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Maths / Coordinate Geometry / Mid Point Formula

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Mid-Point Formula

Coordinates of the mid-point M (x,y) of the segment AB are obtained by taking $m_1 = m_2$ in the section formula:

$x=frac{m_{2}x_{1}+m_{1}x_{2}}{m_{1}+m_{2}};and;y=frac{m_{2}y_{1}+m_{1}y_{2}}{m_{1}+m_{2}}$

$x=frac{m_{1}x_{1}+m_{1}x_{2}}{m_{1}+m_{1}};and;y=frac{m_{1}y_{1}+m_{1}y_{2}}{m_{1}+m_{1}}$

$x=frac{m_{1}(x_{1}+x_{2})}{2m_{1}};and;y=frac{m_{1}(y_{1}+y_{2})}{2m_{1}}$

$large x=frac{1}{2}(x_{1}+x_{2}),$ $large y=frac{1}{2}(y_{1}+y_{2}),$

Illustration: Find the distance of the point (1,2) from the mid-point of the line segment joining the points (6,8) and (2,4).

Solution: Let the points (1,2), (6,8), and (2,4) be denoted by A, B, and C respectively. Let M be the mid-point of BC. Coordinates of M, the mid-point of BC, are given by

$large left ( frac{6+2}{2},frac{8+4}{2} right )=(4,6)$

We have $large AM^{2}=(4-1)^{2}+(6-2)^{2}$

$large =3^{2}+4^{2}=9+16=25$

$large Rightarrow$ $large AM=sqrt{25}=5;units$

Illustration: Find the lengths of medians of the triangle with vertices A(2,2), B(0,2), and C(2,-4).

Solution: Let the coordinates of A, B, and C be (2, 2), (0,2), and (2, -4) respectively. Let D, E, F be the mid-points of BC, CA, and AB respectively.

Then the coordinates of D, E, F are given by

$large Dleft ( frac{0+2}{2},frac{2-4}{2} right )$ , $large large Eleft ( frac{2+2}{2},frac{2-4}{2} right )$ , $large Fleft ( frac{0+2}{2},frac{2+2}{2} right )$

or D (1, -1) E ( 2, -1) , F (1, 2)

$large therefore$ Length of median AD= $large sqrt{(1-2)^{2}+(-1-2)^{2}}=sqrt{10}$

Length of median BE= $large sqrt{(2-0)^{2}+(-1-2)^{2}}=sqrt{13}$

Length of median CF = $large sqrt{(1-2)^{2}+[2-(-4)]^{2}}=sqrt{37}$

$large therefore$ Lengths of medians are $large sqrt{10},sqrt{13},sqrt{37};;units$ .

Illustration: Three consecutive vertices of a parallelogram are A(1,2), B(1,0) C(4,0). Find the fourth vertex D.

Solution: Let coordinates of D be (x,y). As ABCD is a parallelogram, the diagonals AC and BD bisect each other. If M is the mid-point of AC, then coordinates of M are $large left ( frac{1+4}{2},frac{2+0}{2} right )=left ( frac{5}{2},1 right )$

As M is the mid-point of BD also, coordinates of M are $large left ( frac{1+x}{2},frac{0+y}{2} right )=left ( frac{1+x}{2},frac{y}{2} right ).$

We have $large left ( frac{1+x}{2},frac{y}{2} right )=left ( frac{5}{2},1 right ) Rightarrow frac{1+x}{2}= frac{5}{2};and;frac{y}{2}=1$

$large Rightarrow 1+x=5;and;y=2Rightarrow x=4,y=2$

$large therefore$ Coordinates of D are (4,2)

Coordinates of Centroid of Triangle

The centroid of a triangle is the point of concurrence of the medians of a triangle.

If G is the centroid of triangle ABC, then G divides the median AD in the ratio 2:1.

Let D be the mid-point of BC, the coordinates of D are $large left ( frac{x_{2}+x_{3}}{2},frac{y_{2}+y_{3}}{2} right ).$

As G(x,y) divides AD in the ratio 2 : 1

$large x=frac{1(x_{1})+2left ( frac{x_{2}+x_{3}}{2} right )}{2+1}=frac{x_{1}+x_{2}+x_{3}}{3}$

and $large y=frac{1(y_{1})+2left ( frac{y_{2}+y_{3}}{2} right )}{2+1}=frac{y_{1}+y_{2}+y_{3}}{3}$

Thus, the coordinates of the centroid are

$large left ( frac{x_{1}+x_{2}+x_{3}}{3},frac{y_{1}+y_{2}+y_{3}}{3} right )$

Illustration: Find the third vertex of the triangle whose two vertices are (-3,1) and (0,-2) and the centroid is the origin.

Solution: Let the two given vertices be A(-3,1) and B(0,-2). Let the third vertex be C(x,y) . Coordinates of the centroid of $large Delta ABC$ is given by

$large left ( frac{-3+0+x}{3},frac{1-2+y}{3} right )=left ( frac{-3+x}{3},frac{-1+y}{3} right )$ .

As the centroid of $large Delta ABC$ is given to be the origin,

we have $large left ( frac{-3+x}{3},frac{-1+y}{3} right )=(0,0)$

$large Rightarrow ;;-3+x=0,;;-1+y=0$

$large Rightarrow ;;x=3,;;y=1$

Thus, the coordinates of the third vertex are (3,1).

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