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Maths / Coordinate Geometry / Distance Formula

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Distance Formula

If two points A and B have the same y-coordinate then A and B lie on a line parallel to the x-axis and the distance between A and B is the difference between their x-coordinates. See fig. Note that $large AB=left | x_{2}-x_{1} right |$

Similarly, if two points C and D have the same x-coordinate, then C and D lie on a straight line parallel to the y-axis and the distance between them is the difference between their y-coordinates. See fig. Note that $large CD=left | y_{2}-y_{1} right |$

Distance Formula:

The length of the segment AB which joins $large P(x_{1},y_{1})$ and $large Q(x_{2},y_{2})$ is

$large PQ=sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Proof: We plot points P and Q on the Cartesian plane and construct right triangle PQN. Note that N is the point of intersection of the line drawn through P and parallel to the x-axis, and the line through Q and parallel to the y-axis. As PN is parallel to the x-axis, P and N have the same y-coordinates. Thus, y-coordinate of N is $large y_{1}$ .

Next, as QN is parallel to the y-axis, Q and N have the same x-coordinates. Thus, the x-coordinate of N is $large x_{2}$ . $large therefore$ Coordinates of N are $large (x_{2},y_{1}).$

Now, PN = $large left | x_{2}-x_{1} right |$ and QN = $large left | y_{2}-y_{1} right |$

By Pythagoras theorem,

$large PQ^{2}=PN^{2}+QN^{2}$ $large =left | x_{2}-x_{1} right |^{2}+left | y_{2}-y_{1} right |^{2}$

$large =(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

$large Rightarrow ;;PQ=sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Note: Distance of the point P(x,y) from the origin (0,0) is given by $large OP=sqrt{x^{2}+y^{2}}$

Illustration: Find the distance between A(2, -3) and B(-3, -8)

Solution: Here $large x_{1}=2,;;;x_{2}=-3,;;;y_{1}=-3,;;;y_{2}=-8$

We know that $large AB^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$ $large =(-3-2)^{2}+left [ -8-(-3) right ]^{2}$ $large =(-5)^{2}+(-5)^{2}=25+25=50$ $large Rightarrow ;;;;AB=5sqrt{2};;units$

Illustration: If A is the point (x,2), B is (2, -2) and AB = 5 units, find all possible values of x.

Solution: Given A(x,2) and B(2, -2) are the two points. Also AB = 5 $large Rightarrow AB^{2}=25.$

We have,

$large AB^{2}=(2-x)^{2}+(-2-2)^{2}$

$large Rightarrow ;;;;25=x^{2}-4x+4+16Rightarrow x^{2}-4x-5=0$

$large Rightarrow ;;;x^{2}+x-5x-5=0Rightarrow x(x+1)-5(x+1)=0$

$Rightarrow ;;;(x+1)(x-5)=0 Rightarrow either ;x+1=0;or;x-5=0$

$large Rightarrow ;;;;;x=-1,or;5$

Thus, possible values of x are -1 and 5.

Illustration:Find the value of k if the point P(0,2) is equidistant from A(3,k) and B(k,5)

Solution:

$large AP =sqrt{(0-3)^{2}+(2-k)^{2}}$

$large BP=sqrt{(0-k)^{2}+(2-5)^{2}}$

We are given AP = BP

$large Rightarrow$ $large AP^{2}=BP^{2}$

$large Rightarrow$ $large (0-3)^{2}+(2-k)^{2}=(0-k)^{2}+(2-5)^{2}$

$large Rightarrow$ $large 9+4-4k+k^{2}=k^{2}+9Rightarrow 4-4k=0;or;k=1$

Illustration: Find the coordinates of the circumcentre of the triangle whose angular points are A(8,6), B(8, -2) and C(2, -2).

Solution: Let P(x,y) be the circumcentre of $large Delta ABC.$

Then $large AP=BP=CPRightarrow AP^{2}=BP^{2}=CP^{2}.$

As $large AP^{2}=BP^{2},$

we get $large (x-8)^{2}+(y-6)^{2}=(x-8)^{2}+(y+2)^{2}$

$large Rightarrow ;;y^{2}-12y+36=y^{2}+4y+4Rightarrow -16y=-32;or;y=2$

Next, since $large BP^{2}=CP^{2}$

we get $large (x-8)^{2}+(y+2)^{2}=(x-2)^{2}+(y+2)^{2}$

$large Rightarrow$ $large x^{2}-16x+64=x^{2}-4x+4Rightarrow -12x=-60;or;x=5$

Thus, circumcentre of the triangle ABC is (5,2).

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