Please help with this problem?

Question

In the x-y plane, how many lines whose x-intercept is a positive prime number and whose y intercept is a positive integer pass through the point 4,3.

Answer 1

Cool problem, I think that I will give it to my students some day. Let n be the y-int and p be the x-int. Find the slope of the line using the given pt and the x-int. Then write the eqn for the line and solve for n by using the given pt. You then need to use some number theory.

Answer 2

Let:

the point Y = (0, b),
the point X = (a, 0), and
the point W = (m, n),

where:

b, m, and n are positive integers, and
a is prime.

In order for all three points to lie on a straight line, the segments YW and XW must share a common slope:

(b - n) / (0 - m) = (0 - n) / (a - m)
(b - n)(a - m) = mn

Since a, b, m, and n are all integers, we can examine the factors of mn.

For (m, n) = (4, 3),
(b - 3)(a - 4) = 12

With the exponents from the prime factorization of 12 = 2^2 * 3^1, we know there are a total of (2 + 1) * (1 + 1) = 6 (positive) factors, including 1, 2, 3, 4, 6, 12.

To solve for a,
(a - 4) = {-12, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 12}
a = {-8, -2, 0, 1, 2, 3, 5, 6, 7, 8, 10, 16}

And since there is a further restriction for prime,

a = {5, 7}

(b - 3)(a - 4) = 12
(b - 3)({5, 7} - 4) = 12
(b - 3)({1, 3}) = 12
(b - 3) = {12, 4}
b = {15, 7}

Using the segment YX, slope m = (b - 0) / (0 - a) = -b/a, so in the form y = mx + b:

y = (-15/5)x + 15 = -3x + 15
y = (-7/7)x + 7 = -x + 7

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This should be useful for general values for W = (m, n) as well as cases where either a or b must satisfy additional conditions.

Answer 3

We have that

(p,0) and (0,m) are points with m a positive integer and p a positive prime

y - m = [m-0]/[0-p] * ( x - 0)

y = m - m/p x

We want (4,3) on the line.

So 3 = m - m/p * 4

So 3 = m (1- 4/p); 3 = m [p-4]/p
-----------------------
Case 1 m = pn (for some integer n)

3 = (pn) [p-4]/p = n(p-4)

We need 0< p-4 <= 3 with p prime yielding p = 5, 7

p = 5 ; n = 3 (i.e., m = 15)
p = 7 ; n = 1 (i.e., m = 7)
______________________________
Case 2 m != pn thus gcd(p,m) = 1

3 = m (p-4) / p

Since gcd(m,p)=1 then (p-4)/p must be an integer

But this is clearly impossible since 0 < (p-4)/p < 1 for all p > 0

Thus there are no soltuions for case 2

---------------

Thus our solutions are

(5,0), and (0,15) giving y = 15 - 3x

and

(7,0) and (0,7) giving y = 7 - x

There are 2 solutions

EDIT

There is nothing more annoying than a prick coming around copying my solutions... and only the solution not even the procedure... well asides from him getting the pts

Answer 4

try using the prime points on the x intercept to make lines with the point on the y intercept

Answer 5

X:= {all prime numbers}
Y:={All real numbers} You can draw infinite lines through a fixed point all over the XY plane

[Hey Don't take my words for it, think and the make ur decision]

Answer 6

2 ?

Answer 7

No offense, and I know I'm not helping. (I'm not trying to get points), but I think most of the people are on summer vacation, and most of them are probably with blank brains right now. Like me. No offense to you people out there. And besides, I'm not to that math yet, is it Algebra II?

Answer 8

i usually fly Midwest, and you can take as many prime numbers along as you want, as long as they fit in your carry-on.

Answer 9

2 lines.
Their equation are:
y = -3x + 15
and
y = -x + 7

Answer 10

sorry at this time of night that's to complicated