Prove that the midpoint of the hypotenuse of the right triangle with vertices (0, 0), (5, 0), and (0, 7) is equidistant from the three vertices.
AD=CD=BD=
The Mid-point of right triangle is equidistant from all the three vertices.
Given information:
Vertices of right triangle are (0, 0), (5, 0), and (0, 7)
Concept used:
- First, find the value for x on the x-axis.
- Next, find the y-value on the y-axis.
- The point should be plotted at the intersection of x and y.
- Finally, plot the point on your graph at the appropriate spot.
- Then join all the points to make a closed figure.
Calculation:
The given points are A(0, 0), B(5, 0), and C(0, 7)
Plotting the points in the graph -
The mid-point is D(2.5, 3.5).
Using distance formula −
A(0, 0), D(2.5, 3.5)
C(0,7), D(2.5, 3.5)
B(5, 0), D(2.5, 3.5)
AD=CD=BD=
As D is the mid-point of hypotenuse CB and the distance from three vertices to mid-point is same.
So mid-point of right triangle is equidistant from all the three vertices.
Chapter P Solutions
Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition
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