Question

triangle abc is translated 6 units to the right and 1 units up. what are the coordinates of c? \\( \bigcirc \\) c(4, −2) \\( \bigcirc \\) c(4, −0) \\( \bigcirc \\) c(4, 2) \\( \bigcirc \\) c(1, 2)

Explanation:

Step1: Find original coordinates of C

From the graph, point C has coordinates \((-2, -3)\).

Step2: Apply translation (right 6, up 1)

To translate a point \((x, y)\) 6 units right, we add 6 to the x - coordinate. To translate 1 unit up, we add 1 to the y - coordinate.
For \(x\) - coordinate: \(-2 + 6=4\)
For \(y\) - coordinate: \(-3+1 = - 2\)? Wait, no, wait. Wait, looking at the graph again, maybe I misread C's coordinates. Let's re - examine the graph. The triangle: A is at \((-2,-1)\)? Wait, no, the grid: Let's see, the x - axis and y - axis. Let's find C's coordinates correctly. From the graph, point C: let's count the grid. The x - coordinate: looking at the horizontal lines, C is at \(x=-2\)? Wait, no, maybe I made a mistake. Wait, the original triangle: B is at \((-5,-1)\), A is at \((-2,-1)\), and C is at \((-2,-3)\)? Wait, no, when we translate 6 units right and 1 unit up. Wait, maybe my initial coordinate for C is wrong. Wait, let's look at the options. The options have x = 4. So let's recalculate. Suppose original C is \((-2,-3)\)? No, wait, maybe C is at \((-2,1)\)? No, the graph: the y - axis, the red triangle. Wait, maybe C is at \((-2,-3)\)? Wait, no, when we translate 6 units right (add 6 to x) and 1 unit up (add 1 to y). If original C is \((-2,-3)\), then \(x=-2 + 6 = 4\), \(y=-3 + 1=-2\)? But one of the options is \(C'(4,-2)\)? Wait, no, wait the options: \(C'(4,-2)\), \(C'(4,0)\), \(C'(4,2)\), \(C'(1,2)\). Wait, maybe I misread the original C's coordinates. Let's look again. The graph: the y - axis, the triangle has A at \((-2,-1)\), B at \((-5,-1)\), and C at \((-2,-3)\)? No, that can't be. Wait, maybe C is at \((-2,1)\)? No, the vertical line. Wait, maybe the original C is at \((-2,1)\)? No, the options have y - coordinates. Wait, let's try again. Let's assume that the original coordinates of C are \((-2,1)\)? No, that doesn't fit. Wait, the translation is 6 units right (so \(x\) increases by 6) and 1 unit up (so \(y\) increases by 1). Let's suppose original C is \((-2,1)\): \(x=-2 + 6 = 4\), \(y = 1+1=2\). Then \(C'(4,2)\), which is one of the options. Ah, maybe I misread the original y - coordinate of C. Let's re - examine the graph. The triangle: A is at \((-2,-1)\)? No, the grid lines: the y - axis has 0, 2, 4 on the positive side and - 2, - 4 on the negative side. The red triangle: A is at \((-2,-1)\)? No, maybe A is at \((-2, - 1)\), B at \((-5,-1)\), and C at \((-2,-3)\) is wrong. Wait, maybe the original C is at \((-2,1)\)? No, the y - axis: the numbers are 4,2,0,-2,-4. So if C is at \((-2,-3)\), translating 6 right (x = 4) and 1 up (y=-3 + 1=-2) gives (4,-2). But one of the options is \(C'(4,-2)\). Wait, but let's check the options again. The options are:

1. \(C'(4,-2)\)

2. \(C'(4,0)\)

3. \(C'(4,2)\)

4. \(C'(1,2)\)

Wait, maybe my initial coordinate for C was wrong. Let's assume that the original C is at \((-2,-3)\), then translation gives (4,-2). But let's check the translation rule: when you translate a point \((x,y)\) \(h\) units right and \(k\) units up, the new point is \((x + h,y + k)\). So if \(h = 6\) (right) and \(k = 1\) (up), then new \(x=x_{original}+6\), new \(y=y_{original}+1\).

Suppose original C is \((-2,-3)\), then new \(x=-2 + 6 = 4\), new \(y=-3+1=-2\), so \(C'(4,-2)\). But let's check the options. One of the options is \(C'(4,-2)\). But wait, maybe I made a mistake in the original y - coordinate. Wait, maybe the original C is at \((-2,-3)\), then after translation, \(y=-3 + 1=-2\), \(x=-2 + 6 = 4\), so \(C'(4,-2)\). But let's check the other options. If original C is \((-2,1)\), then \(x=-2 + 6 = 4\), \(y=1 + 1=2\), which is \(C'(4,2)\). But which is correct? Let's look at the graph again. The red triangle: A is at \((-2,-1)\), B at \((-5,-1)\), and C at \((-2,-3)\)? No, that would make the triangle a right triangle with base 3 (from x=-5 to x=-2) and height 2 (from y=-1 to y=-3). When we translate 6 units right and 1 unit up, the new coordinates: A' would be \((-2 + 6,-1+1)=(4,0)\), B' would be \((-5 + 6,-1 + 1)=(1,0)\), and C' would be \((-2+6,-3 + 1)=(4,-2)\). Ah, that makes sense. So A is at \((-2,-1)\), so A' is (4,0), B is at \((-5,-1)\), so B' is (1,0), and C is at \((-2,-3)\), so C' is (4,-2). But wait, the option \(C'(4,-2)\) is one of the choices. Wait, but the options also have \(C'(4,0)\) (which is A's translation). Wait, maybe I misidentified C. Let's confirm: in the triangle, A and C have the same x - coordinate (since it's a right triangle), so A is at \((-2,-1)\), C is at \((-2,-3)\) (same x, different y). So translating A: \((-2 + 6,-1+1)=(4,0)\), which is one of the options (\(C'(4,0)\) is A'? No, the question is about C'. So C is at \((-2,-3)\), so C' is \((-2 + 6,-3 + 1)=(4,-2)\).

Step3: Match with options

The calculated coordinates of \(C'\) are \((4,-2)\), which matches the first option.

Answer:

A. \(C'(4, - 2)\)