Work out the equation of the line which has a gradient of 2 and passes through the point (1,4)
1 answer:
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Answer:
The true statements are
1. BD = DB'
3. m∠EFA = 90°
4. The line of reflection, EH, is the perpendicular bisector of BB', AA', and
CC'
Step-by-step explanation:
* Lets explain how to solve the problem
- Reflection is flipping an object over the line of reflection.
- The object and its image have the same shape and size, but the
figures are in opposite directions from the line of reflection
- The objects appear as if they are mirror reflections, with right and left
reversed
- The line of reflection is a perpendicular bisector for all lines joining
points on the figure with their corresponding images
- Look to the attached figure for more understand
* Lets solve the problem
- ΔA'B'C' was constructed using ΔABC and line segment EH, where
EH is the reflection line
- D is the mid-point of BB'
- F is the mid-point of AA'
- G is the mid-point of CC'
* Lets find from the answer the true statements
1. BD = DB'
∵ D is the mid point of BB'
- Point D divides BB' into two equal parts
∴ BD = DB' ⇒ <em>True</em>
2. DF = FG
- It depends on the size of the sides and angles of the triangle
∵ We can't prove that
∴ DF = FG ⇒ <em>Not true</em>
3. m∠EFA = 90°
∵ The line of reflection ⊥ the lines joining the points with their
corresponding images
∴ EH ⊥ AA' and bisect it at F
∴ m∠EFA = 90° ⇒ <em>True</em>
4. The line of reflection, EH, is the perpendicular bisector of BB',
AA', and CC'
- Yes the line of reflection is perpendicular bisectors of them
∴ The line of reflection, EH, is the perpendicular bisector of BB',
AA', and CC' ⇒ <em>True</em>
5. ΔABC is not congruent to ΔA'B'C'
∵ In reflection the object and its image have the same shape and size
∴ Δ ABC is congruent to Δ A'B'C'
∴ ΔABC is not congruent to ΔA'B'C' ⇒ <em>Not true</em>
