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Linear Functions (6.1 – 6.5, 6.7)

 

6.1              The linear function y = mx + b and its graph.

6.2              The straight line: equation of a line passing through a given point with given slope; equation of a line passing through two given points; the general equation ax + by + c = 0; parallel lines; perpendicular lines.

6.3              Intersection of lines: intersection of two lines and the solution of two linear equations in two unknowns; the equation of a line passing through the point of intersection of two given lines.

6.4              Regions determined by lines: linear inequalities.

6.5              Distance between two points and the (perpendicular) distance of a point from a line.

6.6              E. The angle between two lines. (E = extension only)

6.7              The mid-point of an interval

 

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THE LINEAR FUNCTION / COORDINATE GEOMETRY

 

Gradient-Intercept form of a straight line: y = mx + b     m = gradient, b = y-intercept

 

General form of a straight line: ax + by + c = 0

 

Gradient of a line between two points (x1, y1), (x2, y2):   m =  

                                                                                               

 

Equation of line with gradient m, passing through (x1, y1):            m =

                                                                                                y – y1 = m(x – x1)

 

Parallel lines have the same gradient.

 

Equation of a line between two points (x1, y1) & (x2, y2): =

 

Equation of line when x-intercept = a and y-intercept = b:          +  = 1

 

For perpendicular lines: m1m2 = -1

If the gradient of a line is m then the gradient of its normal is -

 

Solving the equations of two straight lines gives their point of intersection.

 

Distance between the two points (x1, y1) & (x2, y2):        d =

 

Mid-point between (x1, y1) & (x2, y2): x =, y =

Dividing (x1, y1), (x2, y2) in the ratio m:n;  x = , y =

Perpendicular distance of a point from a line:

 

 

 

 

 

 

 

 

 

 

 

 

COORDINATE GEOMETRY

 

Consider the straight line AB between the points A(x1,y1) and B(x2,y2)

                       

 

The gradient of the line is given by gradient = rise / run

The rise is y2 – y1 and the run is x2 – x1

Gradient, m =

 
Hence the gradient is  

The gradient is given the symbol m so  

 

 

The length  AB is given by Pythagoras’ theorem.

AB2 = (x2 – x1)2 + (y2- y1)2

Length =

 
AB =

                       

 

 

The mid point of AB has half the rise and half the run of AB

Mid-point =

 
Hence the mid-point of AB is

 

 

Exercise:

Find the gradient, mid-point and length of the lines joining the following pairs of points;   (i) A(1,1), B(5,5)          (ii) C(3,2), D(5,8)  (iii) F(2,5), G(6,11)

Answers:

(i) gradient = 1,             mid-point is (3,3),         length =  units

(ii) gradient = 3 mid-point is (4,5)          length =

(iii) gradient = 1.5         mid-point is (4,8)          length = 

 

COORDINATE GEOMETRY 2

 

Q.1.     Consider the following pairs of points:

(a)    P(4,2)  Q(12,8)

(b)   X(3,3)  Y(7,6)

(c)    R(3,2)   S(9,6)

(d)   A(4,-2)  B(-4,2)

(e)    M(4,8)   N(-1,-4)

 Find

(i)                  The distance between them.

(ii)                The gradient of the line joining them.

(iii)               The mid-point of the line joining them.

(iv)              The equation of the line joining them in the gradient form.

(v)                The equation of the line joining them in the general form.

 

Q.2.     The following equations of straight lines are written in the general form.

            Rewrite them in the gradient form and write down the gradient and y-intercept.

(i)                  3x – y + 3 = 0

(ii)                x – 2y – 4 = 0

(iii)               4x – 5y + 2 = 0

(iv)              5x - 2y + 3 = 0

(v)                3x – 6y = 0

 

Q.3.     Write down the general form of the equation of a straight line with

(i)                  a gradient of 2 and passing through the point (1,1)

(ii)                a gradient of ½ and passing through the point (3, -6)

(iii)               a gradient of -2/3 and passing through the point (4, 0)

(iv)              a gradient of 450 and passing through the point (2, 4)

(v)                a gradient of –3/4 and passing through the origin.

