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Equation of a straight line and graphs GCSE
27/06/2022
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Equation of a straight line 1. Equation of a straight line General form: y = mx + ₁ c ↓ gradient e.g. Example 2y + 4 = 8x 2y = 8x-4 Y = 4x-2 50 Gradient The gradient can be however they eve all = y ² = 2x + 3 m 5²-6 7-3 = 4 5-2 3 gradient = vertical change. horizontal change. - gradient Excemples i (2, 3) (5, 7) =35 3 Lise TUN Ly intercept gradient: 4 y- intercept: - 2 calculated in 3 ways, the Same: 2. (-1,-3) (5,8) 8-(-3) = 5-(-1) 6 6 S 71 3. A is the point with coordinates (2, 10) B is the point with coordinates (5, d) The gradient of line AB is 4 Work out the value of d. 10-d 2-5 10-d =4 3 10-α = - 2 -α = -22 -Finding the equation of a straight line a) gradient 6 which passes through point (2,7) d = 22✓ y - 7 = 6(x-2) y = 7 = 6x-12 y Y-2 Formula: Y-y₁ = m ( x-x₁) H is) gradient - 4 which passes through point (-7, 2) y = 2 = 4(x--7) y-2 = 4 11 it = L 6x- + -4(X+7) -4x-28 4x T 26 Equation of a straight line c) gradient 3 4 Y--1 Y +1 Y +1 Steps H 1. find # 11 MJ MJ 3 m 4 which passes through point (5,-1) (x - 5) 3 (x+5) y = 3x - 19 Y 4 4 2x-15 4 19✓ to follow: → given m=y²-y₁ x2-x1 2. y - y¹= m(x-x₁) 3. rearrange r 3 - Finding the equation of the line from the graph- Steps: 1. Pick two points. 2. m³ Y-YL x₂-x 1. 2. L -6-5-4-3-2-1, 6 5 JO 5 30 y 4. IN -3 -S m = 2 C = 3 (1,5) (-2, -1) 5--1 Ḥ 1--2 -3-2- = 2x + 3/ fo 250 ✓ h 3 2 1. - 0 * * 4 23 456 -5. 22 -3 -6 1 3 5+L 1+2 56 3. c = મું. ′ = mx + C 62 3 (-4,-1) (0, -2) M = Y -1--2 -4 m = -1 4 C=-2 -1 or rise run 4 x =1+2 = 4 -4-0 -2✓ ■ 4...
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Alternative transcript:
Equation of a straight line - Finding the equation of the line from points a) (2, 2) and (6, (8) m=2-18 2-6 L Y - Y₁ = m ( x-x₁) y = 2 = 4(x-2) y-2 Y = 4 Y Y Y 11 b) (-2, 2) (1, - 7) m = 27 2-1 Y 1 4x -2 = 4x - 8 Y - Y₁ = m ( x-x₁) -2=-3(x-(-2)) = -3(x + 2) -30-6 2 = 11 P 3 = 3x ( T - 16 2+7 -2-1 4✓ : = 4 4 9=3 -3 T B c) (3, 1) and (15,9). Give your answer in the form ax + by + c = 0 where a, b and I are integers. M = 13 3y y - y₁ = m(x-x₁) Y-1= 3² (x-3) V مای 3y - 3 = 2 ( x − 3) 3y - 3 = 2x-6 3y-3-2x 3808 +8=2 3 2x +3=0✓ m = 5-7 -8--4 Y - Y₁ Y-5 - d) (-8,5) and (4,7). Give your answer in the form ax + by + C =0 where a b and I are integers. () 5-7-2 -8 +4 = m ( x-x₁) H 1(x--8) 2 2y-10 = x +8 2y - 10 - x = 1 8 -4 2y - 18 - x = 0 - x + 2y - 18 -0 ✓ 2 1. Equation of a straight line - Review Exam-style questions & -2 - O 9 8 7 6 5 4 3 2 1 O 4 -2 -3 -4 Line B Line C o Line D 4 Y Y 2. The equations of fow lines we given below: Line A = 4x + 1 (2,9) x = 2 y = 9 y = 2 R Lexi says the line below has an equation of y=- 2x + 8. Explain we mistake. H (1,4) (20) Y = 11 1 = 4(2)+( 8 + 1 9✓ y + 2x 8 9 +2(2) 9 + 4 = 13 X E : 5 x 1-2 9- 200 9 - 2(2) १-4 3x - 3 9-3(2) = 3 9-6 F3 3 = 3√ = 4 = - 4 -1 4x+8/ which lines go through the point (2, a) ? Line A and Line D. / 7 D 17 3. 508 15 12 (1 10 9 6 7 5 4 wa 3 297 -1 Y 1 > -12 y - y₁ = m(x-x.) y - (8) +8 m = = y + 8 = 2 19 3 -/(x-(-6)) - 12/²2 (x + 6) 3 1-1/2 x X- = -212x 11✓ 5 y - 5 = 5(x - - 3) - 5 = 5(x + 3) Y y = 5= 5x + 15 y = 5x + 20 a) Find the equation of the line. ✓ 4. A line was a gradient of -1/2 and passes though the (-6-8). Find the equation of the line. (-6,-8) (0.16) (4,0) 16-0 0-4 y = -40℃ + 16 ✓ 3 = Xxx =mx+c Y Y Y -15-5 = 20 = 20 = 5 1--3 1+3 4 G 1) Give the y coordinate of the point on the line with an x-coordinate of 8. 32 = -16✓ 16 -4 4 (8) + 16 +16 The point A 1-3, 5) and the point B 11,-15) lie on the line I. Find the equation of the line L. (-3,5) (1, -15) o 4 1 L Equation of a straight line- Review is a 5. Do the points (1,4) (4,10) and (9,20) cie. straight line (1,4) (4, 10) 4-10 = − 6 = 2 -3 Y-y₁ = m(x-x₁) y = 4 = 2 (5c-x) y-4=2x-2 Y = 200 + d ( 9, 20) x=9 Y = 20 = 2x + 2 Y = 2(9) + 2 Y = 18 + 2 = 20 Y .. On straight line / 9 17