The points P \left(-3,0\right), Q \left(-3,-5\right) and R \left(0,-5\right) are the vertices of a right-angled triangle, as shown on the number plane. Find the length of interval PQ.
We use the distance formula. The square of the distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $(x_2-x_1)^2+(y_2-y_1)^2$, so the distance is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
We find that $PQ\sqrt{(-3-(-3))^2+(0-(-5))^2}=\sqrt{0+25}=\boxed{5}$.
The points P (-3,0), Q (-3,-5) and R (0,-5) are the vertices of a right-angled triangle, as shown on the number plane. Find the length of interval PQ.
To find the length of interval PQ, we can use the distance formula.
The distance formula states that the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Using this formula, we find that the length of interval PQ is
$\sqrt{(-3-(-3))^2 + (0-(-5))^2} = \sqrt{(0)^2 + (5)^2} = \sqrt{25} = \boxed{5}$.
The points P (-3,0), Q (-3,-5) and R (0,-5) are the vertices of a right-angled triangle, as shown on the number plane. Find the length of interval PQ. Find the length of interval PR.
To find the length of interval PQ, we can use the distance formula.
The distance formula states that the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Using this formula, we find that the length of interval PQ is
$\sqrt{(-3-(-3))^2 + (0-(-5))^2} = \sqrt{(0)^2 + (5)^2} = \sqrt{25} = 5$.
To find the length of interval PR, we can again use the distance formula. The length of PR is
$\sqrt{(0-(-3))^2 + (-5-(-5))^2} = \sqrt{(3)^2 + (0)^2} = \sqrt{9} = 3$.
Therefore, the length of interval PQ is 5 and the length of interval PR is 3.
The points P (-3,0), Q (-3,-5) and R (0,-5) are the vertices of a right-angled triangle, as shown on the number plane. Find the length of interval PQ. Find the length of interval PR. If the length of PR is denoted by c, use Pythagoras’ theorem to find the value of c to three decimal places.
To find the length of interval PQ, we can use the distance formula: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Plugging in the given coordinates of P and Q, we have
\begin{align*}
PQ &= \sqrt{(-3-(-3))^2 + (0-(-5))^2} \\
&= \sqrt{(0)^2 + (5)^2} \\
&= \sqrt{0 + 25} \\
&= \sqrt{25} \\
&= 5.
\end{align*}
Next, using the distance formula, we find that the length of interval PR is
\begin{align*}
PR &= \sqrt{(0-(-3))^2 + (-5-(-5))^2} \\
&= \sqrt{(3)^2 + (0)^2} \\
&= \sqrt{9 + 0} \\
&= \sqrt{9} \\
&= 3.
\end{align*}
Now, using the Pythagorean Theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides, we have
\begin{align*}
PQ^2 + PR^2 &= c^2 \\
5^2 + 3^2 &= c^2 \\
25 + 9 &= c^2 \\
34 &= c^2 \\
c &= \sqrt{34}.
\end{align*}
To three decimal places, $\sqrt{34} \approx \boxed{5.831}$.
Consider the circle with equation x^2+y^2=25.
The equation $x^2+y^2=25$ represents a circle centered at the origin $(0,0)$ with a radius of $\sqrt{25}=5$.
The general equation for a circle is $(x-a)^2+(y-b)^2=r^2$, where $(a,b)$ is the center of the circle and $r$ is the radius.
In this case, since the equation is $x^2+y^2=25$, we can see that $(a,b)=(0,0)$ and $r=5$. Therefore, the circle is centered at the origin with a radius of 5 units.
Consider the circle with equation x^2+y^2=25. Graph the circle on the coordinate plane below.
The equation $x^2+y^2=25$ represents a circle centered at the origin $(0,0)$ with a radius of 5 units. To graph this circle on the coordinate plane, we can plot points on the circle by substituting different values of $x$ and solving for $y$.
Using $x=0$, we find $y^2=25$ which gives us $y=\pm5$. So, the points $(0,5)$ and $(0,-5)$ are on the circle.
Similarly, using $y=0$, we find $x^2=25$ which gives us $x=\pm5$. So, the points $(5,0)$ and $(-5,0)$ are on the circle.
We can also use other values of $x$ or $y$ to find more points on the circle.
Using these points, we can now plot the circle on the coordinate plane as shown below:
[asy]
size(200);
draw((0,10)--(0,-10),black+linewidth(0.5),EndArrow);
draw((10,0)--(-10,0),black+linewidth(0.5),EndArrow);
draw(Circle((0,0),5),red);
dot((0,5),red);
dot((0,-5),red);
dot((5,0),red);
dot((-5,0),red);
label("$(0,5)$",(0,5),W);
label("$(0,-5)$",(0,-5),W);
label("$(5,0)$",(5,0),N);
label("$(-5,0)$",(-5,0),S);
[/asy]
The red curve represents the circle with equation $x^2+y^2=25$.