Answers

Q.1.

(a)        (i) 10    (ii) Ύ                (iii) (8,5)           (iv) y = Ύ x – 1            (v) 3x – 4y – 4 = 0

(b)        (i) 5      (ii) Ύ                (iii) (5, 4.5)       (iv) y = Ύ x + Ύ           (v) 3x – y + 3 = 0

(c)        (i)             (ii) 2/3              (iii) (6,4)           (iv) y = 2/3 x                (v) 2x – 3y = 0

(d)        (i)          (ii) -½               (iii) (0,0)           (iv) y = -½ x                 (v) x + 2y = 0

(e)        (i) 13   (ii)           (iii) (1.5, 2)     (iv) y = x -                  (v) 12x – 5y - 8 = 0

 

Q.2.     (i) y = 3x + 3                m = 3               y-intercept = 3

(ii) y = ½ x – 2 m = ½              y-intercept = -2

(iii) y = 4/5 x + 2/5       m = 4/5            y-intercept = 2/5

(iv) y = 5/2 x + 3/2       m = 5/2            y-intercept = 3/2

(v) y = ½ x                   m = ½              y-intercept = 0

 

Q.3.     (i)         2x – y – 1 = 0

(ii)                x – 2y – 15 = 0

(iii)               2x + 3y – 8 = 0

(iv)              x – y + 2 = 0

(v)                3x + 4y = 0

COORDINATE GEOMETRY 2 (Cont.)

Q.4. Find the general form of the equation of the line

(i)                  perpendicular to the line 3x - 4y + 8 = 0 and passing through (3,4)

(ii)                perpendicular to the line x – 2y + 8 = 0 and passing through (1,1)

(iii)               parallel to the line 2x – 3y + 12 = 0 and passing through (3,5)

(iv)              parallel to the line 5x + 2y –3 = 0 and passing through (5,2)

(v)                normal to the line 2x = 3y and passing through (2,2)

 

Q.5. Find the perpendicular distance of the point

(i)                  (2,1) from the line 2x – 3y + 4 = 0

(ii)                (3,4) from the line 4x + y + 10 = 0

(iii)               (4,3) from the line 3x + 4y + 1 = 0

(iv)              (12,1) from the line 2x – 5y + 7 = 0

(v)                (4,2) from the line 2x – 3y + 8 = 0

 

Q.6. Find the point of intersection of the following pairs of straight lines.

(i)                  2x + 3y – 8 = 0 and x – 4y + 7 = 0

(ii)                3x + 4y –32 = 0 and 4x – y - 11 = 0

(iii)               2x – y + 7 = 0 and 4x + y + 5 = 0

(iv)              3x – 2y + 8 = 0 and x + 4y + 12 = 0

(v)                2x + y + 2 = 0 and 2x – y – 2 = 0

 

Q.7. The points (3,2), (-1,3) , (2+k, k-14) are collinear. Find the value of k.

 

Q.8. Show that the lines 3x – y – 5 = 0 and x + 3y – 5 = 0 intersect at a right angle.

 

Q. 9. Show that the points (1,2), (-1,3) and (5,0) are collinear.

 

Q.10. Show that the lines joining the points (2,1), (5,5) and (-2,-2) form an isosceles triangle.

 

Q.11. Show that the points A(1.5,4), B(5,1), C(-2,-1), D(-5.5, 2) are the vertices of a parallelogram.

