straight lines ppt

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Presentation Transcript

Slide 3:

Pre-requisite Knowledge Distance Formula Slope Straight Line Drawing

Slide 4:

Review:Distance and Slope Equation of Straight Lines Points of Division Perpendicular and Parallel Lines Intersection of Two Straight Lines Contents

Slide 5:

Equation of Special Lines Two Point Form Point-Slope Form Slope-Intercept Form Intercept Form General Form Back =>

Slide 6:

Topic: Distance and Slope (Review) Back=>

Slide 7:

y1 – y2 distance ? By Pythagoras Theorem, x1 – x2

Slide 9:

(a) Find AB if A=(4,0) and B=(9,a) (Give the answer in terms of a.)

Slide 10:

x1 – x2 y1 – y2 slope ?

Slide 13:

If a line//x-axis slope = 0

Slide 14:

If a line // y-axis slope is undefined

Slide 15:

Back=> End of Topic

Slide 16:

Topic: Point of Division Back=>

Slide 18:

∵ ΔBCD ～ΔCAE

Slide 19:

C(x,y) 4 3 x = 3 x 8 + 4 x 1 3 + 4 y = 3 x 9 + 4 x 2 3 + 4

Slide 20:

What are the coordinates of P ? Ans: P = (2, 4)

Slide 21:

Find the values of a and b

Slide 22:

Solution

Slide 23:

Slide 24:

Find the coordinates of point P

Slide 25:

Mid-Point Formula P is the mid-point of AB

Slide 26:

B (5, -2) (4, 1) Let P = (a, b) & G = (p, q)

Slide 27:

Given : G is the centroid of △ABC

Slide 30:

Let AP : PB = 1 : k Solution

Slide 31:

Topic: Equations of Special Lines Back=>

Slide 32:

(1, 3) (-1, -3) x = -3 y = 3 y = -3 (2,1) y = 1

Slide 33:

(2, 2) (2, 0) (-3, -3) x = -3 x = 2 x = -1 x = -3

Slide 34:

(a, b) L2 L1 P Ans: L1 : x = a L2 : y = b P= (0, b) Find The equations ofL1 and L2; The coordinates of point P.

Slide 35:

y = x y =-2x

Slide 37:

Find the equations of L1 and L2.

Slide 39:

Topic: Two-Point Form Back=>

Slide 40:

Find the equation of L. MAP = MAB

Slide 41:

MBP = MAB

Slide 42:

(a) Find the equation of L. (b) Find the value of b. (c) Find the coordinates of P. L: 7x + 6y + 4 = 0

Slide 43:

(a) Find the equation of the straight line joining (-3, 2) and (2, -1). (b) Does the point (7, -4) lie on the straight line ? (c) State whether the point (3, -2) lies on the straight line or not. L: 3x - 5y + 1 = 0

Slide 44:

(a) Find the equation of the straight line which passes through (0,0) and (-4,-6). (b) If the point A(a,3) lies on L, find a.

Slide 45:

Back=> End of Topic

Slide 46:

Topic: Point-Slope Form Back=>

Slide 47:

Point-slope Form MAB = Slope

Slide 48:

Find the equation of the line which passes through (-1,-5) and has slope -3 : Solution

Slide 49:

(a) Find the equation of L. (b) What is the value of b ? Put B(2, b) into the equation L: x + 3y - 3 = 0

Slide 50:

Find (a) The equation of L. (b) The coordinates of P (c) The coordinates of Q

Slide 51:

Solution.

Slide 53:

Topic: Slope-Intercept Form

Slide 54:

L1 cuts the y-axis at point (0,3) L1 cuts the x-axis at point (-2,0)

Slide 55:

What is the equation of L ?

Slide 56:

(a) Find the equation of the straight line with y-intercept –1 and slope –3 in the slope-intercept form. y=3x1 Slope-intercept Form

Slide 57:

L : kx + 3y – 2k = 0 with slope –2. (a) Find the value of k .

Slide 58:

Ans.

Slide 60:

Topic: Intercept Form Back=>

Slide 61:

MAP = MAB What is the equation of L ?

Slide 62:

Find the equation of L in intercept form. Do the point (4, 6) and (12, 9) lie on L ?

Slide 63:

(a) Convert 7x + 4y + 28 = 0 into the intercept form. (b) What are the x-intercept and y-intercept of the straight line ? x-intercept = -4 and y-intercept = -7

Slide 64:

Find the area of the shaded region. The area of the shaded region is Intercept form

Slide 66:

Solution.

Slide 68:

Topic: General Form Back=>

Slide 69:

Ax + By + C = 0

Slide 71:

What are the slope and the y-intercept of the straight line 4x – 3y + 7 = 0 ?

Slide 72:

Find the equation of L in the general form.

Slide 73:

Find the x-intercept and the y-intercept of the straight line 12x – 7y + 4 = 0.

Slide 75:

Topic: Parallel Lines and Perpendicular Lines Back=>

Slide 76:

If L1 // L2 , then mL1 = mL2 What will happen if Two lines L1 and L2 Are parallel? A FACT to know... Conversely, if mL1 = mL2 Then L1 // L2

Slide 77:

Determine whether L1 // L2 Since m1 = m2= 2, then, L1 is parallel to L2

Slide 78:

Find the equation of L2 mL2 = mL1 = 2

Slide 79:

(a) Find the equation of L2. (b) Does the point (-3, -5) lies on L2 ? L.H.S. = = 3(-3) + (-5) + 15 = 1  R.H.S. Thus, (-3, -5) does not lie on L2

Slide 80:

Find the equation of L2. Step 1: Express L1 into slope intercept form. Step 2: Find the slope of L2 Step 3: Use point-slope form to find L2.

Slide 81:

Steps : 1. Express the given line into slope-intercept form. 2. Find the slope of L1. 3. Use point-slope form to find the equation of the line. Find the equation of the line L1 which is parallel to 3x + 2y – 5 = 0 and passes through (4, -1).

Slide 82:

Solution.

Slide 83:

If L1 ⊥ L2 , then mL1 x mL2 =-1 One more FACT... Conversely, if mL1 x mL2 =-1 Then L1 ⊥ L2

Slide 84:

Find the coordinates of P.(Hint: Let P = (a,0) thus, P = (-0.5, 0) ∵ L1 ⊥ L2 ∴ mL1 x mL2 =-1

Slide 85:

Find the equation of L2. Step 1: Express L1 into slope intercept form. Step 2: Find the slope of L2 Step 3: Use point-slope form to find L2.

Slide 86:

Steps : 1. Express the given line into slope-intercept form. 2. Find the slope of L. 3. Use point-slope form to find the equation of the line. Find the equation of the line L which is perpendicular to 3x - 2y + 6 = 0 and pases through (-4, 3).

Slide 87:

Solution.

Slide 90:

Find the equation of the perpendicular bisector of the line segment joining (3, -5) and (-7, 9). [ Ans.: 5x - 7y + 24 = 0 ] Steps : 1. Find the coordinates of the midpoint.2. Find the slope of the line segment. 3. Find the slope of the perpendicular bisector4. Use point-slope form to find the equation of the line.

Slide 91:

Topic: Point of Intersection

Slide 92:

What are the coordinates of P ? A. P = (-5, -7) B. P = (-5, 7) C. P = (5, -7) D. P = (5, 7) E. P = (7, 5)

Slide 95:

What are the coordinates of P ? A. P = (-5, 7) B. P = (5, 7) C. P = (7, 2) D. P = (7, 13) E. P = (13, 7)

straight lines ppt

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