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Question
If A (-1, 3), B (1, -1) and C (5, 1) are the vertices of a triangle ABC, find the length of the median through A.
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Solution
The distance d between two points `(x_1,y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`
The co-ordinates of the midpoint `(x_m,y_m)` between two points `(x_1, y_1)` and `(x_2, y_2)` is given by,
`(x_m,y_m) = (((x_1 + x_2)/2)"," ((y_1+y_2)/2))`
Here, it is given that the three vertices of a triangle are A(−1,3), B(1,−1) and C(5,1).
The median of a triangle is the line joining a vertex of a triangle to the mid-point of the side opposite this vertex.
Let ‘D’ be the mid-point of the side ‘BC’.
Let us now find its co-ordinates.
`(x_D,y_D) = (((1 + 5)/2)"," ((-1+1)/2))`
`(x_D, y_D) = (3,0)`
Thus we have the co-ordinates of the point as D(3,0).
Now, let us find the length of the median ‘AD’.
`AD = sqrt((-1-3)^2 + (3 - 0)^2)`
`= sqrt((-4)^2 + (3)^2)`
`= sqrt(16 + 9)`
AD = 5
Thus the length of the median through the vertex ‘A’ of the given triangle is 5 units
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By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`