 

ANSWERS

Q.4.     (i)         4x + 3y –24 = 0

(ii)                2x –y - 3 = 0

(iii)               2x – 3y + 9 = 0

(iv)              5x + 2y – 29 = 0

(v)                3x + 2y – 10 = 0

 

Q.5.     (i)  = 1.39 units     (ii) = 6.31 units    (iii) 5 units        

(iv) = 4.83 units               (v) = 2.77 units

 

Q.6. (i) (1,2)                (ii) (4,5)            (iii) (-2,3)         (iv) (-4,-2)        (v) (0,-2)

 

Q.7. k = 13

COORDINATE GEOMETRY – Problems

 

1.            Find the gradient form of equation of the line inclined at 45o to the x-axis and passing through the point (5, 4).

2.            Find the equation in the gradient form of the perpendicular bisector of the line joining the points A(7, 6) and B(9, 2)

3.            What is the general form of the equation of a straight line that cuts off an intercept of 4 on the x-axis and 10 on the y-axis?

4.            Find the general form of the equation of the line parallel to the line 3x -2y +4 = 0 and passing through the point (3, 4)

5.            The points (1, 1), (7, -3) and (k-7, k-2) are collinear. Find the value of k.

6.            What is the perpendicular distance of the point (7,3) from the line y = 2x -4?

7.            What is the perpendicular distance of the point (2,4) from the line 3x –y + 6 = 0.

8.            What is the perpendicular distance between the lines y = ½ x + 3 and y = ½ x + 8.

9.            Find the point of intersection of the lines y -3x +5 = 0 and 2y + 5x – 12 = 0.

10.        Find the equation in general form, of the straight line with gradient  that passes through the point of intersection of the lines 2x – y + 4 = 0 and 3x – 2y + 6 = 0.

11.        A circle of radius 5 units has its centre at the origin. Find the equation of the tangent to the circle at the point (3, 4). (Hint: Find the equation of the radius to (3, 4) first)

12.        The points A(5, y), B(8, 4), C(3, 2) form an isosceles triangle with AB = AC. Find the value of y.

13.        Find the perpendicular distance from the point (5, -3) to the line joining the points

(-2, -6) and (8, 4). Find the coordinates of the point where the perpendicular cuts the line and hence show that it is the perpendicular bisector.

14.    The points (0, -3), (2, 11), (1, 7) and (1, 1) form the vertices of a quadrilateral. Show that the opposite sides of the quadrilateral are parallel and hence it is a parallelogram.

15.    Show that the points (-1, -5), (1, 1) and (4,10) are collinear.

16.    Show that the points P(1,1), Q(3, 7) and R(5, 3) form an isosceles, right angled triangle.

Answers:

1.      y = x-1

2.      y = ½ x

3.      5x + 2y - 20 = 0

4.      3x – 2y – 1 = 0

5.      k = 5

6.       or

7.      or            

8.     

9.      (2, 1)

10.  x – 3y + 2 = 0

11.   3x + 4y – 25 = 0

12.  y = 4 Ό

13.  d = , cuts at (3, -1)

 

COORDINATE GEOMETRY – More Problems

 

1.      Find the gradient form of equation of the line inclined at 135o to the x-axis and passing through the point (-3, -5).

2.      Find the equation in the gradient form of the perpendicular bisector of the line joining the points A(5, -10) and B(9,-12)

3.      What is the general form of the equation of a straight line that cuts off an intercept of -2 on the x-axis and 8 on the y-axis?

4.      Find the general form of the equation of the line parallel to the line 2x -4y +5 = 0 and passing through the point (-1, 5)

5.      The points (3, 1), (1, 2) and (k +7, k-16) are collinear. Find the value of k.

6.      What is the perpendicular distance of the point (4,-2) from the line y = ½ x + 6?

7.      What is the perpendicular distance of the point (-2,-6) from the line 3x –4y - 2 = 0.

8.      What is the perpendicular distance between the lines y = 2x + 5 and y = 2x + 7.

9.      Find the point of intersection of the lines 5x -2y + 20 = 0 and 2x + y – 10 = 0.

10.    Find the equation in general form, of the straight line with gradient  that passes through the point of intersection of the lines 3x + 4y - 8 = 0 and x + 2y + 10 = 0.

