Jaydin draws a triangle on the coordinate grid. Its vertices are at (−4,3) , (−1,−5) , and (−2,6) . He translates this triangle 5 units up and 2 units to the right. What are the coordinates of the lowest vertex on the translated triangle?(1 point)

Question

Responses

(1,0)
left parenthesis 1 comma 0 right parenthesis

(0,1)
left parenthesis 0 comma 1 right parenthesis

(−2,8)
left parenthesis negative 2 comma 8 right parenthesis

(4,−3)

Answer 1

To find the new coordinates of the vertices after translating the triangle, we need to apply the translation to each vertex.

The translation is 5 units up and 2 units to the right. This means we will add 2 to the x-coordinates and 5 to the y-coordinates of each vertex.

1. For the vertex \((-4, 3)\):

   • New x-coordinate: \(-4 + 2 = -2\)
   • New y-coordinate: \(3 + 5 = 8\)
   • New vertex: \((-2, 8)\)

2. For the vertex \((-1, -5)\):

   • New x-coordinate: \(-1 + 2 = 1\)
   • New y-coordinate: \(-5 + 5 = 0\)
   • New vertex: \((1, 0)\)

3. For the vertex \((-2, 6)\):

   • New x-coordinate: \(-2 + 2 = 0\)
   • New y-coordinate: \(6 + 5 = 11\)
   • New vertex: \((0, 11)\)

Now, we identify the lowest vertex among the translated vertices, which are \((-2, 8)\), \((1, 0)\), and \((0, 11)\).

Comparing the y-coordinates:

• y-coordinate of \((-2, 8)\) is 8,
• y-coordinate of \((1, 0)\) is 0,
• y-coordinate of \((0, 11)\) is 11.

The lowest vertex is \((1, 0)\).

Therefore, the coordinates of the lowest vertex on the translated triangle are \(\boxed{(1, 0)}\).