Find all points having an x coordinate of -3 whose distance from the point (-6,5) is 5

1 answer:

3 0

Ok
distance formula

D= $\sqrt{(x1-x2)^2+(y1-y2)^2}$
given
distance between (-6,5) and (-3,y) is 5
5= $\sqrt{(-6-(-3))^2+(5-y)^2}$
5= $\sqrt{(-6+3)^2+(5-y)^2}$
5= $\sqrt{(-3)^2+(5-y)^2}$
5= $\sqrt{9+y^2-10t+25}$
5= $\sqrt{y^2-10t+34}$
square both sides
25= $y^2-10t+34$
minus 25 both sides
$0= y^2-10t+9$
factor
0=(y-9)(y-1)
set to zero
0=y-9
9=y

0=y-1
1=y

the points are (-3,1) and (-3,9)

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Express the function y(t)= 4 sin 2πt + 15 cos 2πt in terms of (a) a sine term only. (b) Determine the amplitude, the period, the

Bond [772]

Answer:

See the step by step solution

Step-by-step explanation:

Consider a more general case of the form $a\sin(x)+b\cos(x)$ . We want to express it as a sine term only by using the following formula $\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$

The trick consists on find a number $\beta \in [0,2\pi]$ such that $\cos(\beta) = a, \sin(\beta)=b$ . But, note that since $\cos^2(\beta)+\sen^2(\beta)=1$ it is most likely that $a^2+b^2neq 1$ . Then, we will use the following equations:

$\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B$

Then, problem turns out to be

$a\sin(x)+b\cos(x)= \sqrt[]{a^2+b^2}(A\sin(x)+B\cos(x)) =\sqrt[]{a^2+b^2}(\sin(x)\cos(\beta)+\sin(\beta)\cos(x))=\sqrt[]{a^2+b^2}\sin(x+\beta)$ ,

where $\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same. Then, a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})$

b) The amplitude is $\sqrt[]{15^2+4^2} = \sqrt[]{241}$ . Since $2\pi rad/s = 1 Hz$ the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.