11.    A circle of radius 10 units has its centre at the origin. Find the equation of the tangent to the circle at the point (8, 6). (Hint: Find the equation of the radius to (8, 6) first)

12.    The points A(2, y), B(5, 8), C(9, 11) form an isosceles triangle with AB = BC. Find the value of y.

13.    Find the perpendicular distance from the point (6, -8) to the line joining the points

(-8, -12) and (2, 6). Find the coordinates of the point where the perpendicular cuts the line and hence show that it is the perpendicular bisector.

14.    The points (-2, -1), (-5, 6), (0, 8) and (3, 1) form the vertices of a quadrilateral. Show that the opposite sides of the quadrilateral are parallel and hence it is a parallelogram.

15.    Show that the points (-3, -19), (1, 1) and (4,16) are collinear.

16.    Show that the points P(4,2), Q(5, 5) and R(7, 3) form an isosceles, right angled triangle.

Answers:

1.   y = -x - 8                     2.  y = 2x – 25        3.   4x - y + 8 = 0

4.   x – 2y + 11 = 0            5.  k = 10                6.   or 

7.  

8.   or

9.   (0, 10)

10. x - 2y - 66 = 0

11. 4x + 3y – 50 = 0

12. y = 4  or 12

13. d =  = , cuts at (-3, -3)

 

COORDINATE GEOMETRY – Even More Problems

 

1.      Find the gradient form of equation of the line inclined at 63o26’6” to the x-axis and passing through the point (2, 3).

2.      Find the equation in the gradient form of the perpendicular bisector of the line joining the points A(-2, -4) and B(3, 6)

3.      What is the general form of the equation of a straight line that cuts off an intercept of 4 on the x-axis and -6 on the y-axis?

4.      Find the general form of the equation of the line parallel to the line -3x +2y - 7 = 0 and passing through the point (-6, -2)

5.      The points (0, 2), (-15, -1) and (k+7, k-3) are collinear. Find the value of k.

6.      What is the perpendicular distance of the point (-6,-4) from the line y = ?

7.      What is the perpendicular distance of the point (4,-2) from the line 2x –3y + 7 = 0.

8.      What is the perpendicular distance between the lines y = 3x + 8 and 2y = 6x - 17.

9.      Find the point of intersection of the lines 2x + 3y +5 = 0 and 3x + 4y +10 = 0.

10.    Find the equation in general form, of the straight line with gradient  that passes through the point of intersection of the lines 2x + 4y - 9 = 0 and 3x + y - 11 = 0.

11.    A circle of radius 13 units has its centre at the origin. Find the equation of the tangent to the circle at the point (5, 12). (Hint: Find the equation of the radius to (5, 12) first)

12.    The points A(-3, y), B(2, -2), C(6, 2) form an isosceles triangle with AB = AC. Find the value of y.

13.    Find the perpendicular distance from the point (-5, 5) to the line joining the points

(-12, -4) and (6, 8). Find the coordinates of the point where the perpendicular cuts the line and hence show that it is the perpendicular bisector.

14.    The points (-4, 6), (1, -2), (-1, 9) and (4, 1) form the vertices of a quadrilateral. Show that the opposite sides of the quadrilateral are parallel and hence it is a parallelogram.

15.    Show that the points (-5, -3), (-1, 0) and (3,3) are collinear.

16.    Show that the points P(-2,11), Q(0, 9) and R(3, 14) form an isosceles, right angled triangle.

Answers:

1.   y = 2x -1                      2.   y = - ½ x + 1Ό                               3.  -2x + 3y - 18 = 0

4.   3x – 2y + 14 = 0          5.   k = 8                                              6. or

7.       or        

8.   or 

9.   (-10, 5)

10. 2x + 8y - 29 = 0

11. 5x + 12y – 169 = 0     

12. y = 7   

13. d = , cuts at (3, -1)