1)D is mid point of side ac and G divides BD in ratio 2:1 the coordinates of G are: - Maths - Coordinate Geometry

Question

1)D is mid point of side ac and G divides BD in ratio 2:1 the
coordinates of G are:
2) if O is origin, then the area of triangleAOB is:
3) The segments joining A(3,2) and B(7,3) is extended along a
straight line till C such that AC is 4 times AB> the coordinates
of C are:
4) find distance between the poiints
(a cos20 + b sin20,0) and (0, a sin20 + b cos20)
5) ABCD is a parallelogram whose diagonals are intersecting at
P(6,4) find coordinates of A and D
6) find the coordinates of the points of trisection of segmentAB
where A(-3,2) and B(9,5)
7) the two opposite vertices of a square are (-1,2) and (3,2)
find the coordinates of the other two vertices
8) find ratio in which the line 3x+4y-9=0 divides the line
segment joining points (1,3) and (2,7)
9) find centroid of triangleABC whose sides are given by
2x+3y=12; x-y=1,y=0

Santosh Kumar · Accepted Answer

4 Let $A = (a \cos 20 ° + b \sin 20 °, 0) and B = (0, a \sin 20 ° + b \cos 20 °) AB = \sqrt{(a \cos 20 ° + b \sin 20 ° - 0)^{2} + (0 ° - a \sin 20 ° - b \cos 20 °)^{2}} = \sqrt{(a \cos 20 ° + b \sin 20 °)^{2} + (a \sin 20 ° + b \cos 20 °)^{2}} = \sqrt{a^{2} \cos^{2} 20 ° + 2 a b \sin 20 ° \cdot \cos 20 ° + b^{2} \sin^{2} 20 ° + a^{2} \sin^{2} 20 °} + \sqrt{2 a b \sin 20 ° \cdot \cos 20 ° + b^{2} \cos^{2} 20 °} = \sqrt{a^{2} (\cos^{2} 20 ° + \sin^{2} 20 °) + b^{2} (\sin^{2} 20 ° + \cos^{2} 20 °)} + \sqrt{4 a b \sin 20 ° \cdot \cos 20 °} = \sqrt{a^{2} + b^{2} + 2 (2 a b) \sin 20 ° \cdot \cos 20 °} = \sqrt{a^{2} + b^{2} + 2 a b \cdot \sin 2 (20 °)} = \sqrt{a^{2} + b^{2} + 2 a b \cdot \sin 40 °}$ 6

Let A (â€“3, 2) and B (9, 5) be the given points Let the points of trisection be C and D Then, AC = CD = BD CB = CD + DB = 2Î» and AC = AC + CD = 2Î» â‡’ AC : CB = Î»: 2Î» = Î» : 2 and AD : DB = 2Î» : Î» = 2:1 So, C divides AB internally in the ratio 1:2, while D divides internally in the ratio 2:1 Thus, the coordinates of C and D are $C [\frac{(1 \times 9) + 2 \times (- 3)}{1 + 2}, \frac{(1 \times 5) + (2 \times 2)}{1 + 2}] = C(1,3) D [\frac{(2 \times 9) + 1 \times (- 3)}{1 + 2}, \frac{(2 \times 5) + 1 \times (2)}{2 + 1}] = D(5,4)$