All categories

4 years ago

Prove: Every point on the perpendicular bisector of a segment is equidistant from the ends of the segment.

Mathematics

2 answers:

sweet [91]4 years ago

8 0

AP=√(a² + b² )
BP=√(a² + b² )

andrew-mc [135]4 years ago

6 0

Answer with explanation:

Here ,let line, x=0 intersect ,segment AB at point M.

So, PM ⊥ AB.

And, AM =BM →→→∵ y axis or segment PM , is Perpendicular bisector of segment AB.

In Δ AMP and ΔB MP

∠AMP = ∠ B MP=90°[→→ Each being 90°]

AM = BM →Line PM , is perpendicular Bisector.

Side MP , is Common.

Δ AMP ≅ ΔB MP→→→[S AS]

So, AP= BP →→[C P C T]

2. We will use distance formula to find AP and BP.

Distance between two points $(x_{1},y_{1}), and ,(x_{2},y_{2})$ is given by $=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2$

$AP=\sqrt{(-a-0)^2+(0-b)^2}\\\\ AP=\sqrt{a^2+b^2}\\\\ BP=\sqrt{(a-0)^2+(0-b)^2}\\\\ BP=\sqrt{a^2+b^2}$

Hence, AP = BP

Where ,Point P, can be located anywhere on y axis.

You might be interested in

The inequality of the most people that can be seated in the auditorium

WINSTONCH [101]

Let x be the number of people that can be seated.

The value of x is some number 0 or greater, such that this number is a whole number (we can't have half a person). This means that $x \ge 0$ which is the same as saying $0 \le x$

The instructions state that we can seat up to 1200 people. This is the max capacity. We can't go over this amount. So we will also have $x \le 1200$ indicating that x can be less than 1200 or equal to 1200.

So putting the two inequalities together, we get $0 \le x \le 1200$

Note: if you haven't learned about compound inequalities yet, then the teacher may only want $x \le 1200$ as the answer.

3 0

3 years ago

I need help wit one & two please help

seropon [69]

23 is less than 25 which means it's closer to 20, so you would round it down to 20. 1.75 is over 0.50, 1/2, so you would round it up to 2. If you multiply 2 and 20 you would get 40.

Number two gets the same explanation just with different numbers.

3 0

3 years ago

If x= eighth root of y and y= z^16 how is x proportionate to z ?

ankoles [38]

$x= \sqrt[8]{y}
\\ y = z^{16}

$

Plug "y" into the first equation.

$x = \sqrt[8]{z^{16}} = (z^{16})^{1/8} = z^{16/8} = z^2$

8 0

3 years ago

A company manufactures 295 toy car each day. How many toy cars dose the company manufacture in 34 days?

exis [7]

295 X 34 = 10030 <== Cars in 34 days

3 0

3 years ago

Read 2 more answers

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

3 years ago