NathalyN
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Find the distance between (3, 8) and (7, -11). Round to the nearest hundredth.

Find the coordinates of the midpoint of JK if J (-9, 5) and K(21, -7).

Find the coordinates of the midpoint of AB if A(-8, 7) and B(-9, 11).

If the midpoint between (8, y) and (-11, 6) is (-1.5, 5), find the value of y.

Find the slope of the line that passes through the points (3, -6) and (6, 12).

Find the value of x so that the line passing through (x, 10) and (-4, 8) has a slope of 2/3

Find the value of y so that the line passing through (9, 0) and (3, y) has a slope of 1/2

Find the slope of the line that passes through the points (8, 7) and (11, 7).

Find the distance between 3 8 and 7 11 Round to the nearest hundredth Find the coordinates of the midpoint of JK if J 9 5 and K21 7 Find the coordinates of the class=
Find the distance between 3 8 and 7 11 Round to the nearest hundredth Find the coordinates of the midpoint of JK if J 9 5 and K21 7 Find the coordinates of the class=
Find the distance between 3 8 and 7 11 Round to the nearest hundredth Find the coordinates of the midpoint of JK if J 9 5 and K21 7 Find the coordinates of the class=
Find the distance between 3 8 and 7 11 Round to the nearest hundredth Find the coordinates of the midpoint of JK if J 9 5 and K21 7 Find the coordinates of the class=

Respuesta :

texaschic101
distance formula : d = sqrt (x2 - x1)^2 + (y2 - y1)^2
(3,8)(-7,11)
d = sqrt (-7 - 3)^2 + (11 - 8)^2
d = sqrt (-10^2) + (3^2)
d = sqrt (100 + 9)
d = sqrt 109
d = 10.44
===============
mid-point formula : ((x1 + x2) / 2 , (y1 + y2) / 2)
(-9,5)(21,-7)

midpoint = ((-9 + 21)/2 , (5 + (-7)/2)
               = ((12/2) , (-2,2))
               = (6, -1) <==
=======================
(-8,7)(-9,11)
midpoint = ((-8 + (-9) / 2, (7 + 11)/2 )
               = (-17/2, 18/2)
               = (-8.5, 9) <==
======================
(8,y)(-11,6).....midpoint = (-1.5/5)
midpoint = ((8 + (-11)/2, (y + 6)/2)
               = (-3/2), (4 + 6)/2
               = (-1.5,5)
               y = 4 <==
======================
slope = (y2 - y1) / (x2 - x1)
(3,-6)(6,12)
slope = (12 - (-6) / (6 - 3) = (12 + 6) / 3 = 18/3 = 6 <==
======================
(x,10)(-4,8)...slope = 2/3

y = mx + b
slope(m) = 2/3
(-4,8)...x = -4 and y = 8
sub and find b
8 = 2/3(-4) + b
8 = -8/3 + b
8 + 8/3 = b
24/3 + 8/3 = b
32/3 = b
equation is : y = 2/3x + 32/3 or 3y = 2x + 32
when y = 10
3(10) = 2x + 32
30 - 32 = 2x
-2 = 2x
-2/2 = x
-1 = x <===
=================
(9,0)(3,y)...slope = 1/2

y = mx + b
slope(m) = 1/2
(9,0)...x = 9 and y = 0
sub and find b
0 = 1/2(9) + b
0 = 9/2 + b
-9/2 = b
equation is : y = 1/2x - 9/2 or 2y = x - 9
sub in 3 for x and solve for y
2y = 3 - 9
2y = -6
y = -6/2
y = -3 <==
=================
(8,7)(11,7)...notice how the y values are the same. This means it is a horizontal line. A horizontal line has a slope of 0.
====================
sorry...not too good with the picture graphs







batolisis

The answers are;

  1. The distance between [tex](3,8)[/tex] and [tex](7,-11)[/tex] is [tex]19.42\text{ (to the nearest hundredth)}[/tex]
  2. The midpoint of the line [tex]JK[/tex] is [tex](6,-1)[/tex]
  3. The midpoint of the line [tex]AB[/tex] is [tex](8.5,9)[/tex]
  4. The value of [tex]y[/tex] is [tex]7[/tex]
  5. The slope of the line that passes through [tex](3,-6)\text{ and }(6,12)[/tex] is [tex]6[/tex]
  6. The value of [tex]x[/tex] is [tex]-1[/tex]
  7. The value of [tex]y[/tex] is [tex]-\frac{3}{2}[/tex]
  8. The slope of the line that passes through [tex](8,7)\text{ and }(11,7)[/tex] is [tex]0[/tex]

In coordinate geometry, given two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex];

  • The distance between them is given by [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
  • The midpoint has the coordinates [tex](\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]
  • The slope of the line passing through the points is the ratio [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

These formulae will be used to solve the following problems:

   1)  Given points [tex](3,8)[/tex] and [tex](7,-11)[/tex], the distance between them is

        [tex]\sqrt{(7-3)^2+(-11-8)^2}\approx 19.42 \text{ (to the nearest hundredth)}[/tex]

   2)  If [tex]J=(-9,5) \text{ and }K=(21,-7)[/tex], the midpoint of the line [tex]JK[/tex] is

         [tex](\frac{-9+21}{2}, \frac{5+(-7)}{2})=(6,-1)[/tex]

   3)  If [tex]A=(-8,7) \text{ and } B=(-9,11)[/tex], the midpoint of the line [tex]AB[/tex] is

         [tex](\frac{-8+(-9)}{2}, \frac{7+11}{2})=(8.5,9)[/tex]

   4)  If the midpoint between [tex](8,y) \text{ and } (-11,6)[/tex] is [tex](-1.5,5)[/tex], then, using the

        midpoint formula for the y-coordinate

        [tex]6=\frac{y+5}{2} \implies y=7[/tex]

   5)  The slope of the line that passes through [tex](3,-6)\text{ and }(6,12)[/tex] is

        [tex]\frac{12-(-6)}{6-3}=6[/tex]

   6)  If the slope of the line that passes through [tex](x,10)\text{ and }(-4,8)[/tex] is [tex]\frac{2}{3}[/tex], then,

        using the slope formula

        [tex]\frac{2}{3}=\frac{8-10}{-4-x}\\\implies \frac{2}{3}=\frac{10-8}{4+x}\\\implies \frac{2}{3}=\frac{2}{4+x}\\\implies x=-1[/tex]

   7)  If the slope of the line that passes through [tex](9,0)\text{ and }(3,y)[/tex] is [tex]\frac{1}{2}[/tex], then,

        using the slope formula

        [tex]\frac{1}{2}=\frac{y-0}{3-9}\\\implies \frac{1}{2}=\frac{y}{-3}\\\implies y=-\frac{3}{2}[/tex]

   8)  The slope of the line that passes through [tex](8,7)\text{ and }(11,7)[/tex] is

        [tex]\frac{7-7}{11-8}=0[/tex], (since the y-coordinates of both points are equal)

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