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Distance Formula With Examples and Applications

The Distance Formula and its Applications 



The Distance Formula and its Applications, Pythagorean Theorem


The Distance Formula is formula for finding the distance between two points on a plane or in space. This formula can be derived from the Pythagorean Theorem. We will prove the formula using Pythagorean theorem and then we do some examples to clarify the concept of the distance formula.


Watch the video on distance formula by Khan Academy


Proof of The Distance Between Two Points

Let us consider two points on two dimensional co-ordinate plane be
P $({{x}_{1}},{{y}_{1}})$ and Q $({{x}_{2}},{{y}_{2}})$

Take two mutually perpendicular lines as the coordinate axis with O as origin. Mark the points P$({{x}_{1}},{{y}_{1}})$ and Q $({{x}_{2}},{{y}_{2}})$. Draw lines PA, QB perpendicular to X-axis from the points P and Q, which meet the X-axis in points A and B, respectively.



Draw lines PC and QD perpendicular to Y-axis, which meet the Y-axis in C and D, respectively. Produce CP

to meet BQ in R. Now

OA = abscissa of P = ${{x}_{1}}$

Similarly, OB = ${{x}_{2}}$, OC = ${{y}_{1}}$ and OD =${{y}_{2}}$

Therefore, we have

PR = AB = OB - OA = ${{x}_{2}}$ - ${{x}_{1}}$ = Change in x

Similarly, QR = QB - RB = QB - PA = ${{y}_{2}}$- ${{y}_{1}}$ = Change in y

Now, using Pythagoras Theorem, in right angled triangle PRQ, we have

$P{{Q}^{2}}$ = $P{{R}^{2}}$ + $R{{Q}^{2}}$

or $P{{Q}^{2}}$ = ${{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}$

Since the distance or length of the line-segment PQ is always non-negative, on taking the positive square root, we get the distance as

$PQ=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

This Formula is known as The Distance Formula.

Corollary : The distance of a point P$({{x}_{1}},{{y}_{1}})$ from the origin (0,0) is given by

$OP=\sqrt{x_{1}^{2}+y_{1}^{2}}$ 

NOTE: There is no change in the outcome of the value of distance if we interchange (swap) the given two points, because the values are squared 

Some useful facts related to distance formula:

1. In questions relating to geometrical figures, take the given vertices in the given order and proceed as indicated.

(i) For an isosceles triangle - We have to prove that at least two sides are equal.

(ii) For an equilateral triangle - We have to prove that three sides are equal.

(iii) For a right -angled triangle - We have to prove that the sum of the squares of two sides is equal to the square of the third side.

(iv) for a square - We have to prove that the four sides are equal, two diagonals are equal.

(v) For a rhombus - We have to prove that four sides are equal (and there is no need to establish that two diagonals are unequal as the square is also a rhombus).

(vi) For a rectangle - We have to prove that the opposite sides are equal and two diagonals are equal.

(vii) For a Parallelogram - We have to prove that the opposite sides are equal (and there is no need to establish that two diagonals are unequal sat the rectangle is also a parallelogram).

2. For three points to be collinear - We have to prove that the sum of the distances between two pairs of points is equal to the third pair of points.

Examples Based on Applications of The Distance Formula



Example 1: Find the distance between the points (5 , -2) and (2, 3).

Solution: Let the points (5, -2) and (2, 3) be denoted by P and Q, respectively.

 
Distance Formula


Then, by distance formula, we obtain the distance PQ as

$PQ=\sqrt{{{(2-5)}^{2}}+{{(3+2)}^{2}}}$.

$=\sqrt{{{(-3)}^{2}}+{{(5)}^{2}}}=\sqrt{34}\,\,unit$.


Example 2: Prove that the points $(1,-1),\left( -\frac{1}{2},\frac{1}{2} \right)$ and (1, 2) are the vertices of an isosceles triangle.

Solution: Let the point (1, -1), $\left( -\frac{1}{2},\frac{1}{2} \right)$and (1, 2) be denoted by P, Q and R, respectively. 

Now by using distance formula we get, 

$PQ=\sqrt{{{\left( -\frac{1}{2}- \right)}^{2}}+{{\left( \frac{1}{2}+1 \right)}^{2}}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$

$QR=\sqrt{{{\left( 1+\frac{1}{2} \right)}^{2}}+{{\left( 2-\frac{1}{2} \right)}^{2}}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$

$PR=\sqrt{{{(1-1)}^{2}}+{{(2+1)}^{2}}}=\sqrt{9}=3$

From the above, we see that PQ = QR 

Hence we can say that the triangle is isosceles.


Example 3: Using distance formula, show that the points (-3, 2), (1, -2) and (9, -10) are collinear.

Solution: Let the given points (-3, 2), (1, -2) and (9, -10) be denoted by A, B and C, respectively. Points A, B and C will be collinear, if the sum of the lengths of two line-segments is equal to the third. 

Using distance formula we get,

Now, $AB=\sqrt{{{(1+3)}^{2}}+{{(-2-2)}^{2}}}=\sqrt{16+16}=4\sqrt{2}$

BC = $\sqrt{{{(9-1)}^{2}}+{{(-10+2)}^{2}}}=\sqrt{64+64}=8\sqrt{2}$

AC = $\sqrt{{{(9+3)}^{2}}+{{(-10-2)}^{2}}}=\sqrt{144+144}=12\sqrt{2}$

Since, AB + BC = $4\sqrt{2}+8\sqrt{2}=12\sqrt{2}=AC$, the, points A, B and C are collinear.



Example 4: Find a point on the X-axis which is equidistant from the points (5, 4) and (-2, 3).

Solution: Since the required point (say P) is on the X-axis, its ordinate will be zero. Let the abscissa of the point be x.

Therefore, coordinates of the point P are (x, 0).

Let A and B denote the points (5, 4) and (-2, 3), respectively.

Since we are given that AP = BP, we have

$A{{P}^{2}}$ = $B{{P}^{2}}$ 

Using distance formula we get,

i.e., ${{(x-5)}^{2}}+{{(0-4)}^{2}}$ = ${{(x+2)}^{2}}+{{(0-3)}^{2}}$

or ${{x}^{2}}$ + 25 - 10x + 16 = ${{x}^{2}}$ + 4 + 4x + 9

or -14x = - 28

or x = 2

Thus, the required point is (2, 0).



Example 5: The vertices of a triangle are (-2, 0) , (2, 3) and (1, -3). Is the triangle equilateral, isosceles or scalene?

Solution: Let the points (-2, 0), (2, 3) and (1, -3) be denoted by A, B and C respectively. 

Using distance formula we get,

AB = $\sqrt{{{(2+2)}^{2}}+{{(3-0)}^{2}}}=5$

BC = $\sqrt{{{(1-2)}^{2}}+{{(-3-3)}^{2}}}=\sqrt{37}$

and AC = $\sqrt{{{(1+2)}^{2}}+{{(-0-0)}^{2}}}=3\sqrt{2}$

Clearly, AB$\ne $BC$\ne $AC.

Therefore, ABC is a scalene triangle.



Example 6: The length of a line-segments is 10. If one end is at (2, -3) and the abscissa of the second end is 10, show that its ordinate is either 3 or -9.

Solution: Let (2, -3) be the point A. let the ordinate of the second end B be y. Then its coordinates will be (10, y). 

Using distance formula we get,

$AB=\sqrt{{{(10-2)}^{2}}+{{(y+3)}^{2}}}=10\,\,(Given)$

or 64 + 9 + ${{y}^{2}}$ + 6y = 100

or ${{y}^{2}}$ + 6y + 73 - 100 = 0

or ${{y}^{2}}$ + 6y - 27 = 0

or (y + 9) (y - 3) = 0

Therefore, y = 9 or y = 3.



Example 7: Show that the points (-2, 5), (3, -4) and (7, 10) are the vertices of a right triangle.

Solution: Let the three points be A(-2, 5), B(3, - 4) and C(7, 10). 


Using distance formula we get,


Then $A{{B}^{2}}={{(3+2)}^{2}}+{{(-4-5)}^{2}}$ = 106

$B{{C}^{2}}={{(7-3)}^{2}}+{{(10+4)}^{2}}$= 212

$A{{C}^{2}}={{(7+2)}^{2}}+{{(10-5)}^{2}}$= 106

We see that

$B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$

212 = 106 + 106

212 = 212

Hence by using Pythagorean Theorem, angle A is 90 degree

Thus, ABC is a right triangle, right angled at A.



Example 8: If the distance of P (x, y) from A (5, 1) and B(-1, 5) are equal, prove that 3x = 2y.

Solution: P(x, y), A (5, 1) and B (-1, 5) are the given points.

AP = BP (Given)

Squaring both sides, $A{{P}^{2}}$ = $B{{P}^{2}}$ 

Using distance formula

or ${{(x-5)}^{2}}+{{(y-1)}^{2}}$= ${{(x+1)}^{2}}+{{(y-5)}^{2}}$

or ${{x}^{2}}$ + 25 - 10x + ${{y}^{2}}$ + 1 -2y -${{x}^{2}}$ - 1 -2x - ${{y}^{2}}$ - 25 + 10y = 0

or -12x + 8y = 0

or 3x = 2y.


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