
[ended]Mathematics Tutoring Session #14 Two Dimensional Rectangular Cartesian on Fri. 4th Mar 2016, 7:00pm  8:00pm
Mathematics
Emmanuel Iwara(Tutor)
04032016 16:01:00 +0000Good day all,
This session has ended for today
Two  Dimensional Rectangular Cartesian is often tested by jamb and other exam. Let quickly take a look at the following.
Distance Formula: Given the two points (x_{1}, y_{1}) and (x_{2}, y_{2}) the distance between these points is given by the formula:
√[(x_{2} x_{1})^{2} +(y_{2} y_{1})^{2}].
This formula is an application of Pythagoras' theorem for right triangles:
Don't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you. The distance will be the same, regardless.
Example 1. Find the distance between the pairs of points; A(3,2) and B(4,6)
Solution
(x_{1} = 3,y_{1} = 2, x_{2} = 4,y_{2} = 6)
Substitute this values into the distance formula we have
√[(43)^{2} + (62)^{2}]
=√[(1)^{2} + (4)^{2}]
=√(1 + 16) = √(17)
Midpoint Formula The point halfway between the endpoints of a line segment is called the midpoint. A midpoint divides a line segment into two equal segments.
The midpoint of the two points(x_{1}, y_{1}) and (x_{2}, y_{2}) is;
[(x_{1} + x_{2}) / 2], [(y_{1} + y_{2}) / 2] , but
The point which divides (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio p:q has coordinates
[(px_{2} + qx_{1}) / p+q), (py_{2} + qy_{1}) / p+q]
Example 2. Find the coordinates of the mid points of the lines joining the pairs of point A(4,2) and B (1,3).
Solution
(x_{1} = 4, x_{2} = 1,y_{1} = 2, y_{2} = 3)
Using the formula(x_{1} + x_{2}) / 2,[(y_{1} + y_{2}) / 2]
we have [( 4 + 1)/2, (32)/2] = [ 5/2, 5/2 ].
Example 3. Find the coordinates of the mid points of the lines joining the pairs of point A(5,8) and B (1,3) in the given ratio 2:3.
Solution
Using the formula[(px_{2} + qx_{1}) / p+q, (py_{2} + qy_{1}) / p+q]
(x_{1} = 5, x_{2} = 1,y_{1} = 8, y_{2} = 3 , p = 2, q =3)
we have [(2 x 1 + 3 x 5)/5 , (2 x 3 + 3 x 8)/5] = [(2 + 15)/5 ,(6 + 24)/5] = [13/5, 6]
The gradient of the line = (change in ycoordinate)/(change in xcoordinate) . We can of course use this to find the equation of the line.
The gradient m of the line (x_{1}, y_{1}),(x_{2}, y_{2}) is define as;
M =(y_{2} y_{1})/(x_{2}  x_{1})
The acute angle θ between two lines AB of gradient m_{1} and CD gradient m_{2} is given by;
tan θ = (m_{2}  m_{1})/ ( 1 + m_{2}m_{1})
Equation of the line is given by y = mx + c, where m is the slope or gradient.
Example 4. Find the gradient of the of lines passing through (3,5) and (2, 4)
Solution
Using the formula M =(y_{2} y_{1})/(x_{2}  x_{1})
M = gradient = [(45) /( 2 (3)] = [(9)/ 1] = 9
Example 5.Find the acute angles between the pairs lines;
[4y + 3x = 2 and x2y =3]
Solution
4y + 3x = 2(1)
x 2y = 3(2)
Dividing Equ(1) by 4 and making y subject formula becomes.
y = 3x/4 + 2/4(3)
Dividing Equ(2) by 2 becomes.
y = x/2 + 3/2
y = x/2 3/2(4)
Comparing equ(3) and equ(4) with y = mx +c, where m is the slope or gradient.
From equ(3) m_{1} = 3/4
From equ(4) m_{2} = 1/2
tan θ = (m_{2}  m_{1})/ ( 1 + m_{2}m_{1})
tan θ = (1/2 + 3/4)/ 1 + (1/2 x 3/4)
Taking L.C. M
tan θ = (2+3/4)/ (1 + 3/8)
tan θ = 10/11
tan θ = 0.90909
using your calculator arctan(0.90909) will give us θ = 47^{o}
Now see if you can try the following past jamb and other exams questions.
1. Find the coordinates of the mid points of the lines joining the pairs of point A(5,4) and B (3,2).
2. Find the acute angles between the pairs lines;
[ y = 5x 2 and y  3x + 1 = 0]
3. Find the gradient of the lines passing through (1,1) and (2, 3)
4. Find the coordinates of the mid points of the lines joining the pairs of point A(10,2) and B (6,4) in the given ratio 3:2.
5. Find the distance between the pairs of points; A(2,7) and B(3,2)
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Niyi olàniyi
1= 4,3 2=arctan(1/8) 3=2 4= am confused with the ratio 5=25
0 04032016 16:46:00 +0000

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