In the standard (x, y) coordinate plane, a triangle has vertices A(1, 2), B(7, 2), and C. … — ACT

In the standard $(x, y)$ coordinate plane, a triangle has vertices $A(1, 2)$ , $B(7, 2)$ , and $C$ . The midpoint of side $AC$ lies on the line $y = 3x - 3$ , and the midpoint of side $BC$ lies on the line $y = -2x + 13$ . What is the distance between point $C$ and the midpoint of side $AB$ ?

A
3
B
4
$\sqrt{5}$ Answer
D
$2\sqrt{2}$
E
$2\sqrt{5}$

Answer

The distance between point

C

and the midpoint of side

AB

\sqrt{5}

The coordinates of point

C(3, 4)

are determined by setting up the midpoint coordinates for sides

AC

and

BC

and substituting them into their respective line equations. The midpoint of

AB

is calculated to be

(4, 2)

. Using the distance formula between

C(3, 4)

and

(4, 2)

yields

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

Step-by-Step Solution

Express the midpoint of side

AC

in terms of the unknown coordinates of point

C(x, y)

and substitute it into the given line equation.

The midpoint of

AC

M_{AC} = \left(\frac{x+1}{2}, \frac{y+2}{2}\right)

. Substituting this into

y = 3x - 3

yields:

\frac{y+2}{2} = 3\left(\frac{x+1}{2}\right) - 3

, which simplifies to

y = 3x - 5

Since the midpoint of

AC

lies on the line

y = 3x - 3

, its coordinates must satisfy the equation of the line.

Express the midpoint of side

BC

in terms of the unknown coordinates of point

C(x, y)

and substitute it into the second given line equation.

The midpoint of

BC

M_{BC} = \left(\frac{x+7}{2}, \frac{y+2}{2}\right)

. Substituting this into

y = -2x + 13

yields:

\frac{y+2}{2} = -2\left(\frac{x+7}{2}\right) + 13

, which simplifies to

y = -2x + 10

Since the midpoint of

BC

lies on the line

y = -2x + 13

, its coordinates must satisfy this equation.

Solve the system of two linear equations to find the coordinates of point

C(x, y)

Equating the two expressions for

y

3x - 5 = -2x + 10 \Rightarrow 5x = 15 \Rightarrow x = 3

. Substituting

x = 3

back into the first equation:

y = 3(3) - 5 = 4

. Thus,

C = (3, 4)

Point

C

must simultaneously satisfy the midpoint constraints on both sides

AC

and

BC

Find the coordinates of the midpoint of side

AB

The midpoint of segment

AB

with endpoints

A(1, 2)

and

B(7, 2)

M_{AB} = \left(\frac{1+7}{2}, \frac{2+2}{2}\right) = (4, 2)

The question asks for the distance between point

C

and the midpoint of

AB

, so we need to determine the coordinates of this midpoint first.

Calculate the distance between point

C(3, 4)

and the midpoint

M_{AB}(4, 2)

using the distance formula.

The distance

d

is:

d = \sqrt{(4-3)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}

The distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

is used to find the straight-line distance between two points in a coordinate plane.

Key Concept

Applying the midpoint and distance formulas within coordinate geometry constraint systems.

Alternative Method

Instead of algebraically solving for the lines of midpoints, one can translate the lines using vectors. The set of possible points

C

when the midpoint of

AC

lies on line

L_1

is a line

L'_1

obtained by dilating

L_1

by a factor of 2 with respect to center

A

. Dilating

y = 3x - 3

from

A(1, 2)

gives the line

y = 3x - 5

. Similarly, dilating

y = -2x + 13

from

B(7, 2)

by a factor of 2 gives

y = -2x + 10

. The intersection of these two dilated lines is point

C(3, 4)

Estimated Time:2m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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