1 Find an equation for the line that passes through the point (2, -5) and ( a ) has slope -3. First step: write down the slope-intercept equation, y = mx + b Next replace m (your slope) with the number -3. y = (-3) x + b Second step: we will substitute our 'x' and 'y' values from the point into the equation. Substitute 2 in for our x value and -5 in for our y value. -5 = -3 (2) + b Third step: Solve for b. We will multiply 3 times 2 (watch out the sign!) and get six and then subtract six for each side. -5 = - 6 + b -5 + 6 = -6 + 6 + b Y-intercept: after simplifying we get 1 for b. 1=0+b 1=b Last step: Take our results and write an equation in slope-intercept form. Answer: y=-3x+1 ( b ) is parallel to the x-axis. A horizontal line always has an equation that looks like y = [some constant] If I know that a horizontal line passes through the point (2,-5), then I know that the value of y is the constant I want, so the equation must be y = -5 So in fact, we can be a little more precise, and say that a horizontal line has the equation y = [the y-value of any point on the line]
( c ) is parallel to the y-axis. and a vertical line always has an equation that looks like x = [some constant] If I know that a vertical line passes through the point, then I need the value of x instead: x=2 So in fact, we can be a little more precise, and say that a vertical line has the equation x = [the x-value of any point on the line] ( d ) is parallel to the line 2 x - 4 y = 3 . - Let m1 be the slope of the line whose equation is to found and m2 the slope of the given line. Rewrite the given equation in slope intercept form and find its slope.
2x−4y=3 −4y=3−2x 3 2x y= − −4 −4 3 2x y=− −[− ] 4 4 3 2x y=− 4 4 3 x y=− 4 2 x 3 y= − 2 4 Slope m2 = (½). Two lines are parallel if and only if they have equal slopes: m1 = m2 = (½). We now use the point slope form to find the equation of the line with slope m1.
y− y 1=m x− x 1 1 y−−5= x−2 2 Which may be written as
1 1 y5= x[ −2] 2 2 1 y5= x1 2 1 y= x−1−5 2 1 y= x−6 2 2. Find an equation for the circle that has center (-1, 4) and passes through the point (3, -2). So given the center and an endpoint, we'll write the standard form of the equation and replace h and k with the values for the center: h = -1 and k = 4
x−h2 y−k 2 =r 2 x−−1 2 y−42=r 2 Notice that when you replace the h and k, you can check to see if it is right by seeing if when you put 1 for x you get 0 and when you put 0 in for y you get 0. Now we have to find the radius. We do this by using point that is on the circle. We replace the x and the y with those coordinates of the circle. (3, -2)
3−−12 −2−42 =r 2 Now solving:
312 −62 =r 2 42 36=r 2 1636=r 2 52=r 2 Finally, write the equation:
[ x−−1]2 y−42 =52 x12 y−42 =r 2 3. Find the center and radius of the circle with equation
x 2 y 2−6x10y9=0 In order to find the center and the radius of the circle, we first rewrite the given equation into the standard form as in the definition. Put all terms with x and x 2 together and all terms with y and y 2 together using brackets. Definition: a circle is the set of points equidistant from a point C(h, k) called the center. The fixed distance r from the center to any point on the circle is called the radius. The standard equation of a circle with center C(h, k) and radius r is as follows: (x - h)2 + (y - k)2 = r2 (x2 – 6 x) + (y2 + 10 y) + 9 = 0 We now complete the square within each bracket: (x2 - 6x + 9) – 9 + ( 10y + 25) - 25 + 9 = 0 (x - 3)2 + ( y - 5)2 - 9 - 25 + 9 = 0 Simplify and write in standard form (x – 3)2 + ( y - 5)2 = 25 (x – 3)2 + ( y - 5)2 = 52 We now compare this equation and the standard equation to obtain
center at C(h, k) = C(3, 5) and radius r = 5 4. Let A (-7, 4) and B (5, -12) be points in the plane. ( a ) Find the slope of the line that contains A and B. By dividing the difference in y-coordinates by the difference in x-coordinates, one can obtain the slope of the line:
y y 2 − y 1 −12−4 −16 −16 4 m= = = = = =− x x 2 − x 1 5−−7 57 12 3 ( b ) Find an equation of the line that passes through A and B. What are the intercepts? We first calculate the slope of the line, but we already had calculate it in incise ( a )
y y 2 − y 1 −12−4 −16 −16 4 m= = = = = =− x x 2 − x 1 5−−7 57 12 3 The we use the slope and any of the two points to write the equation of the line using the point slope form:
y− y 1=m x− x 1 using the first point:
4 y−4=− [ x−−7] 3 4 y−4=− x7 3 4 28 y−4=− x− 3 3 4 −28 y=− x 4 3 3 4 −2812 y=− x 3 3 4 16 y=− x− 3 3 1 y= [−4x−16] 3 3y=[−4x−16] 4x3y=−16 4x3y16=0 And if the equation is in the slope/intercept form, we can easily see what the slope (or x-intercept) and y-intercept are.
