# Math Help - Introduction to analytic geometry?

1. ## Introduction to analytic geometry?

Can anyone assist me in these 2 questions

1. Find all values of a so that the distance between points (A,-9) and (-2A,7) is 20 units. I did this

Square root (7-(-9))^2+(-2a-a)^2

I get square root 256+9A=20 But how would I proceed taking the square root and the squaring 20?

My next question I am not sure how to start

Collinear points lie on the same line. Find the value of K for which the points (15,1)
(-3,8) and (3,k) are collinear.

2. Originally Posted by homeylova223
Can anyone assist me in these 2 questions

1. Find all values of a so that the distance between points (A,-9) and (-2A,7) is 20 units. I did this

Square root (7-(-9))^2+(-2a-a)^2

I get square root 256+9A=20 But how would I proceed taking the square root and the squaring 20?
Which variable are you solving for? A or a? Math is case sensitive!

$20 = \sqrt{(7 - (-9))^2 + (-2a - a)^2}$

$20 = \sqrt{256 + 9a^2}$

Now square both sides and solve the quadratic.

Some advice...Write out the whole line each time, not just part of it. You'll see what you need to do more easily.

-Dan

3. Originally Posted by homeylova223
Collinear points lie on the same line. Find the value of K for which the points (15,1)
(-3,8) and (3,k) are collinear.
Recall the equation for a line: y = mx + b

Three points are colinear if they fall on the same line. So you have three equations:
1 = 15m + b
8 = -3m + b
k = 3m + b

Three equations, three unknowns. (Hint: solve the first two equations for m and b first.)

-Dan

4. Originally Posted by homeylova223
Collinear points lie on the same line. Find the value of K for which the points (15,1)
(-3,8) and (3,k) are collinear.
first find the slope m from the first 2 given points then use $y-y_1=m(x-x_1)$ from this you should be able to see how to get $k$

from this i got $y=\frac{7}{18 }x +\frac{41}{6 }$

now if you plug in 3 for x then y will give the k value