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Andrei [34K]
3 years ago
15

Find the distance between line 11x-60y-15=0 and point (2, 1/60)

Mathematics
2 answers:
vovikov84 [41]3 years ago
8 0

Answer:

The answer to your question is: d = 6 / 61 or 0.1 units

Step-by-step explanation:

Data

Line: 11x - 60y - 15 = 0   A = 11; B = -60; C = -15

Point (2, 1/60)                 x = 2    y = 1/60

Formula

d = | Ax + By + C | / √(A² + B²)

Process

d = | (11)(2) + (-60)(1/60) -15 | / √(11² + (-60²)

d = | 22 -1 - 15 | / √ 121 + 3600

d = | 6 | / √3721

d = | 6 | / 61

d = 6 / 61 or 0.1 units

Neko [114]3 years ago
8 0
Replace x and y

(2, 1/60)
2 is your x and 1/60 is your y. You want to plug those into the equation to get your distance.

11x-60y-15=0

Solving for Y:

11(2) -60y-15=0
22 - 60y - 15 = 0
-22. -22
--------------------------
60y - 47 = 0
+47. +47
---------------------------
60y = 47
---------------
60

y = .78

- do the same for x and graph your point if asked
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x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12

Step-by-step explanation:

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Third equation:

\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

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\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}

Now we use this value for "x" back in equation 1 to solve for "y":

1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}

And finally we solve for the third unknown "z":

\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12

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