Daily Practice 5.6.2015 Q1. Simplify Q2. Evaluate L.I: Today we will be revising over our knowledge of the straight line. Q3. Write in completed square form x 2 + 4x + 7 Q4. State the equation of the line joining (0, 3) and (4, -6) S.C: We will be able to find the gradient from a line given its equation. We will also be able to find the equation of a line given the graph or 2 points. Q5. Factorise 6x 2 + 17x + 12 Gradient Given two points (x 1, y 1 ) and (x 2. y 2 ) m = y 2 - y 1 x 2 - x 1 Calculate the gradient of the lines joining the following points: (a) A(-2, 1) and B(3, -5) (c) E(-3, -4) and F(0, 4) (b) C(2, -7) and D(3, 1) (d) G(2, 7) and H(5, -3) Gradient Calculating the gradient given the equation of a line: Rearrange the line so that it is in the form y = mx + c and then state the gradient State the gradients of the following lines: (a) y = -x + 3 (e) x / 2 + 3y = 5 Example: State the gradient of the line 4x - 2y + 6 = 0 (b) 2x + 4y = 6 (c) 8y - 10 = 2x (d) 3x + 4y - 7 = 0 (f) 2x + y = 4 3 (g) 1 / 3 y - x = -1 Q2. Find the equation of the lines joining the following pairs of points: State the gradients of the following lines: (a) y = -x + 3 (e) x / 2 + 3y = 5 (b) 2x + 4y = 6 (f) 2x + y = 4 3 (c) 8y - 10 = 2x Write each answer in the form Ax + By + C = 0 (a) (-1, 1) and (2, 4) (b) (0, 5) and (-1, 7) (c) (0, 0) and (-1, -3) (d) (4, 2) and (-3, -8) (d) 3x + 4y - 7 = 0 (g) 1 / 3 y - x = -1
State the gradients of the following lines: (a) y = -x + 3 (e) x / 2 + 3y = 5 (b) 2x + 4y = 6 Q2. Find the equation of the lines joining the following pairs of points: Write each answer in the form Ax + By + C = 0 (a) (-1, 1) and (2, 4) (b) (0, 5) and (-1, 7) (c) (0, 0) and (-1, -3) (d) (4, 2) and (-3, -8) (c) 8y - 10 = 2x (f) 2x + y = 4 3 (g) 1 / 3 y - x = -1 (d) 3x + 4y - 7 = 0 Daily Practice 8.6.2015 Q1. State the gradient of the line joining (3, 2) and (5, 4) Q2. Write x 2 8x 2 in completed square form Q3. Solve 1 7cosx 0 = 3 L.I: Today we will be learning how to calculate the gradient using tan. Q4. State the roots of the function y = x 2 + 7x + 12 Q5. Simplify 30x 3 15x 12 Gradient Remember: Gradient Given a right - angled triangle on a coordinate grid The gradient of a vertical line is The gradient of a horizontal line is θ hyp θ adj opp tan θ = opp adj m = v = opp h adj = tan θ The gradient = The tan of the angle that the line makes with the positive direction of the x - axis (m = tan θ )
Gradient using Tan Example: Find the angle that the line joining the pair of points (-4, -2) and (2, 5) makes with the x - axis Gradient Example 2: Find the acute angle that the line y = -3x + 1 makes with the x - axis Perpendicular Gradient a a a b b b 90 0 Ex. 1B Q6 & 7 Daily Practice 9.6.15 Q1. Find the equation of the line perpendicular to y = 3x + 2 that passes through the point (4, 5) Q2. State the gradient of the line 3y + 5x 4 = 0 Q3. State the angle that the line y = 1.5x + 2 makes with the positive direction of the x axis L.I: Today we will be continuing to learn about perpendicular gradient. Q4. Multiply out and simplify (3x 1) 2 + 2
