Question

If p(x,3), q(7,-1) and pq=5 units find the possible value of x

Answer 1

the possible values of x are:

x = 7 + √21

x = 7 - √21

Explanation:

To find the possible value of x, we need to calculate the distance between points P(x, 3) and Q(7, -1) using the distance formula:

d(PQ) = √((x - 7)² + (3 - (-1))²) = 5

Simplifying the equation:

d(PQ) = √((x - 7)² + 4) = 5

Squaring both sides of the equation:

(x - 7)² + 4 = 25

Expanding the equation:

(x - 7)² = 25 - 4

(x - 7)² = 21

Taking the square root of both sides:

x - 7 = ±√21

Now we can solve for x:

x = 7 ± √21