Three vertices of a square are A(-1, -9), B(3, -1) and C(-5, 3). Plot the points. Then find the coordinates of the missing vertex D
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Given :- Three vertices of a square are A(-1, -9), B(3, -1) and C(-5, 3) .
To Find :- find the coordinates of the missing vertex D ?
Concept and formula used :-
- Diagonals of a square are equal in length and bisect each other.
- Mid - Point formula :- if (x,y) is equidistant from (x1,y1) and (x2,y2) than, x = (x1 + x2)/2 and y = (y1 + y2)/2 .
Solution :-
Given that, A(-1, -9), B(3, -1) and C(-5, 3) .
Let coordinates of D are (x,y).
So,
we told above in square , diagonals are same and bisect each other.
Using mid - point formula we get,
→ mid point of AC = mid point of BD
Putting values we get,
→ {(-1 - 5)/2 , (-9 + 3)/2} = {(3 + x)/2 , (-1 + y)/2}
→ (-3, -3) = {(3 + x)/2 , (-1 + y)/2}
Comparing both we get,
→ (3 + x)/2 = (-3)
→ 3 + x = 2 * (-3)
→ 3 + x = (-6)
→ x = (-6) - 3
→ x = (-9) .
and,
→ (-1 + y)/ 2 = (-3)
→ (-1 + y) = 2 * (-3)
→ y - 1 = (-6)
→ y = (-6) + 1
→ y = (-5)
Hence, coordinates of the missing vertex D are {(-9) , (-5)}.
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