document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=554469>Question 554469</A>: <I><SPAN STYLE=\"word-break: break-all;\"> hi, i need your help please... i don't know how to answer these questions because i can't fully imagine them... i hope you can help me because i really want to understand this lesson for the sake of my grades.....\r<BR>\n" );
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document.write( "a) the sum of the distance from a point P to (4,0) and (-4,0) is 9. if the abscissa of P is 1, find its ordinate...\r<BR>\n" );
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document.write( "b)the center of a circle is at (-3,-2). if a chord of length 4 is bisected at (3,1), find the length of the radius...\r<BR>\n" );
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document.write( "THANK YOU!!!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #361853 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=361853\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> to begin with, we need to understand what ordinate and abscissa mean.<BR>\n" );
document.write( "the following link addresses that issue:<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"http://www.blurtit.com/q550444.html\" target = \"_blank\">http://www.blurtit.com/q550444.html</a><BR>\n" );
document.write( "the short answer is:<BR>\n" );
document.write( "abscissa is the x value of the coordinate.<BR>\n" );
document.write( "ordinate is the y value of the coordinate.<BR>\n" );
document.write( "they are giving you the x value of the point P.<BR>\n" );
document.write( "that's the abscissa of 1.<BR>\n" );
document.write( "you have to find the ordinate of the point P.<BR>\n" );
document.write( "that would be the y value of the point P.<BR>\n" );
document.write( "coordinates are given in pairs of (x,y).<BR>\n" );
document.write( "x is the first value of the pair and points to the x-value of the point.<BR>\n" );
document.write( "y is the second value of the pair and points to the y-value of the point.<BR>\n" );
document.write( "a diagram of what i think your triangle will look like is shown below:<BR>\n" );
document.write( "the triangle is labeled APC.<BR>\n" );
document.write( "there is a perpendicular from point P to point D on AC.<BR>\n" );
document.write( "point A is (-4,0)<BR>\n" );
document.write( "point P is (1,y) ***** y is the value we are looking for.<BR>\n" );
document.write( "point C is (4,0)<BR>\n" );
document.write( "point D is (1,0)<BR>\n" );
document.write( "the abscissa of point P is 1 which makes it the x value of the coordinate pair represented by point P.<BR>\n" );
document.write( "you are looking for the ordinate of point P.<BR>\n" );
document.write( "the ordinate is the y value of the coordinate pair.<BR>\n" );
document.write( "i labeled it \"y\" until we could find a value for it.<BR>\n" );
document.write( "you are given that the distance between point A and C is 8.<BR>\n" );
document.write( "you are also given that the abscissa of the point P is 1.<BR>\n" );
document.write( "this means that the vertical line to point P lies on the value of x = 1.<BR>\n" );
document.write( "this means that there are 5 units from point A to D.<BR>\n" );
document.write( "this means that there are 3 units from point D to C.<BR>\n" );
document.write( "you need to find the length of the line from point P to D.<BR>\n" );
document.write( "this is the altitude of the right triangles formed.<BR>\n" );
document.write( "the 2 right triangles formed are APD and CPD.<BR>\n" );
document.write( "all you know is that the sum of the distance from point P to A and P to C is equal to 9.<BR>\n" );
document.write( "in the diagram i labeled the following:<BR>\n" );
document.write( "the length of the line AP is called b.<BR>\n" );
document.write( "the length of the line PC is called a.<BR>\n" );
document.write( "the length of the line PD is called c.<BR>\n" );
document.write( "by using the pythagorean formula, we can find the value of c.<BR>\n" );
document.write( "the pythagorean formula states that the hypotenuse squared is equal to the sum of the legs squared.<BR>\n" );
document.write( "in triangle APD, b is the hypotenuse and 5 and c are the legs.<BR>\n" );
document.write( "in triangle CPD, a is the hypotenuse and 3 and c are the legs.<BR>\n" );
document.write( "our formulas that we start with are:<BR>\n" );
document.write( "5^2 + c^2 = b^2<BR>\n" );
document.write( "3^2 + c^2 = a^2<BR>\n" );
document.write( "we solve for c^2 in both formulas to get:<BR>\n" );
document.write( "c^2 = b^2 - 5^2<BR>\n" );
document.write( "c^2 = a^2 - 3^2<BR>\n" );
document.write( "since they are both equal to c^2, then they are both equal to each other, so we get:<BR>\n" );
document.write( "b^2 - 5^2 = a^2 - 3^2<BR>\n" );
document.write( "we are given that a + b = 9<BR>\n" );
document.write( "from that we can solve for b in terms of a to get:<BR>\n" );
document.write( "b = 9 - a<BR>\n" );
document.write( "we can then substitute for b in the equation of b^2 - 5^2 = a^2 - 3^2 to get:<BR>\n" );
document.write( "(9 - a)^2 - 5^2 = a^2 - 3^2<BR>\n" );
document.write( "we now solve for a.<BR>\n" );
document.write( "we simplify the equation to get:<BR>\n" );
