Which equation would you use to find the distance between the two points a 6 5 B 6 5 C 6 5 D 6 5

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Exercises 1.2
What is the Distance Between Two Points?
Distance Between Two Points Formula
Formula for Distance Between Two Points:
Derivation of Formula for Distance Between Two Points
How to Find Distance Between Two Points?
Distance Between Two Points in Complex Plane
FAQs on Distance Between Two Points
How Do We Calculate the Distance Between Two points in 2D Plane?
How to Find the Distance Between Two points in 3D Plane?
What is the Shortest Distance Between Two Points?
How to Find the Distance Between Two Points Using Pythagorean Theorem?
What is the Distance Formula to Find Distance Between Two Points in Coordinate Geometry?
How to Derive the Formula to Find The Distance Between Two Points?
How to Find the Vertical Distance Between Two Points?
What are Steps to Find Euclidean Distance Between Two Points?

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Given two points $(x_1,y_1)$ and $(x_2,y_2)$, recall that their horizontal distance from one another is $\Delta x=x_2-x_1$ and their vertical distance from one another is $\Delta y=y_2-y_1$. (Actually, the word "distance'' normally denotes "positive distance''. $\Delta x$ and $\Delta y$ are signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs $|\Delta x|$ and $|\Delta y|$, as shown in figure 1.2.1. The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides: $$ \hbox{distance} =\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{(x_2-x_1)^2+ (y_2-y_1)^2}. $$ For example, the distance between points $A(2,1)$ and $B(3,3)$ is $\sqrt{(3-2)^2+(3-1)^2}=\sqrt{5}$.

Figure 1.2.1. Distance between two points, $\Delta x$ and $\Delta y$ positive.

As a special case of the distance formula, suppose we want to know the distance of a point $(x,y)$ to the origin. According to the distance formula, this is $\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$.

A point $(x,y)$ is at a distance $r$ from the origin if and only if $\sqrt{x^2+y^2}=r$, or, if we square both sides: $x^2+y^2=r^2$. This is the equation of the circle of radius $r$ centered at the origin. The special case $r=1$ is called the unit circle; its equation is $x^2+y^2=1$.

Similarly, if $C(h,k)$ is any fixed point, then a point $(x,y)$ is at a distance $r$ from the point $C$ if and only if $\sqrt{(x-h)^2+(y-k)^2}=r$, i.e., if and only if $$ (x-h)^2+(y-k)^2=r^2. $$ This is the equation of the circle of radius $r$ centered at the point $(h,k)$. For example, the circle of radius 5 centered at the point $(0,-6)$ has equation $(x-0)^2+(y- -6)^2=25$, or $x^2+(y+6)^2=25$. If we expand this we get $x^2+y^2+12y+36=25$ or $x^2+y^2+12y+11=0$, but the original form is usually more useful.

Example 1.2.1 Graph the circle $x^2-2x+y^2+4y-11=0$. With a little thought we convert this to $(x-1)^2+(y+2)^2-16=0$ or $(x-1)^2+(y+2)^2=16$. Now we see that this is the circle with radius 4 and center $(1,-2)$, which is easy to graph. $\square$

Exercises 1.2

Ex 1.2.1 Find the equation of the circle of radius 3 centered at:

(answer)

Ex 1.2.2 For each pair of points $A(x_1,y_1)$ and $B(x_2,y_2)$ find (i) $\Delta x$ and $\Delta y$ in going from $A$ to $B$, (ii) the slope of the line joining $A$ and $B$, (iii) the equation of the line joining $A$ and $B$ in the form $y=mx+b$, (iv) the distance from $A$ to $B$, and (v) an equation of the circle with center at $A$ that goes through $B$.

a) $A(2,0)$, $B(4,3)$

d) $A(-2,3)$, $B(4,3)$

b) $A(1,-1)$, $B(0,2)$

e) $A(-3,-2)$, $B(0,0)$

c) $A(0,0)$, $B(-2,-2)$

f) $A(0.01,-0.01)$, $B(-0.01,0.05)$

(answer)

Ex 1.2.3 Graph the circle $x^2+y^2+10y=0$.

Ex 1.2.4 Graph the circle $x^2-10x+y^2=24$.

Ex 1.2.5 Graph the circle $x^2-6x+y^2-8y=0$.

