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Find the equation of a straight line given two points. (Part c is incomplete)

Mathematics

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4 parts
a

y=−x+3y = -x + 3

b

y=12x+132y = \frac{1}{2}x + \frac{13}{2}

c

y=12x+52y = \frac{1}{2}x + \frac{5}{2}

d

y=12x+4y = \frac{1}{2}x + 4

Step-by-step solution

12 steps

To find the equation of a straight line, we first calculate the gradient (mm) using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. Then, we use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) to find the equation.

Part a· 3 steps

Points: $(3, 0)$ and $(1, 2)$

  1. Calculate the gradient.

    Let (x1,y1)=(3,0)(x_1, y_1) = (3, 0) and (x2,y2)=(1,2)(x_2, y_2) = (1, 2).

    m=2−01−3=2−2=−1m = \frac{2 - 0}{1 - 3} = \frac{2}{-2} = -1

  2. Use the point-slope form with one of the points, e.g., (3,0)(3, 0), and the gradient m=−1m = -1.

    y−0=−1(x−3)y - 0 = -1(x - 3)

  3. Simplify to the slope-intercept form y=mx+cy = mx + c.

    y=−x+3y = -x + 3

    The equation of the line is y=−x+3\boxed{y = -x + 3}.

Part b

y=12x+132y = \frac{1}{2}x + \frac{13}{2}

Points: $(1, 7)$ and $(5, 9)$

  1. Calculate the gradient.

    Let (x1,y1)=(1,7)(x_1, y_1) = (1, 7) and (x2,y2)=(5,9)(x_2, y_2) = (5, 9).

    m=9−75−1=24=12m = \frac{9 - 7}{5 - 1} = \frac{2}{4} = \frac{1}{2}

  2. Use the point-slope form with one of the points, e.g., (1,7)(1, 7), and the gradient m=12m = \frac{1}{2}.

    y−7=12(x−1)y - 7 = \frac{1}{2}(x - 1)

  3. Simplify to the slope-intercept form y=mx+cy = mx + c.

    y−7=12x−12y - 7 = \frac{1}{2}x - \frac{1}{2} y=12x−12+7y = \frac{1}{2}x - \frac{1}{2} + 7 y=12x+132y = \frac{1}{2}x + \frac{13}{2}

    The equation of the line is y=12x+132\boxed{y = \frac{1}{2}x + \frac{13}{2}}.

Part c

y=12x+52y = \frac{1}{2}x + \frac{5}{2}

Points: (−1,2)(-1, 2) and (3,4)(3, 4)

  1. Calculate the gradient.

    Let (x1,y1)=(−1,2)(x_1, y_1) = (-1, 2) and (x2,y2)=(3,4)(x_2, y_2) = (3, 4).

    m=4−23−(−1)=23+1=24=12m = \frac{4 - 2}{3 - (-1)} = \frac{2}{3 + 1} = \frac{2}{4} = \frac{1}{2}

  2. Use the point-slope form with one of the points, e.g., (−1,2)(-1, 2), and the gradient m=12m = \frac{1}{2}.

    y−2=12(x−(−1))y - 2 = \frac{1}{2}(x - (-1)) y−2=12(x+1)y - 2 = \frac{1}{2}(x + 1)

  3. Simplify to the slope-intercept form y=mx+cy = mx + c.

    y−2=12x+12y - 2 = \frac{1}{2}x + \frac{1}{2} y=12x+12+2y = \frac{1}{2}x + \frac{1}{2} + 2 y=12x+52y = \frac{1}{2}x + \frac{5}{2}

    The equation of the line is y=12x+52\boxed{y = \frac{1}{2}x + \frac{5}{2}}.

Part d

y=12x+4y = \frac{1}{2}x + 4

Points: $(4, 6)$ and $(10, 9)$

  1. Calculate the gradient.

    Let (x1,y1)=(4,6)(x_1, y_1) = (4, 6) and (x2,y2)=(10,9)(x_2, y_2) = (10, 9).

    m=9−610−4=36=12m = \frac{9 - 6}{10 - 4} = \frac{3}{6} = \frac{1}{2}

  2. Use the point-slope form with one of the points, e.g., (4,6)(4, 6), and the gradient m=12m = \frac{1}{2}.

    y−6=12(x−4)y - 6 = \frac{1}{2}(x - 4)

  3. Simplify to the slope-intercept form y=mx+cy = mx + c.

    y−6=12x−42y - 6 = \frac{1}{2}x - \frac{4}{2} y−6=12x−2y - 6 = \frac{1}{2}x - 2 y=12x−2+6y = \frac{1}{2}x - 2 + 6 y=12x+4y = \frac{1}{2}x + 4

    The equation of the line is y=12x+4\boxed{y = \frac{1}{2}x + 4}.

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