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4 years ago

What is the equation of the line that passes through point (4, 12) and has a intercept of -2

Mathematics

1 answer:

kozerog [31]4 years ago

7 0

Answer:

$\large\boxed{y=\dfrac{7}{2}x-2}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept → (0, b)

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the point (4, 12), and y-intercept b = -2 → (0, -2).

Substitute:

$m=\dfrac{-2-12}{0-4}=\dfrac{-14}{-4}=\dfrac{14:2}{4:2}=\dfrac{7}{2}$

Finally:

$y=\dfrac{7}{2}x-2$

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2 years ago

Q4 Help pleaseeeeeeee

emmasim [6.3K]

a. The general equation for a circle centered at $(a,b)$ with radius $r$ is

$(x-a)^2+(y-b)^2=r^2$

The described circle has equation

$(x+3)^2+(y+2)^2=r^2$

We know the circle passes through the origin. This means that the equation above holds for $x=0$ and $y=0$ . The distance between any point on the circle and its center is the radius, so we can use this fact to determine $r$ :

$(0+3)^2+(0+2)^2=r^2\implies 9+4=13=r^2\implies r=\sqrt{13}$

So the circle's equation is

$(x+3)^2+(y+2)^2=(\sqrt{13})^2=13$

b. If the distance between point B and the center is less than $\sqrt{13}$ , then B lies inside the circle. If the distance is greater than $\sqrt{13}$ , it falls outside the circle. Otherwise, if the distance is exactly $\sqrt{13}$ , then B lies on the circle.