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3 years ago

10

What is the slope of the line? A. -2 B. -1/2 C. 1/2 D. 2

1 answer:

Alexxandr [17]3 years ago

6 0

To find the slope, you can either use the slope equation:
(y2-y1) / (x2-x1)
{I would recommend you remember this because you'll need to know this in the future}

or,

Just count. A slope is rise over run, so you see how far up the slope goes from one point to the other, (1), and then you see how far sideways the slope goes (2).
Going right, the slope is naturally declining downwards, so it's a negative.

The answer is B. -1/2.

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find the standard form of equation of a line perpendicular to 4X plus 5Y equals 40 that includes the point -10, -3

snow_tiger [21]

The equation of line in standard form is 5x - 4y = -38

<em><u>Solution:</u></em>

Given that we have to find the equation of line perpendicular to 4x + 5y = 40 that includes the point (-10, -3)

<em><u>The equation of line in slope intercept form is given as:</u></em>

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Given equation of line is:

4x + 5y = 40

Rearranging to slope intercept form, we get

5y = -4x + 40

$y = \frac{-4}{5}x + \frac{40}{5}\\\\y = \frac{-4}{5}x + 8$

On comparing the above equation with eqn 1,

$m = \frac{-4}{5}$

We know that product of slope of line and slope of line perpendicular to given line is equal to -1

Therefore,

$\frac{-4}{5} \times \text{ slope of line perpendicular to it } = -1\\\\\text{ slope of line perpendicular to it } = \frac{5}{4}$

Now find the equation of line with slope 5/4 and passing through (-10, -3)

Substitute $m = \frac{5}{4}$ and (x, y) = (-10, -3) in eqn 1

$-3 = \frac{5}{4}(-10) + c\\\\-3 = \frac{-25}{2} + c\\\\c = -3 + \frac{25}{2}\\\\c = \frac{-6 + 25}{2}\\\\c = \frac{19}{2}$

Substitute $c = \frac{19}{2}$ and $m = \frac{5}{4}$ in eqn 1

$y = \frac{5}{4}x + \frac{19}{2}$

The standard form of an equation is Ax + By = C

Therefore,

$\frac{5}{4}x - y = -\frac{19}{2}\\\\\frac{5x - 4y}{4} = \frac{-19}{2}\\\\5x - 4y = -38$

Thus the equation of line in standard form is found

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