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It is known that the coordinates of the three vertices of △ ABC are a (- 5,0) B (3, - 3) C (0,2), and the linear equations of height on the BC edge are obtained respectively
The slope of BC is k = (- 3 - 2) / (3 - 0) = - 5 / 3
High on BC = - 1 / k = 3 / 5
The linear equation y - 0 = (3 / 5) (x + 5)
y = 3x/5 + 3
It is known that in △ ABC, vertex a (1,1) B (4,2), vertex C is on the line X-Y + 5 = 0, and the linear equation where the height of BC side is located is 5x-2y-3 = 0
1) Find the coordinates of vertex C
2) Is triangle ABC a right triangle
(1) Let B (4,2) be substituted into the equation: 2 = - 2 * 4 / 5 + B B = 18 / 5. Let B (4,2) be substituted into the equation: y = - 2x / 5 + 18 / 5
In △ ABC, ∠ C = 90 ° AC = BC. Through point C, make a straight line Mn outside △ ABC, am ⊥ Mn in M, BN ⊥ Mn in n. (1) prove: Mn = am + BN; (2) if
Do the conclusions in (1) still hold? Explain the reason
(1) The results show that: C = 90 °, AC = BC, am ⊥ Mn in M, BN ⊥ Mn in N,
∴∠ACM=90°-∠BCN=∠CBN,
∴△ACM≌△CBN(AAS),
∴AM=CN,CM=BN,
∴MN=CN+CM=AM+BN.
(2) This is because Mn = | cm-cn |
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