When finding the area of a triangle on a coordinate plane, you need to identify the base and height, which must be perpendicular to each other. For right triangles, this is easier because two sides already meet at a 90° angle.

First, plot the points: $$S(-2, 3)$$, $$T(1, 3)$$, and $$U(1, -1)$$. Notice that points $$S$$ and $$T$$ have the same y-coordinate (3), so they form a horizontal line. Points $$T$$ and $$U$$ have the same x-coordinate (1), so they form a vertical line. These perpendicular sides meet at point $$T$$, confirming this is a right triangle.

Now calculate the lengths. The horizontal side $$ST$$ has length $$|1 - (-2)| = 3$$ units. The vertical side $$TU$$ has length $$|3 - (-1)| = 4$$ units. Using the formula Area = $$\frac{1}{2} \times \text{base} \times \text{height}$$, we get $$\frac{1}{2} \times 3 \times 4 = 6$$ square units.

Answer choice A incorrectly uses base 4 and height 6, possibly from mixing up the side lengths or including the hypotenuse. Answer choice B uses base 4 and height 4, likely from miscalculating one of the side lengths. Answer choice D uses base 5 and height 4, where 5 appears to be the length of the hypotenuse (which shouldn't be used for area calculations in this method).

Study tip: When finding areas of triangles on coordinate grids, look for horizontal and vertical sides first—they're the easiest to work with as base and height since they're automatically perpendicular.

When you see a coordinate movement problem, you need to track each step carefully by updating the x and y coordinates separately. Think of moving on a grid: right/left changes x-coordinates, while up/down changes y-coordinates.

Starting at point P$$(-4, -1)$$, let's follow each move:

First, "right 6 units" means adding 6 to the x-coordinate: $$-4 + 6 = 2$$. So P moves to $$(2, -1)$$.

Next, "up 3 units" means adding 3 to the y-coordinate: $$-1 + 3 = 2$$. Now P is at $$(2, 2)$$.

Finally, "left 2 units" means subtracting 2 from the x-coordinate: $$2 - 2 = 0$$. P's final position is $$(0, 2)$$.

To find the distance between P's final position $$(0, 2)$$ and Q$$(1, 2)$$, notice both points have the same y-coordinate (both are at height 2). This means they lie on a horizontal line, so the distance is simply the difference in x-coordinates: $$|1 - 0| = 1$$ unit.

Choice A incorrectly calculates P's final position as $$(-2, 2)$$, likely by making an error in the rightward movement. Choice C shows $$(4, 2)$$, probably forgetting the final leftward move. Choice D gives $$(-1, 2)$$, which suggests confusion about the direction of the final move.

Study tip: Always work through movement problems step-by-step, updating coordinates after each move. When finding distance between points with the same x or y coordinate, you can skip the distance formula and just find the difference along the varying coordinate.

When you see a coordinate plane problem involving paths between points, break it down step by step by tracking your movement along each axis separately.

Let's trace this treasure hunt journey. You start at the origin $$(0, 0)$$ and must go to checkpoint $$C(5, 0)$$, then to the treasure at $$(5, -2)$$.

From $$(0, 0)$$ to $$(5, 0)$$: You move 5 units east (positive x-direction) while staying at the same y-coordinate. That's 5 units of distance.

From $$(5, 0)$$ to $$(5, -2)$$: You move 2 units south (negative y-direction) while your x-coordinate stays the same. Since you're going from y = 0 to y = -2, that's a distance of 2 units.

Total distance: 5 + 2 = 7 units.

Answer A incorrectly states you go south 5 units instead of 2 units, leading to a wrong total of 10. Answer C miscalculates the eastward movement as 3 units instead of 5, and the southward movement as 5 instead of 2. Answer D gets both directions wrong, claiming 7 units east and 5 units south.

The correct answer is B: 7 units total, going east 5 units then south 2 units.

Study tip: Always find the distance between coordinates by subtracting the smaller coordinate from the larger one. For horizontal movement, use x-coordinates; for vertical movement, use y-coordinates. Remember that moving to negative numbers still counts as positive distance!

First, find the distance from A to B using the absolute value formula: $$|2 - (-5)| = |7| = 7$$. Point C must be 7 units away from A along the vertical line $$x = -3$$. From A at $$(−3, 2)$$, moving 7 units up gives $$(-3, 9)$$ and moving 7 units down gives $$(-3, -5)$$. Choice B incorrectly calculates one distance as 6 instead of 7. Choice C uses distances of 5 and 5 instead of 7. Choice D uses distances of 3 and 3 instead of 7.

When you see points with the same x-coordinate, you're dealing with a vertical line, which makes finding distances much simpler. This question tests your understanding of the coordinate plane, distance on vertical lines, and quadrant identification.

Since both points M and N have the same x-coordinate (4), they lie on a vertical line. The distance between them is simply the absolute difference of their y-coordinates: $$|y - (-2)| = |y + 2| = 9$$. Since we're told $$y > 0$$, we know $$y + 2 > 2$$, so $$y + 2 = 9$$, which gives us $$y = 7$$. Therefore, point N is at $$(4, 7)$$.