Perpendicular Gradient If two lines are perpendicular, then the product of their gradients is -1. m 1 x m 2 = -1 To find the gradient of the line perpendicular to m 1 just turn the gradient upside down and change the sign. Perpendicular Gradient Examples: 1. Calculate the gradient of the line perpendicular to the line joining A(-1, 4) and B(5, -3) Perpendicular Gradient Examples: 2. State the equation of the line perpendicular to 7x + y + 3 = 0 that passes through (0, 5) Perpendicular Gradient Examples: 3. Given the points B(5, 3), D(3, 7) & C(-1, 5), prove that the line BD is perpendicular to the line CD. Ex. 1D Q1-7 Homework Online due 16.6.2015 (Gradient) Gradient Summary Given two points (x 1, y 1 ) and (x 2. y 2 ) m = y 2 - y 1 x 2 - x 1 The product of perpendicular gradients is -1 Gradient Vertical lines have undefined gradient The gradient is equal to the tan of the angle the line makes with the x - axis (m = tan θ) Parallel lines have equal gradients Horizontal lines have gradient = 0 Daily Practice 10.6.15 Q1. State the gradient of the line 2x + 5y = 3 Q2. State the equation of the line that passes through (0, 8) and is perpendicular to the line y = -x + 2 Q3. Factorise 12x 2 + 11x + 2 Q4. Simplify 3m 2 (2m -4 + m 6 )
Collinearity Points are said to be collinear if they lie on the same straight line. To prove that points are collinear, show that they have the same gradient. C L.I: Today we will be learning about Collinearity and Distance Between Two Points. A B A, B and C are collinear if the gradients of AB and BC are equal and B is common to both AB and BC. Collinearity Examples: Ex. 1B Q1-3 1. A(-2, -5) B(2, -1) and C(8, 5). State whether these points are collinear Distance between two points To find the distance between two points. B(x 2, y 2 ) A (x 1, y 1 ) 2.(i) Calculate the gradient of P(-5, 1) and Q(3, 7) (ii) Which point Z(-2, 5) or S(-1, 4) lies on the line PQ? Distance between two points To find the distance between two points. Example: A Calculate the distance between A(-2, 3) and B(2, 1) B Distance between two points Questions: Find the distance between the following i) (-5, -4) and (2, 1) ii) (0, -3) and (1, 4) iv) (-7, 8) and (2, -6) iii) (0, -2) and (5, 0)
The equation of a line The equation of a line can be written in many forms. The equation of a line Examples: The general equation : Ax + By + C = 0 y = mx + c and y - b = m(x - a) 1. State the equation of a line that passes through (0, -3) and (4, 0) Use when you know the gradient and y - intercept Use when you know the gradient and one point on the line or 2 points on the line 2. State the equation of the line joining (-1, 3) and (2, -5) The equation of a line Showing that a point is on a line... Substitute the point into the line. Example: 1. Show that the point (3, 2) is on the line y = 3x - 7 2. Which of the following points lie on the line with equation 4y - 3x + 16 = 0? a) (8, 2) b) (-4, -9) Please complete questions: Ex. 1E Q8. Ex. 1F Q1. Ex. 1G Q2. (c) (d), Q3 Read through and complete questions at the end of next slides when finished. The equation of horizontal and vertical lines The equation of horizontal and vertical lines (-7, 4) (-2, 4) (3, 4) (9, 4) Look at all the points on this Horizontal Line. What can you say about the x or y coordinates of any point on the line? (4, 8) (4, 4) (4, -2) (4, -6) Look at all the points on this Vertical Line. What can you say about the x or y coordinates of any point on the line?
