Question

determine how many units quadrilateral abcd has been shifted. (1 point) quadrilateral abcd has been shifted □ units to the left and □ units up.

Explanation:

Step1: Find horizontal shift (left)

Take point \( A(1, -1) \) and \( A'(-5, 3) \). Horizontal shift: \( 1 - (-5) = 6 \)? Wait, no—left shift means \( x \)-coordinate decreases. \( A \) has \( x = 1 \), \( A' \) has \( x = -5 \). So shift left: \( 1 - (-5) = 6 \)? Wait, no, the difference is \( 1 - (-5) = 6 \)? Wait, actually, to find how many units left, calculate \( x_A - x_{A'} = 1 - (-5) = 6 \)? Wait, no, left shift: original \( x \) minus new \( x \). Wait, \( A \) is \( (1, -1) \), \( A' \) is \( (-5, 3) \). So horizontal change: \( -5 - 1 = -6 \), meaning 6 units left (since negative x-direction is left).

Step2: Find vertical shift (up)

Vertical shift: \( y_{A'} - y_A = 3 - (-1) = 4 \). So 4 units up.

Wait, let's check another point. Take \( D(6, -3) \) and \( D'(0, 1) \). Horizontal: \( 0 - 6 = -6 \) (6 left). Vertical: \( 1 - (-3) = 4 \) (4 up). Yep, consistent.

Answer:

Quadrilateral \( ABCD \) has been shifted \(\boldsymbol{6}\) units to the left and \(\boldsymbol{4}\) units up.