Ex 1.2.6 Find the standard equation of the circle passing through $(-2,1)$ and tangent to the line $3x-2y =6$ at the point $(4,3)$. Sketch. (Hint: The line through the center of the circle and the point of tangency is perpendicular to the tangent line.) (answer)

Distance between two points is the length of the line segment that connects the two given points. Distance between two points in coordinate geometry can be calculated by finding the length of the line segment joining the given coordinates. Let us understand the formula to find the distance between two points in a two-dimensional and three-dimensional plane.

What is the Distance Between Two Points?

The distance between any two points is the length of the line segment joining the points. There is only one line passing through two points. So, the distance between two points can be calculated by finding the length of this line segment connecting the two points. For example, if A and B are two points and if $\overline{AB}=10$ cm, it means that the distance between A and B is 10 cm.

The distance between two points is the length of the line segment joining them (but this CANNOT be the length of the curve joining them). Note that the distance between two points is always positive.

Distance Between Two Points Formula

The distance between two points using the given coordinates can be calculated by applying the distance formula. For any point given in the 2-D plane, we can apply the 2D distance formula or the Euclidean distance formula given as,

Formula for Distance Between Two Points:

The formula for the distance, $d$, between two points whose coordinates are $(x_1, y_1)$ and $(x_2, y_2$) is:

d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2]

This is called the Distance Formula.

To find the distance between two points given in 3-D plane, we can apply the 3D distance formula, given as,

d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2 + ($z_2$ − $z_1$)2]

Let's learn how to derive this formula next.

Derivation of Formula for Distance Between Two Points

To derive the formula to calculate the distance between two points in a two-dimensional plane, let us assume that there are two points with the coordinates given as, A($x_1, y_1$) B($x_2, y_2$)

Next, we will assume that the line segment joining A and B is $\overline{AB}=d$. Now, we will plot the given points on the coordinate plane and join them by a line.

Next, we will construct a right-angled triangle with $\overline{AB}$ as the hypotenuse.

Applying Pythagoras theorem for the △ABC:

AB2 = AC2 + BC2

d2 = ($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2 (Values from the figure)

Here, the vertical distance between the given points is |$y_2$ − $y_1$|.

The horizontal distance between the given points is |$x_2$ − $x_1$|.

d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2] (Taking square root on both sides)

Thus, the distance formula to find the distance between two points is proved.

Note: In case the two points A and B are on the x-axis, i.e. the coordinates of A and B are ($x_1$, 0) and ($x_2$, 0) respectively, then the distance between two points AB = |$x_2$ − $x_1$|.

Using similar steps and concept, we can also derive the formula to find the distance between two points given in the 3D plane.

How to Find Distance Between Two Points?

The distance between two points using the given coordinates can be calculated with the help of the following given steps:

Note down the coordinates of the two given points in the coordinate plane as, A($x_1, y_1$) and B($x_2, y_2$).
We can apply the distance formula to find the distance between the two points, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2]
Express the given answer in units.

Note: We can apply the 3D distance formula in case the two points are given in 3D plane, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2 + ($z_2$ − $z_1$)2]

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

Solution:

The distance between two points using coordinates can be given as, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2], where ($x_1, y_1$) and ($x_2, y_2$) are the coordinates of the two points.

⇒ d = √[(1 − 1)2 + (5 − 2)2]

⇒ d = 3 units

From the above example, we can also observe that when the x-coordinates of the given points are the same, we can find the distance between the two points by finding the difference between the y-coordinates.

Distance Between Two Points in Complex Plane

The distance between two points in a complex plane or two complex numbers z$_1$ = a + ib and z$_2$ = c + id in the complex plane is the distance between points (a, b) and (c, d), given as,

|z$_1$ − z$_2$| = √[(a − c)2 + (b − d)2]

Related topics:

Euclidean Distance Formula
Geometry
x and y axis

Important Notes on distance between two points:

The distance, d, between two points whose coordinates are $(x_1, y_1)$ and $(x_2, y_2$) is: d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2]
Distance of a point (a, b) from: (i) x - axis is |b|. (ii) y - axis is |a|.
We have used the absolute value signs because distance can never be negative.

Example 1: Find the distance between the two points (2, -6) and (7, 3)

Solution:

Let us assume the given points to be:

$(x_1,y_1)$ = (2, -6)
$(x_2, y_2)$ = (7,3)

The formula to find the distance between two points is:

d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2]
d = √[(7−2)2 + (3−(−6))2]
d = √(52 + 92) d = √(25 + 81)
d = √106

∴ Distance =√106

Example 2: Show that the points (2, -1), (0, 1) and (2, 3) are the vertices of a right-angled triangle.