y=mxb In this form, the slope is m, which is the number in front of x. In our problem, that would have to be -4. The y-intercept is b, which is the constant. In our problem, that would be - (16/3). ( c ) Find the midpoint of the segment AB. Technically, the Midpoint Formula is the following:
x 1 x 2 y 1 y 2 , 2 2 Apply the Midpoint Formula:
−75 4−12 , 2 2
−2 4−12 , 2 2
−1 ,
−8 2
−1 ,−4
( d ) Find the length of the segment AB.
Notice how a right triangle was formed with
AB When working with diagonal segments, use the Distance Formula to determine the length.
d = x 2 − x 1 2 y 2 − y 1 2 The advantage of the Distance Formula is that you do not need to draw a picture to find the answer. All you need to know are the coordinates of the endpoints of the segment. The Distance Formula is really just a coordinate geometry way of writing the Pythagorean Theorem. If you cannot remember the Distance Formula, you can always draw a graph and use the Pythagorean Theorem. When working with diagonal segments, the Pythagorean theorem can be used to determine the length.
It doesn't matter which point you start with. Just start with the same point for reading both the x and y coordinates.
d= 5−−72 −12−42 = 57 2−162 = 12 256 2
= 144256 = 400 =20
( e ) Find an equation of the perpendicular bisector of AB. 1. Find the midpoint of the given segment.
x 1 x 2 y 1 y 2 mp= , 2 2 mp=
−75 4−12 , 2 2
mp=
−2 4−12 , 2 2
mp=−1,
−8 2
mp=−1,−4
2. Find the slope of the given segment.
m=
m=
y2− y1 x2− x1
−12−4 5−−7
m=
−16 57
m=
−16 12
4 m=− 3 3. Determine the slope of the perpendicular bisector knowing that the slopes of perpendicular lines are opposites and reciprocals of each other.
⊥
Line's slope =
3 4
4. Write an equation into point-slope form,
y−k=m x−h , since the slope of the perpendicular bisector and a point (h, k) the bisector goes through is known.
3 y−−4= x−−1 4 5. Solve the point-slope equation for y to get
y=mxb 3 y−−4= x−−1 4 3 y4= x1 4 Distribute the slope value
3 y4= x1 4 3 3 y4= x 4 4 Move the k value to the right side of the equation.
