1. The Distance Between Points
1. Calculate the distance between the points ( -7, -3 ) and ( 16, -3 )
( -7, -3 ) and ( 16, -3 )
x1 = -7 x2 = 16 y1 = -3 y2 = -3
Distance (d) = x2 - x1
= 16 - ( -7 )
= 16 + 7
= 23
2. A is the point ( -2, 4 ) and B ( 3, 1). Calculate the length of the line AB.
( -2, 4 ) and ( 3, 1 )
x1 = -2 x2 = 3 y1 = 4 y2 = 1
Distance (d) = √ ( x2 - x1 )² + ( y2 - y1)²
= √ ( 1 - 4 )² + ( 3 - (-2) )²
= √ ( 9 + 25)
= √ 34
= 5.83
3. Calculate the distance between the points ( 1/2, - 15/4 ) and ( -1, -1 )
Distance (d) = √ ( x2 - x1 )² + ( y2 - y1)²
= √ ( (-1) - 1/2 )² + ( (-1) - (-15/4) )²
= √ (-3/2)² + (11/4)²
= √ 9/4 + 121/16
= √ 36/16 + 121/16
= √ 157/16
= (√157) / 4
2. The Midpoint Formula
1. Calculate the midpoint of the points ( 1, -4 ) and ( 7, 8 )
x1 = 1 x2 = 7 y1 = -4 y2 = 8
Midpoint (m) = ( (x1 + x2)/2 , (y1 + y2)/2 )
= ( ( 1 + 7 )/2 , ( (-4) + 8 )/2 )
= ( 4, 2 )
2. In the diagram below, A ( 9, -2 ) lies on the circumference of the circle with centre C ( 17, 12 ), and the line AB is a diameter of the circle. Find the coordinates of B.
Let ( x, y ) be B.
Midpoint C, ( 17, 12 ) = ( ( x + 9 )/2 , ( y + (-2) )/2 )
( x + 9 ) / 2 = 17 ( y - 2 ) / 2 = 12
x + 9 = 17 x 2 y - 2 = 12 x 2
x + 9 = 34 y - 2 = 24
x = 34 - 9 y = 24 + 2
x = 25 y = 26
Therefore the Midpoint C is ( 25, 36 )
3. Gradients
1. Calculate the gradient of the straight line shown in the diagram below.
m = tan32° = 0.62
2. Find the angle that the line joining P ( -2, -2 ) and Q ( 1, 7 ) makes with the positive direction of the x-axis.
Gradient (Mpq) = ( 7 - (-2) ) / ( 1 - (-2) )
= 9 / 3
= 3
tanX = 3 X = tan -¹ (3) = 71.6
The angle X = 71.6°
3. Find the size of the angle X shown in the diagram below.
Tan x° = gradient(m) = 5
x° = tan-¹ (5)
x° = 78.7
X = 90° - 78.7
X = 11.3°
4. Collinearity
1. Show that the points P ( -6, -1) , Q ( 0, 2 ) and R ( 8, 6 ) are collinear.
Gradient (Mpq) = ( 2 - (-1) ) / ( 0 - (-6) )
= 3 / 6
= 1 / 2
Gradient (Mqr) = ( 6 - 2 ) / ( 8 - 0 )
= 4 / 8
= 1 / 2
Mpq = Mqr so P, Q and R are Collinear.
2. The points A ( 1, -1 ) , B ( -1, k ) and C ( 5, 7 ) are collinear. Find the value of k.
Mab = Mbc
( k - (-1) ) / ( (-1) - 1 ) = ( 7 - k ) / ( 5 - (-1) )
( k + 1 ) / (-2) = ( 7 - k ) / 6
-2 ( 7 - k ) = 6 ( k + 1 )
-14 + 2k = 6k + 6
-20 = 4k
k = -5
5. Gradients of Perpendicular Lines
1. Given that T is the point ( 1, -2 ) and S i s ( -4, 5), find the gradient of a line perpendicular to ST.
Gradient (ST) = ( 5 - (-2) ) / ( (-4) - 1 )
= - ( 7 / 5 )
Perpendicular Gradient (ST) = 5 / 7
2. Triangle MOP has vertices M ( -3, 9 ) , O ( 0, 0 ) and P ( 12, 4 ). Show that the triangle is right-angled.
Gradient (MO) = ( 9 - 0 ) / ( (-3) - 0 )
= -3
Gradient (OP) = ( 4 - 0 ) / ( 12 - 0 )
= 1 / 3
Since the Gradient (MO) x Gradient (OP) = -1 then Gradient (MO) is perpendicular to Gradient (OP) so the triangle MOP is right-angled.