Solution:

Let us assume the given points to be:

A = (2, −1) B = (0, 1)

C = (2, 3)

We will find the distance between every two points using the distance formula.

AB = √[(0−2)2 + (1−(−1))2]
= √[(−2)2 + (2)2] = √(4 + 4)

= √8

BC = √[(2 − 0)2 + (3 − 1)2]
= √[(2)2 + (2)2] = √(4 + 4)

= √8

CA = √[(2 − 2)2 + (3−(−1))2]
= √(02 + 42) = √16

= 4

Now that we know the lengths of all three sides,

AB2 + BC2 = CA2
(√8)2 + (√8)2 = 42 8 + 8 = 16

16 = 16

Thus, A, B and C satisfy the Pythagoras theorem.

So △ABC is a right-angled triangle.

We can prove the same by marking all the coordinates on a graph:

Thus, the given points form a right-angled triangle.

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

Solution:

We know that the x-coordinate of any point on the y-axis is 0.

Hence, we assume the point that is equidistant from the given points to be (0, k). i.e.,

Distance between (0, k) and (-1, 2) = Distance between (0, k) and (2, 3)

√[(−1 − 0)2 + (2 − k)2] = √[(2 − 0)2 + (3 − k)2]

Squaring on both sides,

(−1 − 0)2 + (2 − k)2 = (2 − 0)2 + (3 − k)2
1 + 4 + k2 − 4k = 4 + 9 + k2 − 6k 2k = 8

k = 4

Therefore, the required point is, (0, k) = (0, 4)

∴Required Point = (0, 4)

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FAQs on Distance Between Two Points

The distance between two points is defined as the length of the straight line connecting these points in the coordinate plane. This distance can never be negative, therefore we take the absolute value while finding the distance between two given points.

How Do We Calculate the Distance Between Two points in 2D Plane?

The distance between any two points given in two-dimensional plane can be calculated using their coordinates. Distance between two points A($x_1, y_1$) and B($x_2, y_2$) can be calculated as, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2].

How to Find the Distance Between Two points in 3D Plane?

To calculate the distance between two points in a three-dimensional plane, we can apply the 3D distance formula given as, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2 + ($z_2$ − $z_1$)2], where 'd' is the distance between the two points and ($x_1, y_1, z_1$), ($x_2, y_2, z_2$) are the coordinates of the two points.

What is the Shortest Distance Between Two Points?

The shortest distance between two points can be calculated by finding the length of the straight line connecting both the points. We can apply the distance formula to find this distance depending on the coordinates given in two or three-dimensional plane.

How to Find the Distance Between Two Points Using Pythagorean Theorem?

The distance between two points in the cartesian plane can be calculated by applying the Pythagorean theorem. We can form a right-angled triangle using the line joining the given two points as the hypotenuse. Here the perpendicular and base will be the lines parallel to x and y-axis with one end as one of the given points and the other end as their intersecting point. Using the Pythagoras' theorem, (hypotenuse)2 = (base)2 + (perpendicular)2, we can find the length of the hypotenuse with the help of the given coordinates of two points. This length is equal to the distance between two points.

What is the Distance Formula to Find Distance Between Two Points in Coordinate Geometry?

In coordinate geometry, the distance between two points formula is given as, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2], where, ($x_1, y_1$), ($x_2, y_2$) are the coordinates of the two points. We can apply another formula if the given points liw in 3D plane, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2 + ($z_2$ − $z_1$)2], where 'd' is the distance between the two points and ($x_1, y_1, z_1$), ($x_2, y_2, z_2$) are the coordinates of the two points.

How to Derive the Formula to Find The Distance Between Two Points?

We can apply the Pythagoras theorem to derive the distance between two points formula. We can take the line joining the two points as the hypotenuse of a right triangle formed in the cartesian plane. The length of the hypotenuse can be calculated using the Pythagorean theorem and the given coordinates of two points to derive the distance between two points formula.

How to Find the Vertical Distance Between Two Points?

The vertical distance between two points can be found by calculating the difference of the y coordinates of the two points, i.e., vertical distance between two points, $d_y$ = $y_2 - y_1$, where ($x_1, y_1$), ($x_2, y_2$) are the coordinates of the two points.

What are Steps to Find Euclidean Distance Between Two Points?

The Euclidean distance between two points can be calculated using the following steps,

Note the coordinates of both the given points as ($x_1, y_1$) and ($x_2, y_2$).
Apply the Euclidean distance formula, distance, d = √[($x_2$ − $x_1$)2 + ($y_2$ − $y_1$)2]
Express the given answer in units.