In the standard (x,y) coordinate plane, two opposite vertices of a square are (1, 2) and (… — ACT

In the standard $(x,y)$ coordinate plane, two opposite vertices of a square are $(1, 2)$ and $(4, 6)$ . If all four vertices of the square lie in the first quadrant, what is the $x$ -coordinate of the vertex that is closest to the $y$ -axis?

Answer: 0.5

Answer

The correct answer is

0.5

. The vertex closest to the

y

-axis is

(0.5, 5.5)

, which has an

x

-coordinate of

0.5

The diagonals of a square are perpendicular, equal in length, and bisect each other. Using the given opposite vertices

(1, 2)

and

(4, 6)

, we find the midpoint to be

(2.5, 4)

. The vector between them is

(3, 4)

with length

5

. A perpendicular vector of length

5

(-4, 3)

. Adding and subtracting half of this vector,

(2, -1.5)

, from the midpoint yields the other two vertices:

(0.5, 5.5)

and

(4.5, 2.5)

. Since all four vertices are in the first quadrant, we compare their

x

-coordinates:

1

4

0.5

, and

4.5

. The smallest

x

-coordinate is

0.5

, which represents the vertex closest to the

y

-axis.

Step-by-Step Solution

Find the midpoint of the given diagonal.

The midpoint is

M(2.5, 4)

The diagonals of a square bisect each other at their common midpoint.

Determine the vector representing the given diagonal

AB

and its length.

\vec{AB} = (3, 4)

and its length is

5

The vector is found by subtracting coordinates:

(4 - 1, 6 - 2) = (3, 4)

, and its length is

\sqrt{3^2 + 4^2} = 5

Find a perpendicular vector of the same length to represent the other diagonal.

A perpendicular vector is

(-4, 3)

The dot product of

(3, 4)

and

(-4, 3)

3(-4) + 4(3) = 0

, and its length is

\sqrt{(-4)^2 + 3^2} = 5

Calculate the coordinates of the other two vertices of the square.

The vertices are

C(0.5, 5.5)

and

D(4.5, 2.5)

The vertices are located at

M \pm \frac{1}{2}\vec{CD}

, which gives

(2.5, 4) \pm ( -2, 1.5 )

Determine which of the four vertices is closest to the

y

-axis and identify its

x

-coordinate.

The vertex closest to the

y

-axis is

C(0.5, 5.5)

, and its

x

-coordinate is

0.5

The distance to the

y

-axis is the

x

-coordinate of the point. Comparing the

x

-coordinates

1

4

0.5

, and

4.5

, the smallest value is

0.5

Key Concept

Properties of diagonals of a square on a coordinate plane, including midpoint and perpendicularity.

Alternative Method

Instead of using vectors, one can set up a system of equations. Let

(x, y)

be one of the unknown vertices. Since it forms a right isosceles triangle with the midpoint

(2.5, 4)

and has distance

2.5

from it along a line with slope

-3/4

, we can write the equation of the line as

y - 4 = -0.75(x - 2.5)

and use the distance formula

(x - 2.5)^2 + (y - 4)^2 = 2.5^2

to solve for

x

and

y

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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