त्रिभुज
1). केन्द्रक: माध्य का अंतर्बिंदु
Questions
Q. शीर्ष के निर्देशांक A(x1, y1), B(x2, y2), C(x3, y3) वाले त्रिभुज का अंत:केंद्र ज्ञात कीजिये
Solution:
By geometry, we know that BD/DC = AB/AC (since AD bisects ÐA).
The lengths of the sides AB, BC and AC are c, a and b respectively, then BD/DC = AB/AC = c/b.
Coordinates of D are (bx2+cx3/b+c, by2+cy3/b+c)
IB bisects DB. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.
Let the coordinates of I be (x, y).
Then x = ax1+bx2+cx3/a+b+c, y = ay1+by2+cy3/a+b+c.
Q. यदि (0, 1), (1, 1) और (1, 0) एक त्रिभुज की भुजाओं के मध्य बिंदु हैं, तो इसका अंत: केंद्र ज्ञात कीजिये
Solution:
Let A(x1, y1), B(x2, y2) and C(x3, y3)be teh vertices of a triangle.
x1 + x2 = 0, x2 + x3 = 0, x3 + x1 = 0
y1 + y2 = 0, y2 + y3 = 0, y3 + y1 = 0.
Solving these equations, we get A(0, 0), B(0, 2) and C(2, 0).
Now, a = BC = 2√ 2, b = CA = 2 and c = AB = 2.
Thus, incentre of the triangle ABC is (2-√ 2, 2-√ 2).
Q. यदि किसी त्रिभुज की भुजाओं के मध्यबिंदु (0, 4), (6, 4) और (6, 0) हैं, तो त्रिभुज के शीर्ष, केन्द्रक और परिधि ज्ञात करें।
Solution:
Let points A (x1, y1), B (x2, y2) and C (x3, y3) be vertices of ΔABC.
x1 + x3 = 0 , y1 + y3 = 8
x2 + x3 = 12 , y2 + y3 = 8
x1 + x2 = 12, y1 + y2 = 0
Solving we get A (0, 0), B (12, 0) and C (0, 8)
Hence ΔABC is right angled triangle. ∠A = π/2
Circumcentre is midpoint of hypotenuse which is (6, 4) itself and centroid
(x1+x2+x3)/3 , (y1+y2+y3)/3 = (4 , 8/3)
