Write the slope-intercept form of the line that has a slope of 2 and intersects the line, 2x - 3y = 6 at x = 3.

1 answer:

7 0

To write the equation of a line, we need 1) a point on the line 2) it's slope For the point, we just substitute in the equation of the line intersecting it with x=3 (because they share this point) Therefore, 2*3-3y=6 --> That gives y=0 Then the point is (x1,y1)=(3,0) We already have the slope : m=2 Now we should use this relation for the required line equation (y-y1) / (x-x1) = m Then (y-0) / (x-3) = 2 Then y = 2*x - 6 And that is the required form where 2 is the slope and -6 is the intercept

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What is the center of the hyperbola whose equation is (y+3)^2/81-(x-6)^2/89=1?

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We have been given an equation of hyperbola $\frac{(y+3)^2}{81}-\frac{(x-6)^2}{89}=1$ . We are asked to find the center of hyperbola.

We know that standard equation of a vertical hyperbola is in form $\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where point (h,k) represents center of hyperbola.

Upon comparing our given equation with standard vertical hyperbola, we can see that the value of h is 6.

To find the value of k, we need to rewrite our equation as:

$\frac{(y-(-3))^2}{81}-\frac{(x-6)^2}{89}=1$

Now we can see that value of k is $-3$ . Therefore, the vertex of given hyperbola will be at point $(6,-3)$ and option D is the correct choice.