Question

There is a point $A$ with positive coordinates such that the sum of the coordinates of $A$ is $14$. If the $x$-coordinate of $A$ is $a$, then point $P$ is at $(3a, a^2+13a-11)$. If the slope of a line passing through $A$ and $P$ is $7$, find $a$.

Answer 1

5

Explanation:

Given:

A = (a, 14-a)

P = (3a, a^2 +13a -11)

the slope of AP is 7

a > 0

Find:

a

Solution:

The slope of AP is ...

m = (Py -Ay)/(Px -Ax)

7 = (a^2 +13a -11 -(14 -a))/(3a -a)

14a = a^2 +14a -25

25 = a^2

a = √25 = 5 . . . . . the positive solution

The value of 'a' is 5.

_____

Check

The point A is (a, 14-a) = (5, 9).

The point P is (3a, a^2 +13a -11) = (15, 79)

The slope of AP is (79 -9)/(15 -5) = 70/10 = 7.