parallel lines as shown in our illustration above.

Question 25

Sample Problem:The equation of line A is x – 2y = 10. On the other hand, the slope of line B is (k + 1)x + y = 15. What must be the value of k so that the lines A and B are parallel?

Answer 25

Solution:

Lines A and B will be parallel if and only if their slopes are equal.

Let us take the respective slopes of each line.

Computing for the slope of line A:

x – 2y = 10

y = ½x – 5

Thus, the slope of Line A is ½.

Computing for the slope of line B:

(k + 1)x + y = 15

y = -(k + 1)x + 15

Thus, the slope of line B is -(k + 1).

Now, the slope of lines A and B must be equal. Hence, we have:

½ = -(k + 1)

We can express the equation above as:

½ = -k – 1Using distributive property

Let us now solve for the value of k:

½ = -k – 1

2(½ ) = 2(-k – 1)Multiplying both sides of the equation by the LCD (which is 2)

1 = 2(-k – 1)

1 = -2k – 2Distributive property

2k = -1 – 2Transposition method

2k = -3

k = -3/2Division property of equality

Therefore, the value ofkmust be -3/2

Question 26

If two linear equations have slopes that are negative reciprocal of each other, then the graph of these linear equations would show

Answer 26

Perpendicular Lines

Question 27

if two lines are perpendicular, then their slopes are

Answer 27

negative reciprocal of each other.

Question 28

Sample Problem:Line Z is perpendicular with a line whose equation is 3x – 5y = 15. Determine the slope of Line Z.

Answer 28

Recall that if two lines are perpendicular, then their slopes are negative reciprocal of each other.

Thus, the slope of line Z must be a negative reciprocal of the slope of3x – 5y = 15.

Let us compute for the slope of3x – 5y = 15:

3x – 5y = 15

y = ⅗x – 5

As we can see, the slope of3x – 5y = 15is ⅗. The negative reciprocal of ⅗ is -5/3.

Thus, the slope of line Z is -5/3.

Question 29

Example 1:Graph the linear function f(x) = ½ x + 5.

Answer 29

Solution:If we input a value ofxinto the function, we will obtain an output value forf(x). Since,y = f(x)in a function,then we’ll be able to come up with a coordinate (x, y). Take note that we can input any real number value forxsince thedomain of a linear functionis the set ofreal numbers.

Let us try to inputx = 2:

f(x) = ½ x + 5

f(2) = ½ (2) + 5We input x = 2 to the function

f(2) = 1 + 5

f(2) = 6

Atx = 2, the value off(x)is 6. Since,y = f(x), theny = 6. Therefore, we have a pair of values:x = 2andy = 6. This means that we have the coordinate (2, 6).

We already have one coordinate forf(x) = ½ x + 5. Now, let us look for another coordinate by inputting another value ofxinto the function.

Let us try to inputx = 4:

f(x) = ½ x + 5

f(4) = ½ (4) + 5We input x = 4 to the function

f(4) = 2 + 5

f(4) = 7

Atx = 4, the value off(x)is 7. Sincey = f(x),theny = 7. Therefore, we have a pair of values:x = 4andy = 7. This means that we have the coordinate (4, 7).

Now, we have two coordinates: (2, 6) and (4, 7). We can finally plot these points in the coordinate plane and form a straight line. This straight line is the linear functionf(x) = ½ x + 5.

Question 30

Example 2:The graph of the linear function f(x) = 3x – 9 passes through the points (3, a) and (b, 15). What are the values of a and b?

Answer 30

Solution:Iff(x) = 3x – 9passes through the point (3, a), then this means that if we inputx = 3to the function, we will obtain a value off(x)which is equal toa. Thus, to find the value ofa, we just simply substitutex = 3to the given linear function:

f(x) = 3x – 9

f(3) = 3(3) – 9We input x = 3 to the function

f(3) = 9 – 9

f(3) = 0

Thus, the value ofamust be 0. The point should be (3, 0).

On the other hand, iff(x) = 3x – 9passes through the point (b, 15), then this means that if we inputx = bto the function, we will obtain a value off(x)which is equal to 15. Thus, to find the value ofb, we substitute x = b and f(x) = 15to the given linear function.

f(x) = 3x – 9

15 = 3b – 9We substitute f(x) = 15 and x = b.

