Sample Problems on How to calculate the distance between Two Points?

Problem 1: Find the distance between the two points with the coordinates given as, A(1,5) and B (2,7).

Solution:

Let (x₁, y₁) be (2,7) and (x₂, y₂) be (1,5).

The distance d between the points : [Tex]d = \sqrt{[(x_2 – x_1)^2 + (y_2 – y_1)^2]}[/Tex]

[Tex]\implies \sqrt{[(2-1)^2 + (7-5)^2]}[/Tex]

[Tex]\implies \sqrt{[1^2 + 2^2]}[/Tex]

[Tex]\implies \sqrt{(1+4)} = \sqrt{5} \ units[/Tex]

The distance between the two points is √5 units.

Problem 2: Find the distance between the two points with the coordinates given as, P (2,-6,2) and Q(7, 3, 1).

Solution:

Let (x₁, y₁, z₁) be P (2,-6,2) and (x₂, y₂, z₂) be Q (7,3,1).

The distance d between the points P and Q : [Tex]d = \sqrt{[(x_2 – x_1)^2 + (y_2 – y_1)^2+(z_2-z_1)^2]}[/Tex]

[Tex]\implies \sqrt{(7-2)^2+(3-(-6))^2+(1-2)^2)}[/Tex]

[Tex]\implies \sqrt{ 5^2+9^2+(-1)^2}[/Tex]

[Tex]\implies \sqrt{25+81+1}[/Tex]

[Tex]\implies \sqrt{107}\ units[/Tex]

Problem 3: Prove that the vertices of a right-angled triangle are the points (3, 4), (7, 4), and (3, 8).

Solution:

Let us say the given points be:

P = (3, 4)

Q = (7, 4)

R = (3, 8)

Now, we will find each vertices of the right-angle triangle by distance formula.

[Tex]AB=\sqrt{(7-3)^2+(4-4)^2}=\sqrt{4^2}=4units  [/Tex] 

[Tex]BC=\sqrt{(3-7)^2+(8-4)^2}=\sqrt{(-4)^2+4^2}=4\sqrt{2}units[/Tex]

[Tex]AC=\sqrt{(3-3)^2+(8-4)^2}=\sqrt{4^2}=4units[/Tex]

As we know the length of the sided of the right-angled triangle, by Pythagoras Theorem;

AB² + AC² = BC²

4²+4²=(4√2​)²

16+16 = 32⟹32 = 32

This proves that ABC is a right-angle triangle.

Distance Between Two Points in Complex Plane

The distance between two points in a complex plane or two complex numbers z₁​=a+ib and z₂​=c+id in the complex ⟹1−2k=9+4k plane is the distance between points (a, b) and (c, d), given as,

[Tex]| z_2 – z_1 |= \sqrt{[(a − c)^2 + (b − d)^2]}[/Tex]

Problem 4: Find the distance between the two complex numbers z₁ ​= 2−5i and z₂ ​= 7+7i

Solution:

Here, we have two complex numbers  z₁ = 2-5i and z₂ = 7+7i.

The distance between these complex numbers is equidistance to the two points in the plane, with coordinates, (2,-5) and (7,7).

Thus, distance between the two points is

 [Tex]\implies \sqrt{(7-2)^2+(7-(-5))^2}[/Tex]

[Tex]\implies \sqrt{5^2+12^2}\implies \sqrt{25+144}[/Tex]

[Tex]\implies \sqrt{169}\implies13\ units[/Tex]

Hence, the distance between two complex numbers z_1=2-5i and z_2=7+7i is 13 units.

Problem 5: A complex number ω is 6 units apart from z₁ = -3 – i and 6 units apart from z₂ = 3 + 5i. Check whether the triangle formed by ω, z₁, z₂ is right – angled or not.

Solution:

There are 3 complex numbers ω, z₁, z₂.

As we know the distance between ω and z₁ is 6 units and distance between ω and z₂ is 6 units.

Given,  ω, z₁ = 6 units

ω, z₂ = 6 units

Now, we will find the distance between z₁ and z₂ by using distance formula.

[Tex]z_1z_2=\sqrt{(3-(-3)+(5-(-1))}[/Tex]

[Tex]\implies \sqrt{6^2+6^2}  [/Tex] [Tex]\implies \sqrt{36+36}\implies \sqrt{72}[/Tex]

[Tex]\implies 6\sqrt{2} units.[/Tex]

By Pythagoras Theorem, we have;

(z₁​z₂​)²=(ωz₁​​)²+(ωz₂)​²

[Tex]\implies (6\sqrt{2})^2=6^2+6^2[/Tex]

[Tex]\implies 72=36+36\implies 72=72[/Tex]

Hence, we conclude that the given triangle is right-angle triangle.

Problem 6: Find a point on the x-axis that is equidistant from the points (1, -2) and (-2, -3).

Solution:

We know that any point on the x-axis has an y-coordinate of 0. As a result, we consider the point equidistant from the provided points to be (k,0). i.e., Distance between ( k,0) and (1, -2) = Distance between (k, 0) and (-2, -3).

 [Tex]\implies \sqrt{[(1-k)^2 + (-2-0)^2]} = \sqrt{[(-2-k)^2 + (-3-0)^2]}[/Tex]

[Tex]\implies  12 + k^2 -2k + 4 = 4+ k^2 +4k + 9  [/Tex]

[Tex]\implies 1-2k= 9+ 4k    [/Tex]

\implies   -4k-2k= 9-1

[Tex]\implies   -6k=8  [/Tex]

 [Tex]  \implies  k=\frac{-4}{3}[/Tex]

Therefore, the required point is (k, 0) = [Tex] (\frac{-4}{3}, 0).[/Tex]