A2. Lines and circles 1. Straight lines 2. Midpoint and distance between two points 3. Circles 4. Intersections of circles and lines 5. Trigonometry 6. Parametric equations of lines and circles

1. Straight lines Slope or Gradient: e.g. The slope of the line through the points A(2,-3) and B(5,6) is m = (6 + 3)/(5 – 2) = 3 (x 1,y 1 ) (x 2,y 2 ) x y run rise

Definition: A linear equation has the form y = m x + c. Its graph is a straight line such that m is the slope of the line. c is the vertical intercept (where the graph crosses the y-axis or the value of y when x = 0). x y m > 0 c x y c m < 0

e.g. y + 3x = 7 y - 3x + 2 = 0 y = -3x + 7 y = 3x – 2 y = m x x y m = -3 < 0 7 x y -2-2 m = 3 > 0 x y m = 1/2 m = 1 m = 2 m = -1 m = -2 m = -1/2

 An equation of the line having a slope m and passing through the point A(x 1,y 1 ) is y – y 1 = m(x – x 1 ) e.g. If m = 2 and A(-1,3), then the equation is y – 3 = 2(x + 1), in the normal form y = 2x + 5  An equation of the line that passes through the two points A(x 1,y 1 ) and B(x 2,y 2 ) can be obtained as follows: e.g. If A(-1,3) and B(2,9), then m = (9 - 3)/(2 + 1) = 2, y – 3 = 2(x + 1) or y = 2x + 5

 A horizontal line has the equation y = c (Here the slope m = 0)  A vertical line has the equation x = b (Here the slope is unbounded, ∞) x y y = 0 y = 4 y = x y x = 4 4 x = 0 -3 x = -3

Example: Find equations for the following straight lines: (i)Horizontal and through the point (-9,5). (ii)Vertical and through the point (-9,5). (iii)Through the points (1,2) and (3,-1). (iv)Vertical intercept is 3 and horizontal intercept is 2. (i) y = 5. (ii) x = -9. (iii) The slope is m = (-1 - 2)/(3 - 1) = -3/2, therefore, y - 2 = -3/2(x – 1), thus y = -1.5x (iv) An equation of the line through the points (0,3) and (2,0) can be obtained as follows: m = (0 - 3)/(2 - 0) = -3/2, y = -1.5x + 3 x y Vertical intercept y-intercept Horizontal intercept x-intercept 3 2

 Parallel and perpendicular lines Product of slopes = -1 Parallel lines have the same slope e.g. The lines y = 3x + 4 and y = 3x - 5 are parallel and the lines y = -3x + 5 and y = (1/3)x – 8 are perpendicular. x y y = 2x y = 2x +3 y = 2x -1 y x Slope = m Slope = -1/m

Example: Find equations for the following straight lines: (i)Through the point (-2,2) and parallel to the line 3x+2y=5. (ii)Through the point (1,2) and perpendicular to the line 2x+4y=1. (iii)Through the point A(1,-2) and perpendicular to the line from A to B(2,2). (i) y = (-3/2)x + 5/3, the slope of the parallel line is m = -3/2, y – 2 = (-3/2)(x + 2) or y = -1.5x – 1. (ii) y = (-1/2)x + (1/4), thus the slope of the perpendicular line is m = 2, y – 2 = 2(x - 1) or y = 2x. (iii) The slope of the line AB is (2 + 2)/(2 - 1) = 4, thus the slope of the perpendicular line is m = -1/4, therefore, y + 2 = -1/4(x – 1) or y = -0.25x – 1.75.

 Intersection of two lines Example: Find the point at which the line y = 2x + 4 meets the line y = -3x – 1. So the point of intersection of the two lines is (-1,2).

2. Midpoint and distance between two points The midpoint M of the line segment between A and B and the distance AB between A and B are (x1,y1)(x1,y1) (x2,y2)(x2,y2) A B M

Example: Let A(1,-2) and B(4,2) be two points in the plane. Find: (i)The distance between A and B. (ii) The midpoint of the line AB.

3. Circles The circle with center C(a,b) and radius r has equation (x – a) 2 + (y – b) 2 = r 2 e.g. x 2 + y 2 = 9 has center (0,0) and radius 3. (x + 3) 2 + (y – 2) 2 = 4 has center (-3,2) and radius 2. The circle with center (2,0) and radius 5 has the equation (x – 2) 2 + y 2 = 25. r C(a,b) y x

Example: Show that the following equation represents a circle and find its center and radius x 2 + 6x + y 2 - 8y + 20 = 0 By completing the square, we have (x + 3) 2 – 9 + (y – 4) 2 – = 0 (x + 3) 2 + (y – 4) 2 = 5 The center is (-3,4) and the radius is. Note: Completing the square:

Example: Find an equation of the circle that passes through the point P(2,3) and has center C(-1,2). An equation of the circle is (x + 1) 2 + (y – 2) 2 = r 2 The point (2,3) satisfy the circle equation, therefore (2 + 1) 2 + (3 – 2) 2 = r 2

Example: Find an equation of the circle that passes through the points A(-3,2), B(-1,4) and C(1,2). Let M and N be the midpoints of the line segments AB and BC, respectively. The slopes of the line segments AB and BC are 1 and -1, respectively, therefore, the perpendicular bisectors MD and ND are y – 3 = -1(x + 2), y = -x + 1 and y – 3 = 1(x – 0), y = x + 3. Two perpendicular bisectors MD and ND meet at (-1,2) which is the center of the circle. The radius of the circle is the distance between D and any of A, B or C. x y A B C M N D

4. Intersections of circles and lines Leads to Leads to Leads to quadratic equation quadratic equation quadratic equation with two solution with one solution with no solution

Example: Find the points, if any, at which the circle (x + 7) 2 + (y – 2) 2 = 80 intersects the line y – 2x + 4 = 0. The point of intersection is (1,-2).

Note: c1c1 c2c2 L = c 1 – c 2

Note: C C C

5. Trigonometry x y θ P(x,y)P(x,y) θ x = cos θ y = sin θ opposite adjacent hypotenuse 1

 Some important identities:  Some important sine and cosine

All + sin + cos - sin - cos - sin - cos + e.g.

6. Parametric equations of lines and circles  Parametric equations of the line L passing through the point P(x 1,y 1 ) and having slope m are x = t + x 1, y = mt + y 1 e.g. A line with slope m = 4 and through the point P(-1,3) has parametric equations x = t - 1, y = 4t + 3 Example: Find parametric equations for the line which passes through the two points A(1,-5) and B(4,7).

 The circle with center (a,b) and radius r has parametric equations x = a + r cosθ, y = b + r sinθ e.g. Center (-1,3) and radius 4 x = cosθ, y = sinθ A3