T Madas T Madas 8 y Plot the

8 y Plot the equation y = x – 3 -12 6 4 2

y Plot the equation y =8 3 x – 8 -12 x 0 1

The same technique can be used when the gradient and the x – intercept

y 8 Plot the line with gradient -2 and x-intercept 1 -12 6 4

Write down the gradient and the y - intercept of the line y –

Write down the gradient and the y - intercept of the line 4 x

8 y Plot the line with x–intercept 10 6 and y–intercept -3 -12 4

8 y Plot the line with x–intercept 10 What is the equation of this

8 y Plot the equation x – y = 3 -12 6 4 2

Rearrange these equations and hence write the gradient and y-intercept of these lines Next

If a line has gradient m then any line parallel to it, also has

If a line has gradient m then any line perpendicular to it, has gradient

ABCD is a rectangle. A is the point (0, -1). C is the point

y A (0, 15) The two lines shown have equations 2 x + y

The two lines shown have equations 2 x + y = 15 and x

The line with equation 6 x – 5 y = 4 is drawn in

Two points P and Q are marked on the grid below. 1. Mark on

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8 y Plot the equation y = x – 3 -12 6 4 2 -10 -8 -6 -4 -2 2 -2 4 6 8 10 x -4 x 0 1 2 3 y -3 -2 -1 0 4 1 -6 -8 © T Madas

8 y 6 4 2 -10 -8 -6 -4 -2 2 -2 7 – x Plot the equation y = x 0 y 7 1 6 2 5 3 4 4 3 4 6 8 10 x -4 -6 -8 © T Madas

y Plot the equation y =8 3 x – 8 -12 x 0 1 2 3 y -8 -5 -2 1 4 4 6 4 2 -10 -8 -6 -4 -2 2 -2 4 6 8 10 x -4 -6 -8 © T Madas

-12 8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 Plot the line with gradient 1 and y-intercept 2 © T Madas

-12 8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 Plot the line with gradient 2 and y-intercept 3 © T Madas

-12 8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 Plot the line with gradient -3 and y-intercept -1 © T Madas

-12 8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 Plot the line with gradient ¼ and y-intercept -1 © T Madas

The same technique can be used when the gradient and the x – intercept is given © T Madas

-12 8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 Plot the line with gradient 1 and x-intercept -4 © T Madas

y 8 Plot the line with gradient -2 and x-intercept 1 -12 6 4 2 -10 -8 -6 -4 -2 2 -2 4 6 8 10 x -4 -6 -8 © T Madas

Write down the gradient and the y - intercept of the line y – 2 x = 4 Gradient = 2 y-intercept = 4 Write down the gradient and the y - intercept of the line y + 3 x = 6 Gradient = -3 y-intercept = 6 © T Madas

Write down the gradient and the y - intercept of the line 4 x – y = 1 Gradient = 4 y-intercept = -1 Multiply whole by. R. H. S -1 The alternativethe approach is toexpression move the -y to the © T Madas

Write down the gradient and the y - intercept of the line 4 x + 2 y = 1 Gradient = -2 y-intercept = ½ Write down the gradient and the y - intercept of the line 3 x – 8 y = -2 Gradient = ⅜ y-intercept = ¼ © T Madas

8 y Plot the line with x–intercept 10 6 and y–intercept -3 -12 4 2 -10 -8 -6 -4 -2 2 -2 4 6 8 10 x -4 -6 -8 © T Madas

8 y Plot the line with x–intercept 10 What is the equation of this line? 6 and y–intercept -3 -12 4 2 -10 -8 -6 -4 -2 2 -2 4 6 8 10 x -4 -6 -8 © T Madas

8 y Plot the equation x – y = 3 -12 6 4 2 -10 -8 -6 -4 -2 2 -2 4 6 8 10 x -4 -6 -8 © T Madas

8 y 6 4 2 -10 -8 -6 -4 -2 2 -2 4 6 8 10 x Plot the equation x -4 +y=7 -6 -8 © T Madas

8 y 6 4 2 -10 -8 -6 -4 -2 2 4 -2 6 8 10 x -4 + 3 y = 12 Plot the equation 2 x -6 -8 © T Madas

Rearrange these equations and hence write the gradient and y-intercept of these lines Next plot in the same diagram the lines 1 & 2 using a table of values Next plot in another graph the lines 3 & 4 without a table of values using gradient and intercept considerations Finally plot in a separate graph line 6 using intercept considerations only © T Madas

-12 8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 Look at the gradients of these parallel lines © T Madas

If a line has gradient m then any line parallel to it, also has a gradient m Write the equation of a line parallel to Write the equation of a line passing through (0, 3) and parallel to © T Madas

8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 these perpendicular lines Look at the gradients of © T Madas

If a line has gradient m then any line perpendicular to it, has gradient Write the equation of a line perpendicular to Write the equation of a line passing through (0, 5) and perpendicular to © T Madas

