Question

the distance between a+7i and -8+4i is √130 units the positive value of a is what?

Answer 1

From Pythagoras's theorem the distance d between any two points on a graph is

$d^2=x^2+y^2$

where x is the vertical distance and y is the horizontal distance.

Now, our points lie on the complex plane, but the Pythagoras's theorem still works because it is a fundamental theorem of geometry which works in every situation.

The vertical distance between the two points is

$7i-4i=3i$

Remember that on the complex plane the imaginary numbers lie on the vertical axis.

The horizontal distance is

$a-(-8)=a+8$

Now from the Pythagoras's theorem, the distance d between the two points is d

$d^2=(a+8)^2+(3)^2$

But we are told that this distance d is √130; therefore, we have

$130=(a+8)^2+(3)^2$$\Rightarrow130=(a+8)^2+9$

Now we just need to solve for a. To do this, we subtract 9 from both sides of the equation and then take the square root of both sides to get:

$a+8=\pm11$

which gives us

$a=3,$$a=-19.\text{ }$

Since the desired value of a is a positive number, our answer is a = 3.