-3b = -15 – 9Transposition method

-3b = -24

-3b⁄-3 = -24⁄-3Dividing both sides by -3

b = 8

Hence, the value ofbmust be 8 and the point should be (8, 15).

Question 31

if the coefficient of the leading term of a quadratic function is positive (a > 0), then its graph is a parabola that opens

Answer 31

upward

Question 32

if the coefficient of the leading term of a quadratic function is negative (a < 0), then its graph is a parabola that opens

Answer 32

downward

Question 33

Example:Determine whether the graph of the function f(x) = x – 3x – x2opens upward or downward.

Answer 33

Solution:The leading term of the given quadratic function is-x2and the sign of its coefficient is negative. Thus, its graph opens downward.

Question 34

The extreme point of the parabola is called its

Answer 34

Vertex

Question 35

The vertex of the parabola is either the highest or lowest point of parabola.If the parabola opens upward, then the vertex is the

Answer 35

Lowest point

Question 36

if the parabola opens downward, then the vertex is the

Answer 36

Highest point

Question 37

Example:The vertex of the function f(x) = -x²– 2x + 4 is (-1, 5). Determine the minimum or maximum value of the function.

Answer 37

Solution:The given function has a leading term that is negative. This means that its graph is a parabola that opens downward. This implies that the y-coordinate of its vertex is the maximum value of the function.

The y-coordinate of the vertex of the functionf(x) = -x²– 2x + 4is 5. This means that the maximum value of the function is 5.

This implies that the value of the function, whatever you will substitute forx, will never exceed 5.

Question 38

How To Determine the Vertex of a Parabola

Answer 38

Method 1: Calculating the vertex using the formulas for its coordinates. If the given quadratic function is in general form, we can calculate its vertex using the formula below:

We usehto refer to thexcoordinate of the vertex whilekto refer to theycoordinate of the vertex.

If thekin the formula looks familiar, that’s because it’s the very same thing we use to determine therange of the quadratic function. The y-coordinate of the vertex tells us the maximum or minimum value of the function. It sets the boundary for which values ofyare included in the functio

Method 2: Using the vertex form of a quadratic function

We can transform a quadratic function in general form into its vertex form to obtain its vertex. The vertex form of a quadratic function is shown below:

For instance,f(x) = (x – 1)2+ 3is a quadratic function in vertex form. We haveh = 1andk = 3. Thus, the vertex of this function is (1, 3).

f(x) = (x + 1)2– 10is another example of a quadratic function in vertex form. We haveh = -1andk = -10. Thus, the vertex of this function is (-1, -10).

If the given quadratic function is in its vertex form, we can easily identify the values ofhandkwhich enable us to determine the vertex of the function quickly.

However, if the function is in general form, we need to transform it first into its vertex form.

Question 39

Example:Determine the vertex of f(x) = x²– 2x + 4 using the formula for the coordinates of the vertex.

Answer 39

The vertex of the parabola is (1, 3).

From the vertex, we can also identify the range off(x) = x²– 2x + 4.

Based on the computation above, we have obtainedk = 3. Since the function has a positive leading term, then the range of the function is the set of all real numbers greater than or equal to 3 or R = {y | y ≥ 3}.

Question 40

We use thecompleting the squaremethod to transform a quadratic function ingeneral form into its vertex form. Here’s how to do it:

Answer 40

Isolate the constant term from the terms withxvariable.Make the coefficient of the quadratic term equal to 1 by factoring.Add to both sides of the equation the square of half of the coefficient of the linear term.Express the perfect quadratic trinomial formed into asquare of binomial.Isolate f(x) from other quantities.

Question 41

Example:Transform the quadratic function f(x) = x²– 2x + 4 into its vertex form and determine the function’s vertex.

Answer 41

Thus, the vertex form of the given quadratic function isf(x) = (x – 1)² + 3.

Question 42

a vertical line that divides the parabola into two mirror parts.

Answer 42

The Axis of Symmetry of a Parabola

Question 43

the vertical line that passes throughhor the x-coordinate of the vertex.

