All categories

3 years ago

The distance between the two points (4,2) and (8,k) is 5 units. Find value(s)of k

1 answer:

5 0

D^2 = (x2 - x1)^2 + (y2 - y1)^2
d^2 = (8 - 4)^2 + (k - 2)^2
d = 5 so
5^2 = 4^2 + k^2 - 4k + 4
25 = 16 + k^2 - 4k + 4
25 = 20 + k^2 - 4k
0 = k^2 - 4k - 5
0 = (k - 5)(k + 1)
k has 2 answers
k - 5 = 0
k = 5
k + 1 = 0
k = - 1

The two answers are
(8, 5) and (8, -1) <<<< answer

You might be interested in

If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o

almond37 [142]

If the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$

⇒ $y=\sqrt{4ax}$

$x= 4a$ ,

Area of the region $=\frac{256}{3}$ Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve $= \[ \int_{0}^{4a} y \,dx \]$

$\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

$\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

To know more about Area bounded by the curve click here

brainly.com/question/13252576

#SPJ2

8 0

2 years ago

Read 2 more answers

WILL GIVE BRAINLIEST

dalvyx [7]

Answer:

D. 8pi>24 cause this is the right answer

7 0

3 years ago

Geometry Select the correct answer from each drop-down menu. Given that , ° and °.

Alexxx [7]

$\measuredangle BAC = 29°$

Answer:

$\measuredangle BOC =58°$

Step-by-step explanation:

$\measuredangle BDC = 29°...(Given) \\\because \measuredangle BAC = \measuredangle BDC\\(\angle 's \: subtended \: by\: same\: arc) \\\huge \purple {\boxed {\therefore \measuredangle BAC = 29°}} \\\\$

Since, angle subtended on the center of the circle is twice the angle subtended on the circumference of the circle.

$\therefore \measuredangle BOC =2\times \measuredangle BDC\\\therefore \measuredangle BOC =2\times 29°\\\huge \red {\boxed {\therefore \measuredangle BOC =58°}} \\$