document.write( "81 - 18a + a^2 - 5^2 = a^2 - 3^2<BR>\n" );
document.write( "we subtract a^2 from both sides of the equation and we add 5^2 to both sides of the equation to get:<BR>\n" );
document.write( "81 - 18a + a^2 - a^2 - 5^2 + 5^2 = a^2 - a^2 - 3^2 + 5^2<BR>\n" );
document.write( "we combine like terms to get:<BR>\n" );
document.write( "81 - 18a = -3^2 + 5^2<BR>\n" );
document.write( "we simplify further to get:<BR>\n" );
document.write( "81 - 18a = -9 + 25<BR>\n" );
document.write( "we combine like terms again to get:<BR>\n" );
document.write( "81 - 18a = 16<BR>\n" );
document.write( "we subtract 81 from both sides of this equation to get:<BR>\n" );
document.write( "- 18a = 16 - 81<BR>\n" );
document.write( "we combine like terms to get:<BR>\n" );
document.write( "- 18a = - 65<BR>\n" );
document.write( "we divide both sides of this equation by -18 to get:<BR>\n" );
document.write( "a = 65/18 which simplifies to:<BR>\n" );
document.write( "a = 3.611111111<BR>\n" );
document.write( "sinca a + b = 9, this means that:<BR>\n" );
document.write( "a = 3.611111111<BR>\n" );
document.write( "b = 5.388888889<BR>\n" );
document.write( "we now have a value for a and b and we can use those values to solve for c using the pythagorean formula again.<BR>\n" );
document.write( "the result of that operation is that:<BR>\n" );
document.write( "c = 2.010005835<BR>\n" );
document.write( "we rounde all answers to the nearest hundredth or whatever rounding is required to get:<BR>\n" );
document.write( "a = 3.61<BR>\n" );
document.write( "b = 5.39<BR>\n" );
document.write( "c = 2.01<BR>\n" );
document.write( "the value of c is the length of the line PD which also becomes the y value of the coordinate of P.<BR>\n" );
document.write( "the coordinate of P becomes (1,2.01)<BR>\n" );
document.write( "the diagram of your triangles formed is shown below:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2012/jan111.jpg\" alt = \"$$$$\" /><BR>\n" );
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document.write( "your second problem starts here<BR>\n" );
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document.write( "the center of your circle is at (-3,-2)<BR>\n" );
document.write( "your chord is bisected at (3,1)<BR>\n" );
document.write( "the length of chord is 2.<BR>\n" );
document.write( "you want to find the length of the radius of your circle.<BR>\n" );
document.write( "the attached diagram shows what i think is the relationship you are looking for.<BR>\n" );
document.write( "the radius of a circle can form the perpendicular bisector of any chord in that circle.<BR>\n" );
document.write( "this is one of the theorems in geometry that you should probably have already studied.<BR>\n" );
document.write( "this is because the radii of the circle form an isosceles triangle with the chord as shown in the diagram.<BR>\n" );
document.write( "that isosceles triangle is ABC in the diagram.<BR>\n" );
document.write( "it is an isosceles triangle because the sides of the triangle are the radii of the circle.<BR>\n" );
document.write( "since you have an isosceles triangle, then the altitude of that triangle is the perpendicular bisector of the base of that triangle which is the chord.<BR>\n" );
document.write( "in your problem, the center of the circle is at (-3,-2) and the bisecting point on the chord is (3,1).<BR>\n" );
document.write( "this means that you can draw a line from the center of the circle to that point and the line formed is the perpendicular bisector of the triangle.<BR>\n" );
document.write( "that perpendicular bisector forms 2 right triangles which are triangle ADB and ADC.<BR>\n" );
document.write( "the base of each of those triangle is 2.<BR>\n" );
document.write( "the length of the perpendicular bisector which is the line AD is given by the equation:<BR>\n" );
document.write( "AD = sqrt (x^2 + y^2) where x and y are the coordinate points of point A and point D.<BR>\n" );
document.write( "the formula becomes:<BR>\n" );
document.write( "AD = sqrt (-3 - 3)^2 + -2 - 2)^2 which becomes:<BR>\n" );
document.write( "AD = sqrt ((-6)^2 + (-4)^2) which becomes:<BR>\n" );
document.write( "AD = sqrt (9 + 16) which becomes:<BR>\n" );
document.write( "AD = sqrt(25) which becomes:<BR>\n" );
document.write( "AD = 5<BR>\n" );
document.write( "the length of AD is equal to 5 as shown in the diagram.<BR>\n" );
document.write( "the length of AB is found using the pythagorean formula of:<BR>\n" );
document.write( "(AB)^2 = 5^2 + 2^2 which becomes:<BR>\n" );
document.write( "(AB)^2 = 25 + 4 which becomes:<BR>\n" );
document.write( "(AB)^2 = 29<BR>\n" );
document.write( "take square root of both sides of this equation to get:<BR>\n" );
document.write( "AB = sqrt(29).<BR>\n" );
document.write( "that's your answer.<BR>\n" );
document.write( "if you solve for AC, you will find that it's value is also sqrt(29).<BR>\n" );
document.write( "diagram for your second problem is shown below:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2012/jan112.jpg\" alt = \"$$$$\" />\r<BR>\n" );
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document.write( " </SPAN>\n" );
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