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3 years ago

Which equations are perpendicular to the line that passes through (2,1) and (3,4)

Mathematics

1 answer:

REY [17]3 years ago

5 0

9514 1404 393

Answer:

y=(-1/3)x+6
y=(-1/3)x+10

Step-by-step explanation:

The slope of the given line can be found from the slope formula:

m = (y2 -y1)/(x2 -x1)

m = (4 -1)/(3 -2) = 3/1 = 3

The slope of the perpendicular line is the opposite reciprocal of this:

perpendicular slope = -1/m = -1/3

Choices (c) and (d) have x-coefficients that are -1/3. Those are the perpendicular lines.

y=(-1/3)x+6
y=(-1/3)x+10

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2 years ago

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

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Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

<em>Hence, circle Q has a smaller radius</em>

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3 years ago