To determine the quadrant, remember that Quadrant I contains points where both x and y are positive. Since N is at $$(4, 7)$$, and both coordinates are positive, N is in Quadrant I.

Option A incorrectly calculates $$y = -11$$ by subtracting 9 from -2 instead of solving the distance equation properly. Point $$(4, -11)$$ would indeed be in Quadrant IV, but this isn't point N.

Option B gives a completely different point $$(-5, 4)$$ that doesn't even lie on the vertical line $$x = 4$$ where N must be located.

Option D also provides an incorrect point $$(-4, -2)$$ that changes the x-coordinate and doesn't satisfy our distance requirement.

Study tip: When points share the same x-coordinate (vertical line) or y-coordinate (horizontal line), distance equals the absolute difference of the other coordinate. Always check your quadrants: I (+,+), II (-,+), III (-,-), IV (+,-).

When you see a rectangle with given vertices on a coordinate plane, you need to find the lengths of the sides by calculating distances between adjacent points, then use the perimeter formula.

To find the side lengths, look at the coordinates systematically. From $$A(2, 1)$$ to $$B(2, -3)$$, notice the x-coordinates stay the same (both are 2), so this is a vertical line. The length is the difference in y-coordinates: $$|1 - (-3)| = 4$$ units. From $$B(2, -3)$$ to $$C(-1, -3)$$, the y-coordinates stay the same (both are -3), so this is a horizontal line. The length is $$|2 - (-1)| = 3$$ units. Since opposite sides of a rectangle are equal, you have two sides of length 4 and two sides of length 3. The perimeter is $$2(4) + 2(3) = 14$$ units.

Choice A incorrectly calculates one side as 5 instead of 4, likely from an arithmetic error when finding $$|1 - (-3)|$$. Choice B claims both sides are 3 units, missing that the vertical sides are actually 4 units long. Choice C states the sides are 2 and 3, probably confusing the coordinate differences with actual distances or making calculation errors.

Study tip: When finding distances on a coordinate plane, if two points share the same x-coordinate, subtract the y-coordinates. If they share the same y-coordinate, subtract the x-coordinates. Always take the absolute value to get a positive distance, and remember that rectangles have two pairs of equal opposite sides.

This question tests quadrant identification by plotting points and checking coordinate signs. R(7,1) in I, S(-2,9) in II (x negative, y positive), T(-6,-4) in III, U(4,-8) in IV; thus, Quadrant II is S(-2,9), choice D. This demonstrates all quadrants without distance elements. Errors could stem from sign confusion, like mistaking IV for II. Practice with mixed signs reinforces quadrant rules. Plotting helps visualize the plane's division. Real-world uses include navigation systems assigning zones.

This question tests graphing points in all four quadrants and identifying their locations without calculating distances. Graphing: plot ordered pair (x,y) by moving x horizontally from origin (left if negative, right if positive), then y vertically (down if negative, up if positive), marks point in appropriate quadrant (signs determine which: I both +, II x− y+, III both −, IV x+ y−). For points A(3,5) in Quadrant I (both positive), B(-4,2) in II (x negative, y positive), C(-3,-6) in III (both negative), D(2,-4) in IV (x positive, y negative), matching choice A. Correct graphing involves checking signs carefully: positive x right, negative left, positive y up, negative down. Common errors include reversing signs (e.g., thinking (-4,2) is in IV) or confusing quadrants (mixing II and IV). Practice plotting with all sign combinations ensures understanding of the full coordinate plane. Real-world applications include mapping locations on grids like murals or games.

This question tests graphing points in all four quadrants and calculating vertical distances on a grid representing city blocks (|y₂-y₁| when same x). Graphing: plot ordered pair (x,y) by moving x horizontally from origin (left if negative, right if positive), then y vertically (down if negative, up if positive), marks point in appropriate quadrant (signs determine which: I both +, II x− y+, III both −, IV x+ y−). For (−3,2) and (−3,-5), same x=-3, vertical distance |2-(-5)|=|7|=7 blocks, matching choice B. Correct calculation uses absolute value for positive distance along vertical street. Common errors: no absolute value (2-(-5)=7, but reverse gives -7 claimed as distance), or arithmetic mistake like |2+5|=7 but misadding. Distance tips: for same x, |y difference|; real-world like navigating city grids vertically.

This question tests calculating horizontal distances between points with same y-coordinate using absolute value. Points with same y are on a horizontal line, distance = |x₂ - x₁|, always positive. For R(-4,3) and S(2,3) on y=3, distance = |2 - (-4)| = |6| = 6 units, matching choice B. Correct calculation uses x-difference with absolute value. Errors like |2 + (-4)| = | -2 | =2, or using y (|3-3|=0), or no absolute value. Distance steps: check same y, use |x₂ - x₁|. Real-world application: measuring east-west distances on maps.