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In the Coordinate/Euclidean geometry, the distance formula is basically an algebraic expression that is used to find the distance between the two points whose coordinates are given. Distance Formula is a variant of the ‘Pythagoras Theorem’ because the formula of the Pythagoras theorem of geometry is used to derive this formula. Before we move on to the derivation part, let’s recall the following two elements of a graph:
- Abscissa/X-coordinate: The distance of a point from the Y-axis is known as its X coordinate or ‘Abscissa’.
- Ordinate/Y-coordinate: The distance of a point from the X-axis is known as its Y coordinate or ‘Ordinate’.
Distance Formula in Mathematics v/s Physics
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- Distance, in physics, is the actual length of the path traveled by an object whereas distance, in math, refers to the length of the shortest line segment joining two points.
- The distance formula used in coordinate geometry is entirely different from the distance formula that is used in physics.
- The distance formula of physics is used to calculate the total distance of the path traveled between the starting and the endpoint with the help of formula (Distance=Speed*Time) while the distance formula used in coordinate geometry calculates the distance between the starting and the endpoint if the coordinates of those two points are given.
The video below explains this:
Coordinate Geometry Detailed Video Explanation:
Distance Formula
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The coordinates of the two points between which the distance has to be calculated could lie on either the X-axis or Y-axis. To calculate the distance between a point (A) and (B) with coordinates (x1, y1) and (x2, y2) respectively in the XY plane, the distance formula is given by:
\(AB=\sqrt{[(x_{2}-x_{1})^{_{2}}]}\) OR
\(D=\sqrt{[(x_{2}-x_{1})^{_{2}}+(y_{2}-y_{1})^{_{2}}]}\)
where D is the distance between points A and B
Derivation Of Distance Formula
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Let A and B be the two points in the Cartesian plane with coordinates (xA, yA) and (xB, yB) respectively. Through points, A and B draw two lines parallel to both the X and Y axis according to the given figure.
Pythagorean-theorem
Point C is the intersection point of the two lines drawn from points A and B.
Thus ΔABC is a right-angled triangle formed by joining points A, B, and C.
In ΔABC, AC is the Base, BC is the perpendicular and AB is the hypotenuse (distance between points A and B).
Hence, by applying Pythogoras theorem in the right angled ΔABC,
AC2 + BC2 = AB2
\(AB^{2}=\sqrt{[(x_{B}-x_{A_{}})^{_{2}}+(y_{B}-y_{A})^{_{2}}]}\)
\(AB=\sqrt{[(x_{B}-x_{A_{}})^{_{2}}+(y_{B}-y_{A})^{_{2}}]}\)
Thus, the distance formula between points A and B is:
\(AB=\sqrt{[(x_{B}-x_{A_{}})^{_{2}}+(y_{B}-y_{A})^{_{2}}]}\)
Similarly, the distance of point P, having coordinates (x,y), from the origin is given by:
\(D=\sqrt{({x}_{2}+{y}_{2})}\)
Sample Questions
Ques: Find the distance between the two points A and B with coordinates (-5, 7) and (-1, 3) respectively. (2 mark)
Soln: Accordingly, xA = -5 , xB = -1 , yA = 7 , yB = 3
By applying distance formula, we get:
\(AB=\sqrt{[(x_{B}-x_{A_{}})^{_{-2}}+(y_{B}-y_{A})^{_{2}}]}\)
\(AB=\sqrt{[(-1-(-5))^{_{2}}+(3-7)^{_{2}}]}\)
\(AB=\sqrt{(16+16)}\)
\(AB=4\sqrt{2}\)
Ques: Determine whether the given points (1, 5), (2, 3), and (-2, -11) are collinear. (2 marks)
Soln To determine whether the given three points are collinear (lie on the same line) or not, we will first calculate the distance between these three points and then will evaluate whether the largest distance is equal to the sum of the other two distances or not.
According to the question, let the coordinates of the point A, B, and C be (1, 5), (2, 3), and (-2, -11) respectively. Hence according to the distance formula;
\(AB=\sqrt{[(2-1)^{2}+(3-5)^{2}]}=\sqrt{(1+4)}=\sqrt{5}\).
\(AC=\sqrt{(-2-1)^{2}+(-11-5)^{2}}=\sqrt{(9+256)}=\sqrt{(265)}\)
Since, AB + BC is not equal to AC, hence the given three points are not collinear.
Ques: Show that the points (1, 7), (4, 2), (–1, –1), and (– 4, 4) are the vertices of a square. (3 marks)
Soln: To prove that the points A, B, C, and D are the vertices of a square, we need to prove:
AB=BC=CD=DA (sides of the square)
AC=BD (diagonals of a square)
According to the distance formula:
\(AB=\sqrt{(4-1)^{2}+(2-7)^{2}}=\sqrt{(9+25)}=\sqrt{(34)}\)Units
\(BC=\sqrt{(-1-4)^{2}+(-1-2)^{2}}=\sqrt{(25+9)}=\sqrt{(34)}\)Units
\(CD=\sqrt{(-4-(-1))^{2}+(4-(-1))^{2}}=\sqrt{(9+25)}=\sqrt{(34)}\)Units
\(AD=\sqrt{(-4-1)^{2}+(4-7)^{2}}=\sqrt{(25+9)}=\sqrt{(34)}\)Units
\(AC=\sqrt{(1+1)^{2}+(7+1)^{2}}=\sqrt{(4+64)}=\sqrt{(68)}\)Units
\(BD=\sqrt{(4+4)^{2}+(2-4)^{2}}=\sqrt{(64+4)}=\sqrt{(68)}\)Units
Since, AB=BC=CD=DA and AC=BD, therefore, points A, B, C and D are the vertices of a square.
Ques: If the point P(k,0) divides a line, that joins points A(2, -2) and B(-7, 4) in the ratio 1:2, what is the value of k. (CBSE 2020) (4 marks)
Soln: According to the given question:
\(\frac{PA}{PB}=\frac{1}{2}\)
By Distance Formula,
\(PA=\sqrt{[(2-k)^{2}+(-2-0)^{2}]}=\sqrt{[4+(2-k)^{2}]}\)
Similarly,
\(PB=\sqrt{[(-7-k)^{2}+(4-0)^{2}]}=\sqrt{[16+(-7-k)^{2}]}\)
Therefore, \(\frac{PA}{PB}=\frac{\sqrt{[4+(2-k)^{2}]}}{\sqrt{[16+(-7-k)^{2}]}}=\frac{1}{2}\)
\(\sqrt{4}\times \sqrt{[4+(2-k)^{2}]}=\sqrt{[16+(-7-k)^{2}]}\)
4\(\times\) (4 + (2 – k)2) = 16 + (-7 – k)2
4\(\times\) (8 - 4k + k2) = 16 + (49 + 14k + k2)
32 – 16k + 4k2 = 65 +14k + k2
32 - 65 - 16k- 14k + 4k2 - k2 = 0
3k2 - 30k- 33 = 0
k2 – 10k – 11 = 0
k2 + k - 11k - 11 = 0
k (k + 1)- 11 (k + 1) = 0
Therefore k = – 1 or k = 11
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