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jekas [21]
3 years ago
11

Parallel and perpendicular lines ​

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0

Answer:

A. Slope = -³/2

B. y = -³/2x - 6

C. ³/2x + y = -6

Step-by-step explanation:

A. Slope of the line would be the negative reciprocal of the slope of 2x - 3y = 12, which is written in standard form. Rewrite in slope-intercept form, y = mx + b, where m = slope.

Thus:

2x - 3y = 12

-3y = -2x + 12

y = -2x/-3 + 12/-3

y = ⅔x - 4

The slope of 2x - 3y = 12 is therefore ⅔.

The slope of the line perpendicular to 2x - 3y = 12 would be the negative reciprocal of its slope, ⅔ which is:

-³/2

Therefore, the slope of the line perpendicular to 2x - 3y = 12 is -³/2

B. To find the equation of the line in slope-intercept form, first find the equation in point-slope form using the point given (-6, 3) and slope of the line, -³/2.

Substitute a = -6, b = 3, and m = -³/2 into y - b = m(x - a).

Thus:

y - 3 = -³/2(x - (-6))

y - 3 = -³/2(x + 6)

Rewrite in slope-intercept form, y = mx + b

Multiply both sides by 2

2(y - 3) = -3(x + 6)

2y - 6 = -3x - 18

2y = -3x - 18 + 6

2y = -3x - 12

y = -3x/2 - 12/2

y = -³/2x - 6

C. Rewrite y = -³/2x - 6 in standard form.

y = -³/2x - 6

³/2x + y = -6

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\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}

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\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}

Multiply through everything on the right side by <em>b/d</em> to get

\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}

and <em>a/b</em> < <em>c/d</em> tells us that

\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}

Then

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and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

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