The equation of horizontal and vertical lines The equation of a vertical line is of the form x = a The equation of a horizontal line is of the form y = b x = a Vertical lines have an undefined gradient Horizontal lines have a gradient of zero y = b The equation of horizontal and vertical lines Find the equation of the following lines: 1) Parallel to the x axis and passing through the point (3, 2) 2) Perpendicular to the x axis and passing through the point (7, 5) 3) Passing through the points ( 3, 8) and ( 3, 5) 4) Passing through the points ( 7, 2) and (4, 2) 5) Passing through (3, 4) and perpendicular to the line passing through the points (1, 3) and ( 6, 3). Daily Practice 12.6.2015 Q1. State the equation of the line that passes through (-1, 3) and is parallel to the line y = 5x - 1 Q2. State the equation of the line that passes through (0, 7) and is perpendicular to the line y = -1x + 2 Q3. Write in completed square form x 2 + 12x - 2 The Mid-point The mid - point is half-way between two given points on a line L.I: Today we will be learning about midpoint and perpendicular bisectors. A (0, 3) B (2, 5) To find it you are finding half-way between the x - coordinates and halfway between the y - coordinates. Midpoint =
The Mid-point Examples: Find the mid-point between the following pairs of coordinates a) A(1, 3) & B(7, 15) b) C(-4, 9) & D(6, -13) The Mid-point Questions: Find the mid-point between each pair of points: 1) A(2, 3) & B(8, 11) 2) C(1, -8) & D(-9, -12) 3) E(-3, 12) & F(5, - 2) Perpendicular Bisectors A perpendicular bisector is a line that bisects another line (cuts in half) at right angles. (Remember in a circle - the radius is a perpendicular bisector of a chord) CD is the perpendicular bisector of the line AB A To find the equation of the perpendicular bisector of a line: 1. Find the midpoint of the line. 2. Find the perpendicular gradient. C D B Perpendicular Bisectors Example: Find the equation of the perpendicular bisector of the line joining C(-8, 5) and D(2, -3) Perpendicular Bisectors Some of the shapes that you can find perpendicular bisectors in: Circle Rhombus Kite Diagonals bisect at right angles. Questions: Find the equations of the perpendicular bisectors of the following 1) A(2, 8) and B(4, 6) 2) Q(0, 3) and D(5, 4) 3) L(-4, 6) and M(-2, -2) 4) S(5, -7) and T(6, 4)
Perpendicular Bisectors Daily Practice 15.6.2015 The perpendicular bisectors of the sides of a triangle are concurrent (intersect at a single point) and this point of intersection is called the circumcentre of the triangle. The circumcentre is the centre of the circle which surrounds the triangle. The three sides of the triangle are chords in the circle. Q1. State the equation of the line that passes through (-1, 2) and is perpendicular to the line y = -x + 2 Q2. State the gradient of the line y - 3x + 2 = 0 Q3. Simplify 3m(2m 2 - m 1/2 ) Q4. State the angle that the line y = 0.5x - 3 makes with the positive direction of the x-axis Altitudes An altitude of a triangle is a line drawn from a vertex perpendicular to the opposite side. Every triangle has 3 altitudes. L.I: Today we will be learning how to calculate the equation of an altitude in a triangle. Homework Due Tomorrow! To find the equation of the altitude... 1. Work out the gradient of the line opposite the vertex and state the perpendicular gradient. 2. Use the coordinate of the vertex to state the equation. D Altitude Example: The vertices of a triangle are P(5, 2) Q(-3, 0) and R(3, -4). Find the equation of the altitude from R. Q P Altitude The 3 altitudes of a triangle are concurrent and the point of intersection is called the orthocentre (this can also be located outside the triangle). D Ex. 1K Q1. b, d, e R D Q2, 3, 4 Page 15 D orthocentre
Daily Practice 16.6.2015 Q1. Calculate the current value of a house that was worth 162 000 and depreciated by 5.4% in its first year and appreciated by 3.7% in its second Q2. L.I: Today we will be learning how to state the equation of a median in a triangle. Homework Due! Q3. Calculate the distance between the points A(-3, 5) and (2, -1) Median A median of a triangle is a line drawn from a vertex to the mid-point of the opposite side. Median Line CD is a median from C M (mid-point of AC) M (mid-point of BC) To find the equation of the median.. 1. Work out the midpoint of the opposite side. D mid-point of AB 2. Use the midpoint and the coordinate of the vertex to find the gradient and hence the equation. The medians of a triangle are concurrent M mid-point of AB and the point at which they meet is called the centroid. Median Example: In the triangle ABC, find the equation of the median from B for a triangle with vertices A(-4, 1), B(-2, 4) and C(4, 1) Daily Practice 17.6.2015 Q1. State the equation of the line that passes through ( 2, 4) and (4, 5). Express your answer in the form Ax + By + C = 0 Q2. Write x 2 + 16x 2 in completed square form Q3. Simplify the expression Q4. Pg. 20 Q10,5
Point of Intersection L.I: Today we will be learning how to find the point of intersection of two lines. We will also be practising some exam style questions on the Straight Line. The point of intersection of two or more lines can be found by using simultaneous equations. (The lines may need to be rearranged first) Examples: Find the point where the lines 2x - 3y - 13 = 0 and 3x + 5y + 9 = 0 meet Exam Questions on The Straight Line