Lines, Angles, & Triangles

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SAT Math › Lines, Angles, & Triangles

Questions 1 - 10

Points A(2, -5) and B(-6, 1) are translated by the vector (-4, 3) and then reflected across the x-axis to A' and B'. What is the midpoint of segment A'B'?

(6, -1)

(-6, 1)

(-6, 5)

(-6, -1)

Explanation

After translating, A and B become (-2, -2) and (-10, 4); reflecting across the x-axis gives (-2, 2) and (-10, -4), whose midpoint is (-6, -1). The other choices come from skipping the reflection, doing transformations in the wrong order, or reflecting across the wrong axis.

Point Q is at (-2, 5). After a dilation centered at the origin with scale factor -2, followed by a translation by the vector (3, -1), what are the coordinates of Q'?

(-2, -8)

(-1, 9)

(7, -11)

(7, -9)

Explanation

Dilating by -2 sends (-2, 5) to (4, -10), then translating by (3, -1) yields (7, -11). The distractors use the wrong order, ignore the negative scale, or misapply the translation signs.

Point P has coordinates (4, -7). After reflection across the line $y=x$, what are the coordinates of its image P'?

(-4, 7)

(7, -4)

(-7, 4)

(4, -7)

Explanation

Reflection across $y=x$ swaps the coordinates, giving (-7, 4). The other choices reflect across an axis or leave the point unchanged.

What is the distance between the points (3, -1) and (-5, 6) in the coordinate plane?

$\sqrt{113}$

$\sqrt{15}$

$\sqrt{128}$

Explanation

By the distance formula, $\sqrt{(3-(-5))^2+(-1-6)^2}=\sqrt{8^2+(-7)^2}=\sqrt{113}$. The other options come from omitting the square root, subtracting squares, or using equal differences.

In the coordinate plane, what is the midpoint of the segment with endpoints (-4, 6) and (2, -2)?

(-2, 4)

(-1, 4)

(-1, 2)

(2, -1)

Explanation

The midpoint is $\left(\frac{-4+2}{2},\frac{6+(-2)}{2}\right)=(-1, 2)$. The distractors come from not dividing by 2, swapping coordinates, or averaging incorrectly.

What is the distance between the points (-4, 1) and (2, -5)?

$8\sqrt{2}$

$6\sqrt{2}$

$6\sqrt{3}$

Explanation

Distance is $\sqrt{(2 - (-4))^2 + (-5 - 1)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{72} = 6\sqrt{2}$. Other choices reflect Manhattan distance or arithmetic errors.

Point A at (3, -4) is reflected across the y-axis to A', and point B at (-5, 2) is translated by the vector (5, 1) to B'. What is the slope of line A'B'?

$\frac{3}{7}$

$-\frac{7}{3}$

$-\frac{3}{7}$

$\frac{7}{3}$

Explanation

A' = (-3, -4) and B' = (0, 3), so slope $=\frac{3-(-4)}{0-(-3)}=\frac{7}{3}$. The other choices are the reciprocal or have incorrect signs.

Point A at (-3, 5) is reflected across the y-axis. What are the coordinates of its image?

(-3, -5)

(3, -5)

(5, -3)

(3, 5)

Explanation

Reflection across the y-axis maps $(x,y)$ to $(-x,y)$, so the image is (3, 5). The other choices correspond to reflecting over the x-axis, swapping coordinates, or reflecting over both axes.

What is the distance between the points (-3, 4) and (5, -2)?

$\sqrt{28}$

Explanation

Distance $=\sqrt{(5-(-3))^2+(-2-4)^2}=\sqrt{8^2+(-6)^2}=\sqrt{100}=10$. Other options use the sum of differences, subtract squares, or only one coordinate difference.

What is the midpoint of the segment with endpoints (5, -1) and (1, 7)?

(2, 3)

(3, -3)

(3, 3)

(6, 6)