-12 8 y 6 4 2 -10 -8 -6 -4 -2 2 4 6 8 10 -2 x -4 -6 -8 Work out the equation of this line © T Madas

ABCD is a rectangle. A is the point (0, -1). C is the point (0, 5). The equation of the straight line through A and D is y = 2 x – 1 (a) Find the equation of the straight line through B and C (b) Find the equation of the straight line through C and D BC C (0, 5) D AD both lines have gradient 2 AD : y = 2 x – 1 BC : y = 2 x + 5 B A (0, -1) © T Madas

ABCD is a rectangle. A is the point (0, -1). C is the point (0, 5). The equation of the straight line through A and D is y = 2 x – 1 (a) Find the equation of the straight line through B and C (b) Find the equation of the straight line through C and D (c) Find the co ordinates of point D C (0, 5) D AD : y = 2 x – 1 CD AD CD : y = B CD has gradient equal to +5 A (0, -1) © T Madas

ABCD is a rectangle. A is the point (0, -1). C is the point (0, 5). The equation of the straight line through A and D is y = 2 x – 1 (a) Find the equation of the straight line through B and C (b) Find the equation of the straight line through C and D (c) Find the co ordinates of point D AD : y = 2 x – 1 C (0, 5) D CD : y = +5 2 x – 1 = +5 2 x B A (0, -1) } =5+1 =6 © T Madas

ABCD is a rectangle. A is the point (0, -1). C is the point (0, 5). The equation of the straight line through A and D is y = 2 x – 1 (a) Find the equation of the straight line through B and C (b) Find the equation of the straight line through C and D (c) Find the co ordinates of point D AD : y = 2 x – 1 CD : y = C (0, 5) D +5 } If B A (0, -1) © T Madas

y A (0, 15) The two lines shown have equations 2 x + y = 15 and x + 2 y = 12 1. Write down the equation of AB 2. Find the coordinates of point E 3. Calculate the area of OBEC C(0, 6) E 0 B(7. 5, 0) D(12, 0) x © T Madas

The two lines shown have equations 2 x + y = 15 and x + 2 y = 12 1. Write down the equation of AB 2. Find the coordinates of point E 3. Calculate the area of OBEC 1 2 } } Sub into either equation 2 E (6, 3) © T Madas

y A (0, 15) The two lines shown have equations 2 x + y = 15 and x + 2 y = 12 1. Write down the equation of AB 2. Find the coordinates of point E 3. Calculate the area of OBEC C(0, 6) 3 3 0 6 7. 5 E (6, 3) B(7. 5, 0) D(12, 0) x © T Madas

y A (0, 15) The two lines shown have equations 2 x + y = 15 and x + 2 y = 12 1. Write down the equation of AB 2. Find the coordinates of point E 3. Calculate the area of OBEC C(0, 6) E (6, 3) 0 B(7. 5, 0) D(12, 0) x © T Madas

The line with equation 6 x – 5 y = 4 is drawn in the grid below (a) Rearrange the equation 6 x – 5 y = 4 making y the subject (b) If the point (9, k ) lies on the line, find the value of k y 4 3 2 1 -2 -1 0 -1 -2 1 2 3 x i. e. k = 10 © T Madas

The line with equation 6 x – 5 y = 4 is drawn in the grid below (c) Shade on the grid the region whose coordinates satisfy the inequalities: y > 0, x > 0, 6 x – 5 y < 4 and x + y < 3 (d) What point inside the region has integer coordinates? (1, 1) 6 x – 5 y < 4 Try the point (0, 0) y 4 -2 -1 3 0<4 So (0, 0) belongs to the region 2 x +y < 3 1 Try the point (0, 0) 0 -1 -2 1 2 3 x 0<3 So (0, 0) belongs to the region Also y > 0 and x > 0 © T Madas

Two points P and Q are marked on the grid below. 1. Mark on the grid as point M, the midpoint of PQ. 2. Work out the gradient of PQ. 3. Construct the perpendicular bisector of PQ by considering gradients of lines. y 12 The gradient of PQ is 11 2 4 = 3 6 10 9 8 We can also use the formal definition of the gradient: P (2, 7) 7 6 5 dif in y dif in x 3 – 7 = 8 – 2 gradient = M 4 4 3 Q (8, 3) 6 2 1 0 x 1 2 3 4 5 -4 6 2 = 3 = 6 7 8 9 10 11 12 © T Madas

Two points P and Q are marked on the grid below. 1. Mark on the grid as point M, the midpoint of PQ. 2. Work out the gradient of PQ. 3. Construct the perpendicular bisector of PQ by considering gradients of lines. y 12 The gradient of PQ is 11 the gradient of a line which is 3 perpendicular to PQ is 2 10 9 8 P (2, 7) 7 2 4 = 3 6 M 5 2 4 Q (8, 3) 3 2 1 0 x 1 2 3 4 5 6 7 8 9 10 11 12 © T Madas