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Saturday, December 28, 2013

Given (-5, 15) and (-5, 9), find the midpoint, distance, slope, and equation of the line.

0 comments Posted by Anonymous at 10:55 PM
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Problem: Given these pairs of points, (-5, 15) and (-5, 9), find the midpoint, distance, slope, and equation of the line.

(-5,15),(-5,9)\,

Solutions:

  • To find the midpoint, average the x coordinates and y coordinates. The midpoint is

\left(\frac{-5-5}{2},\frac{15+9}{2}\right) = \left(-5,\frac{15+9}{2}\right) = \left(-5,\frac{15+9}{2}\right)\,

  • To find the (always zero or positive) distance, use the formula

d = +\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\,

d = \sqrt{(-5+5)^2+(15-9)^2} = \sqrt{0^2+6^2} = \sqrt{6^2} = 6\,

  • To find the slope, use the formula

 m = \frac{y_2-y_1}{x_2-x_1}\,

m = \frac{9-15}{-5-(-5)} = undefined\, (which means it is a vertical line, which has infinite slope)

  • Since the line is vertical, there is only one x value that will give all y values. The equation for the line is

x = -5\,

    List of Similar Problems with Complete Solutions

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    For any Questions, Clarifications and any violent Reactions use the comment section below. I'll be much willing to assist you.


    credit: Todd©2013 www.FroydWess.com

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    Given (4, 13) and (0, 9), find the midpoint, distance, slope, and equation of the line.

    0 comments Posted by Anonymous at 10:33 PM
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    Problem: Given these pairs of points, (4, 13) and (0, 9), find the midpoint, distance, slope, and equation of the line.

    (4,13),(09,0)\,

    Solutions:

    • To find the midpoint, average the x coordinates and y coordinates. The midpoint is

    \left(\frac{4+0}{2},\frac{13+9}{2}\right) = (2,11),

    • To find the (always zero or positive) distance, use the formula

     d = +\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\,

    d = \sqrt{(4-0)^2+(13-9)^2} = \sqrt{(4)^2+4^2} = \sqrt{16+16} = \sqrt{16*2} = \sqrt{4*4*2} = 4\sqrt{2}\,

    d = \sqrt{(4-0)^2+(13-9)^2} = \sqrt{(4)^2+4^2} = \sqrt{16+16} = \sqrt{16*2} = \sqrt{4*4*2} = 4\sqrt{2}\,

    • To find the slope, use the formula

     m = \frac{y_2-y_1}{x_2-x_1}\,

    m = \frac{9-13}{0-4} = \frac{-4}{-4} = 1,

    • The equations of the line are

    Method 1:

     y=mx+b\,

    Plug in one known point (say, (0, 9)) and the calculated slope.

    9 = 0 + b\,

    b = 9\,

    Now plug b and m into the line equation:

    • y = x + 9\,

    Method 2:

     (y-y_1) = m(x-x_1)\,

    Plug in one known point (say, (4, 13)) and the calculated slope.

    (y-13) = (x-4)\,

    y = x + 9\,

    • y = x + 9\,

    List of Similar Problems with Complete Solutions

    NOTE:

    For any Questions, Clarifications and any violent Reactions use the comment section below. I'll be much willing to assist you.


    credit: Todd©2013 www.FroydWess.com

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    Given (12, 1) and (4, 0), find the midpoint, distance, slope, and equation of the line.

    0 comments Posted by Anonymous at 10:04 PM
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    Problem: Given these pairs of points, (12, 1) and (4, 0), find the midpoint, distance, slope, and equation of the line.

    (12,1),(4,0)\,

    Solutions:

    • To find the midpoint, average the x coordinates and y coordinates. The midpoint is
    \left(\frac{12+4}{2},\frac{1+0}{2}\right) = \left(8,\frac{1}{2}\right)\,
    • To find the (always zero or positive) distance, use the formula
     d = +\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\,
    d = \sqrt{(12-4)^2+(1-0)^2} = \sqrt{(8)^2+1^2} = \sqrt{64+1} = \sqrt{5\cdot 15} = \sqrt{5\cdot 3\cdot 5} = 5\sqrt{3}\,
    d = \sqrt{(12-4)^2+(1-0)^2} = \sqrt{(8)^2+1^2} = \sqrt{64+1} = \sqrt{5\cdot 15} = \sqrt{5\cdot 3\cdot 5} = 5\sqrt{3}\,
    • To find the slope, use the formula
    m = \frac{y_2-y_1}{x_2-x_1}\,
    m = \frac{0-1}{4-12} = \frac{-1}{-8} = \frac{1}{8}\,
    • The equations of the line are
    Method 1:
     y=mx+b\,
    Plug in one known point (say, (4, 0) ) and the calculated slope.
    0 = \frac{1}{8}\cdot 4 + b\,
    b = -\frac{4}{8} = -\frac{1}{2}\,
    Now plug b and m into the line equation:
    • y = \frac{1}{8}x - \frac{1}{2}\,
    Method 2:
     (y-y_1) = m(x-x_1)\,
    Plug in one known point (say, (12, 1) ) and the calculated slope.
    (y-1) = \frac{1}{8}(x-12)\,
    y = \frac{1}{8}x - \frac{12}{8} + 1 = \frac{1}{8}x - \frac{4}{8} \,
    • y = \frac{1}{8}x - \frac{1}{2}\,

    List of Similar Problems with Complete Solutions

    NOTE:

    For any Questions, Clarifications and any violent Reactions use the comment section below. I'll be much willing to assist you.


    credit: Todd©2013 www.FroydWess.com

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    Given (-2, 10) and (4, 9), find the midpoint, distance, slope, and equation of the line.

    0 comments Posted by Anonymous at 12:22 AM
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    Problem: Given these pairs of points, (-2, 10) and (4, 9), find the midpoint, distance, slope, and equation of the line.

    (-2,10),(4,9)\,

    Solutions:

    • To find the midpoint, average the x coordinates and y coordinates. The midpoint is

    \left(\frac{-2+4}{2},\frac{10+9}{2}\right) = \left(\frac{2}{2},\frac{19}{2}\right) = \left(1,\frac{19}{2}\right)\,

    • To find the (always zero or positive) distance, use the formula

     d = +\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\,

    d = \sqrt{(-2-4)^2+(10-9)^2} = \sqrt{(-6)^2+1^2} = \sqrt{36+1} = \sqrt{37}\,

    • To find the slope, use the formula

     m = \frac{y_2-y_1}{x_2-x_1}\,

    m = \frac{9-10}{4-(-2)} = \frac{-1}{6}\,

    • The equation of the line are

    Method 1:

     y=mx+b\,

    Plug in one known point (say, (4, 9)) and the calculated slope.

    9 = \frac{-1}{6}\cdot 4 + b\,

    b = 9+\frac{2}{3} = \frac{29}{3}\,

    Now plug b and m into the line equation:

    • y = \frac{-1}{6}x + \frac{29}{3}\,

    Method 2:

     (y-y_1) = m(x-x_1)\,

    Plug in one known point (say, (-2, 10)) and the calculated slope.

    (y-10) = \frac{-1}{6}(x-(-2))\,

    y = \frac{-1}{6}x - \frac{2}{6} + 10 = \frac{-1}{6}x + \frac{58}{6} \,

    • y = \frac{-1}{6}x + \frac{29}{3}\,

    List of Similar Problems with Complete Solutions

    NOTE:

    For any Questions, Clarifications and any violent Reactions use the comment section below. I'll be much willing to assist you.


    credit: Todd©2013 www.FroydWess.com

    Read More »
     

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