Question

find the number of points on a straight line between points -4,11 to 16,-1 whose coordinates are positive integers

Answer 1

Answer:

Number of points = 2

Step-by-step explanation:

Equation of the line passing through two points $(x_1, y_1)$ and $(x_1, y_1)$ is

$\boxed{y-y_1=\frac{(y_2-y_1)}{(x_2-x_1)} (x-x_1)}$

Thus, the equation of the line passing through (-4, 11) and (16, -1) is

$y-11=\frac{(-1-11)}{(16+4)} (x+4)$

$\implies y-11=\frac{(-12)}{(20)} (x+4)$

$\implies y-11=\frac{-3}{5} (x+4)$

$\implies 5y-55=-3x-12$

$\implies 3x+5y=43$

This line touches the x-axis at

3x = 43 or x = 14.33

In between x = 0 to x = 14.33, the positive integral values of x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

The line touches the y-axis at

5x = 43 or y = 8.6

In between y = 0 to y = 8.6, the positive integral values of y = 1, 2, 3, 4, 5, 6, 7, 8

For y = 1

3x + 5 = 43 ⇒ x = 12.66 (not an integer)

For y = 2

3x + 10 = 43 ⇒ x = 11 (which is an integer)

For y = 3

3x + 15 = 43 ⇒ x = 9.33 (not an integer)

For y = 4

3x + 20 = 43 ⇒ x = 7.67 (not an integer)

For y = 6

3x + 30 = 43 ⇒ x = 4.33 (not an integer)

For y = 7

3x + 35 = 43 ⇒ x = 2.67 (not an integer)

For y = 8

3x + 40 = 43 ⇒ x = 1 (which is an integer)

Therefore, there are two points, whose coordinates are positive integers.

Answer 2

Answer:

The equation of the line is y−11=(11+1−4−16)(x+4)=−35(x+4). This means 3x+5y=43.

Clearly (1,8) is a point on the line such that x and y are both positive integers, and other such points are found by increasing x by 5 and decreasing y by 3. We then see that the only other such points are (6,5) and (11,2).

Step-by-step explanation: