In the standard coordinate plane, two opposite vertices of a square are and . If all four vertices of the square lie in the first quadrant, what is the -coordinate of the vertex that is closest to the -axis?
Cevap: 0.5
Cevap
The correct answer is . The vertex closest to the -axis is , which has an -coordinate of .
The diagonals of a square are perpendicular, equal in length, and bisect each other. Using the given opposite vertices and , we find the midpoint to be . The vector between them is with length . A perpendicular vector of length is . Adding and subtracting half of this vector, , from the midpoint yields the other two vertices: and . Since all four vertices are in the first quadrant, we compare their -coordinates: , , , and . The smallest -coordinate is , which represents the vertex closest to the -axis.
Adım Adım Çözüm
Anahtar Kavram
Properties of diagonals of a square on a coordinate plane, including midpoint and perpendicularity.
Alternatif Yöntem
Instead of using vectors, one can set up a system of equations. Let be one of the unknown vertices. Since it forms a right isosceles triangle with the midpoint and has distance from it along a line with slope , we can write the equation of the line as and use the distance formula to solve for and .
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