Answer 43

axis of symmetry of a quadratic function

Question 44

Example:Determine the axis of symmetry of the quadratic function f(x) = -x²– 2x + 4.

Answer 44

Solution:

Using the formula forhof the vertex:

Therefore, the axis of symmetry of the given function is the linex = -1.

Question 45

points in the graph of the quadratic function that touches the x-axis.

Answer 45

X-intercepts

Question 46

Example:Determine the x-intercepts of the graph of f(x) = x2– 7x + 6.

Answer 46

Solution:

Setf(x) = 0:

0 = x² – 7x + 6

We can rewrite the equation above using the symmetric property of equality as:

x² – 7x + 6 = 0

We can solve the equation above by factoring:

x² – 7x + 6 = 0

(x – 6)(x – 1) = 0By factoring

x – 6 = 0 x – 1 = 0Equate each factor to 0

x1= 6 , x2= 1

The x-intercepts of the quadratic function are (6, 0) and (1, 0).

Question 47

Example:Graph f(x) = x2– 7x + 6.

Answer 47

Solution:Since the coefficient of the leading term of the given quadratic function is positive, we are expecting that the graph of this function is a parabola that opens upward.

Now, let us determine the vertex of this function:

From our computation, the vertex of this parabola is (7/2, -25/4).

Meanwhile, the x-intercepts of this function were already calculated in our example above. We have obtained (6, 0) and (1, 0) as the x-intercepts.

We look for other points of the parabola by substituting some arbitrary values ofxto the function:

From our computation above, we will use the points (-1, 14) and (7, 6) to help us graph the functions.

Plot the vertex, x-intercepts, and other points of the parabola. Finally, connect them to form a parabola.

The parabola above is the graph off(x) = x2– 7x + 6.

Question 48

Circle is a geometric shape that consists of points that are equidistant or have the same distance from a certain point. That point is called the

Answer 48

Center point

Question 49

the distance of the points in the circle from the center is called the

Answer 49

Radius

Question 50

If a circle has the origin as its center, the standard form of the equation of the circle is:

Answer 50

x²+ y²= r²

Question 51

Example 1:Determine the center and radius of the circle x2+ y2= 25.

Answer 51

Solution:The center of this circle is at (0, 0) and its radius is √25 = 5.

Question 52

Example 2:What are the center and the radius of the circle x2+ y2= 32?

Answer 52

Solution:The center of this circle is at (0, 0) and its radius is √32 or 4√2.

Question 53

Meanwhile, if the center of the circle is not located in the origin and instead in a point in the coordinate plane, then the equation of the circle becomes:

Answer 53

(x – h)2+ (y – k)2= r2

Question 54

Example 1:What is the center of the circle
(x – 8)²+ (y – 1)²= 12?

Answer 54

Solution:We haveh = 8andk = 1. Thus, the center is (8, 1). On the other hand, the radius is √12 or 2√3.

Question 55

Example 2:Determine the center and the radius of (x + 4)2+ (y + 3)2= 49.

Answer 55

Solution:We haveh = -4andk = 3. Thus, the center is (-4, 3). On the other hand, the radius is √49 = 7.

Question 56

1) In which quadrant of the coordinate plane can you locate (-5, -7)?
a) QI
b) QII
c) QIII
d) QIV

Answer 56

C

Question 57

2) Determine the slope of the line 3x + 2y = 1
a) -3/2
b) 3/2
c) -⅔
d) ⅔

Answer 57

A

Question 58

3) What are the intercepts of the line 5x - 2y = 10?
a) (2, 0) and (0, -5)
b) (2, -5) and (-2, 5)
c) (-5, 0) and (0, 2)
d) (0, -2) and (5, 0)

Answer 58

A

Question 59

4) Line A is perpendicular to Line B. Line B is defined by the equation 2x - y = 5. What is the
slope of line A?
a) 2
b) -2
c) ½
d) -½

Answer 59

D

Question 60

5) The Linear equation y = ¼x - 2 passes through the point ( k, -1). Determine the value of k.
a) -1
b) 2
c) -3
d) 4

Answer 60

D