3 3 y4= x 4 4 3 3 y= x −4 4 4 3 3−16 y= x 4 4 3 13 y= x− 4 4
1 y= 3x−13 4 4y=3x−13 −3x4y=−13 [−3x4y=−13]−1 3x−4y=13
( f ) Find and equation of the circle for which AB is a diameter. The center of the circle is the midpoint of the line segment making the diameter AB. The midpoint formula is used to find the coordinates of the center C of the circle. x coordinate of C =
y coordinate of C =
x 1 x 2 −75 −2 = = =−1 2 2 2 y 1 y 2 4−12 4−12 −8 = = = =−4 2 2 2 2
The radius is half the distance between A and B:
1 r= x 2 − x 1 2 y 2 − y 1 2 2 1 2 2 r= 5−−7 −12−4 2 1 2 2 r= 57 −16 2 1 r= 122−162 2 1 r= 144256 2 1 r= 400 2 1 r= 20 2 r =10 The coordinate of C and the radius are used in the standard equation of the circle to obtain the equation:
The standard equation of a circle with center C(h,k) and radius r is as follows: (x - h)2 + (y - k)2 = r2
x−−12 y−−4 2=102 2
2
x1 y4 =100
5. Sketch the region in the xy-plane defined by the equation or inequalities.
(a)
−1 y3
(b)
∣x∣4
Interval notation:
y
∣y∣2 −4 x4 −2 y2
( c ) Do not appear in the exam complete. ( d ) Do not appear in the exam complete. ( e ) Do not appear in the exam complete.
(f)
9x 2 16y 2 =144
a) We first write the given equation in standard form by dividing both sides of the equation by 144.
1 [9x 216y 2=144] 144 9x 2 16y 2 144 = 144 144 144 9x 2 16y 2 =1 144 144
by the division of fractions law:
because
a 1 = b b a 1 1 1 a = = b b b a a
So,
9x 2 16y 2 =1 144 144 x2 y2 =1 144 144 9 16
x2 y2 =1 16 9 We now identify the equation obtained with one of the standard equation
x2 y 2 2 2 =1 a b and we can say that the given equation is that of an ellipse with
x2 y 2 2 =1 2 4 3 a = 4 and b = 3 (NOTE: a >b) . Set y = 0 in the equation obtained and find the x intercepts.
x2 2 =1 4 Solve for x,
x 2 =42 x=± 42 x=± 16 x=±4 Set x = 0 in the equation obtained and find the y intercepts.
y2 =1 2 3
Solve for y,
y 2=32 x=± 32 x=± 9 x=±3
b) We need to find the coordinates of the foci, so we need to find c first. c2 = a2 – b2 a and b were found in part a) c2 = 42 – 32 c2 = 7 Solve for c. c = ± (7)1/2 The foci are F1 (0, (7)1/2) and F2 (0, - (7)1/2) c) To find the length of the major and minor axes, remember the thoery: The length of the major axis is 2a, and the length of the minor axis is 2b. So, The major axis length is given by 2a = 8. The minor axis length is given by 2b = 6. d) At last we ketch the graph of the equation. First Locate the x and y intercepts, find extra points if needed and sketch.
Bibliography 1- http://www.terragon.com/tkobrien/algebra/topics/equationofline/equationofline.html 2- http://mathforum.org/library/drmath/view/63110.html 3- http://www.analyzemath.com/line/Tutorials.html 4- http://facstaff.gpc.edu/~mhall/online/math99/circle/circle.htm 5- http://www.analyzemath.com/CircleEq/Tutorials.html 6- http://en.wikipedia.org/wiki/Slope 7- http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut15_slope.htm 8- http://www.purplemath.com/modules/midpoint.htm 9- http://www.regentsprep.org/regents/math/geometry/GCG3/Ldistance.htm 10- http://www.algebralab.org/studyaids/studyaid.aspx?file=Algebra2_10-4.xml 11- http://www.analyzemath.com/CircleEq/Tutorials.html 12http://cda.morris.umn.edu/~mcquarrb/BA/Resources/9.7_Absolute_Value_Equations_Inequalities.pdf 13- http://www.mathwarehouse.com/geometry/circle/equation-of-a-circle.php 14- http://www.purplemath.com/modules/syseqgen5.htm 15- http://www.analyzemath.com/EllipseProblems/EllipseProblems.html