# Thread: [SOLVED] College Algebra: the distance formula

1. ## [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.

2. 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 ...

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

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

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

x-intercepts ...

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

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

know what to do now?

3. Originally Posted by skeeter
1. distance formula ...

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

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

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

x-intercepts ...

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

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

know what to do now?
For two I got it.

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

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

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

For #1 I'm still stuck.

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

start by squaring both sides ...

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

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

$\displaystyle 64 = (y+2)^2$

$\displaystyle \pm 8 = y+2$

$\displaystyle -2 \pm 8 = y$

5. Originally Posted by skeeter
$\displaystyle 10 = \sqrt{(9-3)^2 + (y+2)^2}$

start by squaring both sides ...

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

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

I get it up to here ^

$\displaystyle 64 = (y+2)^2$

$\displaystyle \pm 8 = y+2$

$\displaystyle -2 \pm 8 = y$
I don't know how to do it from that point on.

6. 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 .

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

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

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

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