# [SOLVED] College Algebra: the distance formula

• February 24th 2009, 05:39 PM
Ashley189
[SOLVED] College Algebra: the distance formula
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.
• February 24th 2009, 05:48 PM
skeeter
Quote:

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?
• February 24th 2009, 06:01 PM
Ashley189
Quote:

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

For #1 I'm still stuck.
• February 24th 2009, 06:06 PM
skeeter
$10 = \sqrt{(9-3)^2 + (y+2)^2}$

start by squaring both sides ...

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

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

$64 = (y+2)^2$

$\pm 8 = y+2$

$-2 \pm 8 = y$
• February 24th 2009, 06:26 PM
Ashley189
Quote:

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.
• February 24th 2009, 10:31 PM
Quote:

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
• February 25th 2009, 06:01 AM
Ashley189
Now I got it. Didn't know that +- was two separate equations that's the part that got me confused.