[SOLVED] College Algebra: the distance formula

**Ashley189** · February 24th 2009, 04:39 PM

Solve the problem.

Find all the points having an x-coordinate of 9 whose distance from the point (3,-2) is 10.

A) (9,13), (9,-7)
B) (9, -12), (9,8)
C) (9,6),(9,-10)
D) (9,2), (9,-4)

List the intercepts for the graph of the equation.

x^2+15x+54

A) (6,0), (9,0),(0,54)
B) (0,6), (0,9), (54,0)
C) (0,-6), (0,-9), (54,0)
D) (-6,0), (-9,0), (0,54)

This is off a test review (doesn't count only to study), so I have the answers. I just need help solving them. If someone could please help me out.

**skeeter** · February 24th 2009, 04:48 PM

Originally Posted by Ashley189

Solve the problem.

Find all the points having an x-coordinate of 9 whose distance from the point (3,-2) is 10.

A) (9,13), (9,-7)
B) (9, -12), (9,8)
C) (9,6),(9,-10)
D) (9,2), (9,-4)

List the intercepts for the graph of the equation.

x^2+15x+54

A) (6,0), (9,0),(0,54)
B) (0,6), (0,9), (54,0)
C) (0,-6), (0,-9), (54,0)
D) (-6,0), (-9,0), (0,54)

This is off a test review (doesn't count only to study), so I have the answers. I just need help solving them. If someone could please help me out.

1. distance formula ...

$10 = \sqrt{(9-3)^2 + (y+2)^2}$

solve for y ... you should get two solutions.

2. y-intercept ... evaluate $y = x^2 + 15x + 54$ for $x = 0$ .

x-intercepts ...

$x^2 + 15x + 54 = 0$

$(x + 9)(x + 6) = 0$

know what to do now?

**Ashley189** · February 24th 2009, 05:01 PM

Originally Posted by skeeter

1. distance formula ...

$10 = \sqrt{(9-3)^2 + (y+2)^2}$

solve for y ... you should get two solutions.

2. y-intercept ... evaluate $y = x^2 + 15x + 54$ for $x = 0$ .

x-intercepts ...

$x^2 + 15x + 54 = 0$

$(x + 9)(x + 6) = 0$

know what to do now?

For two I got it.

$x^2 + 15x + 54 = 0$

$(x + 9)(x + 6) = 0$

x+9=0 or x+6=0
x=-9 x=-6

So, the answers D.

For #1 I'm still stuck.

**skeeter** · February 24th 2009, 05:06 PM

$10 = \sqrt{(9-3)^2 + (y+2)^2}$

start by squaring both sides ...

$100 = (9-3)^2 + (y+2)^2$

$100 = 36 + (y + 2)^2$

$64 = (y+2)^2$

$\pm 8 = y+2$

$-2 \pm 8 = y$

**Ashley189** · February 24th 2009, 05:26 PM

Originally Posted by skeeter

$10 = \sqrt{(9-3)^2 + (y+2)^2}$

start by squaring both sides ...

$100 = (9-3)^2 + (y+2)^2$

$100 = 36 + (y + 2)^2$

I get it up to here ^

$64 = (y+2)^2$

$\pm 8 = y+2$

$-2 \pm 8 = y$

I don't know how to do it from that point on.

**mathaddict** · February 24th 2009, 09:31 PM

Originally Posted by Ashley189

I don't know how to do it from that point on.

Not sure if this is what you want to know .

$-2\pm8=y$ , there will be 2 equations .

$-2+8=y$ or $-2-8=y$

There will be 2 values of y ie 6 and -10

**Ashley189** · February 25th 2009, 05:01 AM

Now I got it. Didn't know that +- was two separate equations that's the part that got me confused.

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