1/secA-tanA - 1 /secA+tanA <br>So the answer comes 2tanA/sec^2 A-tan^2A <br>how ??.

The given line segment has a midpoint at (3, 1).

Katena32 [7]

Answer:

$y=\frac{1}{3}x$

Step-by-step explanation:

The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)

Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through $(x_1,y_1)\ and\ (x_2,y_2)$ is:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Therefore the line passing through (2, 4) and (4, -2) has an equation:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10$

Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10

Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:

$m_1m_2=-1$ .

Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:

$m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}$

Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:

$\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x$

the equation, in slope-intercept form, of the perpendicular bisector of the given line segment is $y=\frac{1}{3}x$

7 0

3 years ago

Read 2 more answers

Find the length of the missing side. If necessary, round to the nearest tenth. side A = 38 side b =b side c = 34

ohaa [14]

A) 51, assuming this is a *right* triangle:

38^2+34^2 = 50.99^2, rounding 51!!

5 0

3 years ago

Read 2 more answers

Among all pairs of numbers whose difference is 20, find a pair whose product is as small as possible. what is the minimum produc

Elza [17]

Answer:

-100

-10, 10

Step-by-step explanation:

Let the smaller number be x.

Then the larger number is x + 20.

The product is x(x + 20).

Now you can write the function

y = x(x + 20)

y = x^2 + 20x

Take the first derivative of y with respect to x.

y' = 2x + 20

Set the first derivative equal to zero to find the x value for the minimum value of the function.

2x + 20 = 0

2x = -20

x = -10

The minimum value of the function occurs at x = -10. -10 is one of the two numbers.

y = x^2 + 20x

For x = -10,

y = (-10)^2 + 20(-10)

y = 100 - 200

y = -100

The minimum value of the product is -100.

x = -10

x + 20 = -10 + 20 = 10

The numbers are -10 and 10.

If you have not learned derivatives yet, then plot the function

y = x^2 + 20x

Now look at the graph and find the minimum y value and the x value at which it occurs.

The minimum y value is -100. That is the minimum product you are looking for.

The x value of the minimum function value is x = -10.

Then x + 20 = -10 + 20 = 10.

The numbers are -10 and 10.

4 0

3 years ago

Original price: 60 markup:15% what is retail price

sdas [7]

60x1.15=69
Retail price is $69

7 0

3 years ago

Is this a function or not?

Hoochie [10]

Answer:

ummm it think so

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

1/secA-tanA - 1 /secA+tanA So the answer comes 2tanA/sec^2 A-tan^2A how ??.​

1/secA-tanA - 1 /secA+tanA So the answer comes 2tanA/sec^2 A-tan^2A how ??.