In a regular seven-sided polygon, how many diagonals can be drawn from one vertex?
1 answer:
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Answer:
<u>Slope-intercept form</u>: y = -5x - 8
<u>Point-slope form</u>: y - 2 = -5(x + 2)
Step-by-step explanation:
Given the equation, 5y = x - 4, which passes through point (-2, 2):
Transform the given equation into its <u>slope-intercept form</u>, y = mx + b.
In order to do so, divide both sides by 5 to isolate y:
5y = x - 4
⇒ This is the slope-intercept form of 5y = x - 4.
Next, we must determine the equation of the line that is perpendicular to .
<u>Perpendicular lines</u> have <em>negative reciprocal</em> slopes. This means that if we multiply the slopes of two lines, their product will equal to -1.
In other words, if the slope of the given equation is m₁, and the slope of the other line perpendicular to the given linear equation is m₂, then: m₁ × m₂ = -1.
- Slope of the given equation: m₁ = ⅕
If we multiply these two slopes:
- m₁ × m₂ = -1
Now that we have identified the slope of the other line that is perpendicular to 5y = x - 4, we must determine the y-intercept of the <u>other line</u>.
- The <u>y-intercept</u> is the point on the graph where it crosses the y-axis, for which it is the value of "y" when its corresponding x-coordinate equals to zero (0).
- Thus, the standard coordinates of the y-intercept is (0, <em>b</em>), for which its y-coordinate is the value of "<em>b</em>" in the slope-intercept form, y = mx + b.
Using the <u>slope</u> of the other line, m₂ = -5, and the given point, (-2, 2), substitute these values into the slope-intercept form to find the value of the y-intercept, <em>b</em>:
y = mx + b
2 = -5(-2) + b
2 = 10 + b
Subtract 10 from both sides to isolate b:
2 - 10 = 10 - 10 + b
-8 = b
The equation of the other line that is perpendicular to 5y = x - 4 is:
Linear Equation that is perpendicular to 5y = x - 4 in slope-intercept form:
<h3>⇒ y = -5x - 8 </h3><h2>Rewrite the Equation in Point-slope Form:</h2>
The <u>point-slope form</u> is: y - y₁ = m(x - x₁)
In order to rewrite y = -5x - 8 in its point-slope form, we must substitute the value of the given point, (-2, 2) into the point-slope form:
y - y₁ = m(x - x₁)
y - 2 = -5[x - (-2)]
y - 2 = -5(x + 2) ⇒ This is the <u>point-slope form</u> of the line that is perpendicular to 5y = x - 4.
Answer:
The equation that says P is equidistant from F and the y-axis is .
(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.
Step-by-step explanation:
Let and
, where
is a point that is equidistant from F and the y-axis. The following vectorial expression must be satisfied to get the location of that point:
(1)
If we know that and
, then the resulting vectorial equation is:
The equation that says P is equidistant from F and the y-axis is .
If we know that ,
and
, then the coordinates for three points that are equidistant from F and the y-axis:
(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.