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draw the image of \\( \\triangle abc \\) under a dilation whose center …

Question

draw the image of \\( \triangle abc \\) under a dilation whose center is \\( a \\) and scale factor is 4.

Explanation:

Step1: Identify the effect of dilation on points

Since the center of dilation is \(A\), point \(A\) remains unchanged. For points \(B\) and \(C\), we use the scale - factor \(k = 4\).

Step2: Calculate the new lengths of \(AB\) and \(AC\)

If we consider the vector from \(A\) to \(C\) (length \(AC=2.0\)) and from \(A\) to \(B\) (using the right - triangle with legs \(2.0\) and \(3.0\)). But in a coordinate - grid based dilation (assuming \(A\) as the center), if we think of the displacement from \(A\) to \(C\) and \(A\) to \(B\) in terms of grid units.
Let's assume \(A\) is at a fixed point. If we consider the line segments \(AC\) and \(AB\), after dilation:
The new length of \(AC'\) (where \(C'\) is the image of \(C\)): \(AC'=k\times AC\). Since \(k = 4\) and \(AC = 2.0\), \(AC'=4\times2.0 = 8.0\)
The new length of \(AB'\) (where \(B'\) is the image of \(B\)): Using the right - triangle \(ABC\) with \(AC = 2\) and \(BC=3\), by the Pythagorean theorem \(AB=\sqrt{2^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}\approx3.6\). After dilation \(AB'=k\times AB\approx4\times3.6 = 14.4\). But if we consider the horizontal and vertical displacements from \(A\) to \(C\) (vertical displacement \(= 2\) units) and from \(A\) to \(B\) (horizontal displacement \(=3\) units from the right - triangle \(ABC\) shown).
The vertical displacement from \(A\) to \(C\) is \(2\) units. After dilation, the new vertical displacement from \(A\) to \(C'\) is \(4\times2=8\) units.
The horizontal displacement from \(A\) to \(B\) (relative to the right - triangle \(ABC\)) is \(3\) units. After dilation, the new horizontal displacement from \(A\) to \(B'\) (in the direction from \(A\) to \(B\)) is \(4\times3 = 12\) units.

Answer:

To draw the image:

  1. Keep point \(A\) in its original position.
  2. Extend the line segment \(AC\) beyond \(C\) such that the new length \(AC'\) is \(4\) times the original length \(AC\).
  3. Extend the line segment \(AB\) beyond \(B\) such that the new length \(AB'\) is \(4\) times the original length \(AB\).
  4. Connect points \(A\), \(B'\), and \(C'\) to form \(\triangle AB'C'\).