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What are the x- and y-coordinates of points on that lie on the y- and x-axis?
Graph:
Right triangle/Isosceles triangle?: Right
Side lengths: DE=m, EF=n, DF=sqrt(m^2+n^2)
Slope: m_(DE)=0, m_(EF)=Undefined, m_(DF)=- nm
Midpoints: M_(DE)( m2,n), M_(EF)(m, n2), M_(DF)( m2, n2)
Any point that's on the x-axis has a y-coordinate of 0. Similarly, any point on the y-axis has an x-coordinate of 0. Therefore, it must be that &F(m, 0) is on the $x-$axis &D( 0,n) is on the $y-$axis. Let's plot these points in a coordinate plane and draw DF.
The third point has the y-coordinate n and the x-coordinate m. This means FE has to be vertical and DE has to be horizontal. Since vertical and horizontal sides are perpendicular, we are dealing with a right triangle.
Substitute ( m,0) & ( 0,n)
Subtract terms
Put minus sign in front of fraction
Substitute ( m,0) & ( 0,n)
Subtract terms
Calculate power
To find the midpoint of any side we can use the Midpoint Formula.
Side | Points | M(x_1+x_2/2,y_1+y_2/2) | Midpoint |
---|---|---|---|
DE | ( m,n), ( 0,n) | M(m+ 0/2,n+ n/2) | M(m/2,n) |
EF | ( m,n), ( m,0) | M(m+ m/2,n+ 0/2) | M(m,n/2) |
DF | ( m,0), ( 0,n) | M(m+ 0/2,0+ n/2) | M(m/2,n/2) |
Let's summarize the midpoints: M_(DE)(m/2,n), M_(EF)(m,n/2), M_(DF)(m/2,n/2)