6. The Equation of a Straight Line
1. Find the equation of the line with gradient 1/3 passing through the point ( 3, -4 ).
m = 1/3 a = 3 b = -4
y - b = m ( x - a )
y - (-4) = 1/3 ( x - 3 )
y + 4 = 1/3 ( x - 3 )
3y + 12 = 1 ( x - 3 )
3y = x - 3 - 12
3y = x - 15
x - 3y - 15 = 0
2. Find the equation of the line passing through A ( 3, 2 ) and B ( -2, 1 )
x1 = 3 x2 = - 2 y1 = 2 y2 = 1
Gradient (AB) = ( y2 - y1 ) / ( x2 - x1 )
= ( 1 - 2 ) / ( (-2) - 3 )
= -1 / -5
= 1 / 5
m = 1/5 a = 3 b = 2
y - b = m ( x - a )
y - 2 = 1/5 ( x - 3 )
5y - 10 = 1 ( x - 3 )
5y = x - 3 + 10
x - 5y + 7 = 0
3. Find the equation of the line passing through ( -3/5, 4 ) and ( -3/5, 5 ).
x1 = -3/5 x2 = -3/5 y1 = 4 y2 =5
Gradient (M) = ( y2 - y1 ) / ( x2 - x1 )
= ( 5 - 4 ) / ( (-3/5) - (-3/5) )
= 1 / 0
= undefined
so equation is x = -3/5
4. Find the gradient of the line with equation 3x + 2y + 4 = 0
3x + 2y + 4 = 0
For y = mx + c
y = (-3/2x) - 4/2
m = -3/2
5. The line through points A ( 3, -3 ) and B has equation 5x - y - 18 = 0. Find the equation of the line through A which is perpendicular to AB.
5x - y - 18 = 0
y = 5x - 18
m = 5
perpendicular m = -1/5
A ( 3, -3 ) a = 3 b = -3 m = -1/5
y - b = m ( x - a )
y - (-3) = -1/5 ( x - 3 )
5y + 15 = -1 ( x - 3 )
5y + 15 = -x + 3
x + 5y + 12 = 0
7. Medians
1. Triangle ABC has vertices A ( 4, -9 ) , B ( 10, 2 ) and C ( 4, -4 ). Find the equation of the median from A.
Midpoint BC = ( ( x1 + x2 ) / 2 ) + ( ( y1 + y2 ) / 2 )
= ( ( 10 + 4 ) / 2 ) + ( ( 2 + (-4) ) / 2 )
= ( 14 / 2 ) + ( (-2) / 2 )
= ( 7, -1 )
Gradient (AM) = ( (-1) - (-9) ) / ( 7 - 4 )
= 8 / 3
m = 8/3 a = 4 b = -9
y - b = m ( x - a )
y - (-9) = 8/3 ( x - 4 )
3y + 27 = 8x - 32
8x - 3y - 59 = 0
8. Altitudes
1. Triangle ABC has vertices A ( 3, -5 ) , B ( 4, 3 ) and C ( -7, 2 ). Find the equation of the altitude from A.
B ( 4, 3 ) and C ( -7, 2 )
x1 = 4 x2 = -7 y1 = 3 y2 = 2
Gradient (BC) = ( y2 - y1 ) / ( x2 - x1 )
= ( 2 - 3 ) / ( (-7) - 4 )
= (-1) / (-11)
= 1/11
Gradient perpendicular = - 11
m = -11 a = 3 b = -5
y - b = m ( x - a )
y - (-5) = -11 ( x - 3 )
y + 5 = -11x + 33
11x + y - 28 = 0
9. Perpendicular Bisectors
1. A is the point ( -2, 1 ) and B is the point ( 4, 7 ). Find the equation of the perpendicular bisector of AB.
x1 = -2 x2 = 4 y1 = 1 y2 = 7
Midpoints = ( ( x1 + x2 )/2 , ( y1 + y2 )/2 )
= ( ( -2 + 4 )/2 , ( 1 + 7 )/2 )
= ( 1, 4 )
Gradient (AB) = ( y2 - y1 ) / ( x2 - x1 )
= ( 7 - 1 ) / ( 4 - (-2) )
= 6/6
= 1
Perpendicular Gradient = -1
m = -1 a = 1 b = 4
y - b = m ( x - a )
y - 4 = -1 ( x - 1 )
y - 4 = -x + 1
x + y - 4 = 0
10. Intersection of Lines
1. Find the point of intersection of the lines 2x - y + 11 = 0 and x + 2y - 7 = 0.
2x - y = -11 .... x2
so
4x - 2y = -22 ADD x + 2y = 7
5x = -15
x = -3
Substiture x = -3 into x + 2y = 7
-3 + 2y = 7
2y = 7 + 3
2y = 10
y = 4
The point of intersection is ( -3, 5 )
2. Triangle PQR has vertices P ( 8, 3 ) , Q ( -1, 6 ) and R ( 2, -3 )
a.) Find the equation of altitude QS.
x1 = 8 x2 = 2 y1 = 3 y2 = -3
Gradient (PQ) = ( y2 - y1 ) / ( x2 - x1 )
= ( (-3) -3 ) / ( 2 - 8 )
= -6 / -6
= 1
Perpendicular gradient = -1
m = -1 a = -1 b = 6
y - b = m ( x - a )
y - 6 = -1 ( x + 1 )
y - b = - x - 1
x + y - 5 = 0
b.) Find the equation of median RT.
T = ( 8 + (-1) )/2 , ( 3 + 6 )/2
= ( 7/2 , 9/2 )
Gradient (RT) = ( (-3) - 9/2 ) / ( 2 - 7/2 )
= (-15/2) / ( - 3/2)
= -15 / -3
= 5
m = 5 a = 2 b = -3
y - b = m ( x - a )
y - (-3) = m ( x - 2)
5x - y - 13 = 0
c.) Hence find the coordinates of M.
x += y = 5 ADD 5x - y = 13
6x = 18
x = 3
Substitute x = 3 into x + y = 5
3 + y = 5
y = 2
Point of intersection (M) is ( 3, 2 )
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