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What do point O and point J have in common? How about H and J?
Graph:
Right triangle/Isosceles triangle?: Isosceles
Side lengths: AC=2h, AB=hsqrt(2), BC=hsqrt(2)
Slope: m_(AC)=0, m_(AB)=1, m_(AB)=- 1
Midpoints: M_(AC)(h,0), M_(AB)( h2, h2), M_(BC)( 3h2, h2)
The only point where we know both coordinates is A(0,0). Let's start by plotting this point.
Additionally, we know that C(2h, 0) is on the x-axis because it's y-coordinate is 0. Let's mark this point as well. Since both A and C have the same y-coordinate, the side between these points will be horizontal.
As for the third point, B(h,h), we know it's x-coordinate is in the middle of AC which means â–ł ABC is an isosceles triangle. Additionally, the y-coordinate is h units above the x-axis.
To find the slope of the sides, we use the Slope Formula. Note that horizontal lines have a slope of 0 so we do not have to determine the slope of AC.
Side | Points | y_2-y_1/x_2-x_1 | m |
---|---|---|---|
AB | ( h,h), ( 0,0) | h- 0/h- 0 | 1 |
BC | ( h,h), ( 2h,0) | h- 0/h- 2h | - 1 |
To find the length of the sides, we can use the Distance Formula. However, we do not need to use the formula for the horizontal side as it's the absolute value of the difference between the endpoints' x-coordinates. AC: |2h-0|=2h The remaining sides we have to calculate with the Distance Formula.
Distance | Points | sqrt((x_2-x_1)^2+(y_2-y_1)^2) | d |
---|---|---|---|
AB | ( h,h), ( 0,0) | sqrt(( h- 0)^2+( h- 0)^2) | hsqrt(2) |
BC | ( 2h,0), ( h,h) | sqrt(( 2h- h)^2+( 0- h)^2) | hsqrt(2) |
Let's summarize the length and slope: ll Slope & Distance m_(AC)=0 & d_(AC)=2h m_(AB)=1 & d_(AB)=hsqrt(2) m_(BC)=- 1 & d_(BC)=hsqrt(2)
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 |
---|---|---|---|
AC | ( 2h,0), ( 0,0) | M(2h+ 0/2,0+ 0/2) | M(h,0) |
AB | ( h,h), ( 0,0) | M(h+ 0/2,h+ 0/2) | M(h/2,h/2) |
BC | ( 2h,0), ( h,h) | M(2h+ h/2,0+ h/2) | M(3h/2,h/2) |
Let's summarize the midpoints: M_(AC)(h,0), M_(AB)(h/2,h/2), M_(BC)(3h/2,h/2)