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The incentre of triangle with vertices \( \Large \left(1,\ \sqrt{3}\right),\ \left(0,\ 0\right)\ and\ \left(2,\ 0\right) \) is:

A) \( \Large \left(1,\ \frac{\sqrt{3}}{2}\right) \)
B) \( \Large \left(\frac{2}{3},\ \frac{1}{\sqrt{3}}\right) \)
C) \( \Large \left(\frac{2}{3},\ \frac{\sqrt{3}}{2}\right) \)
D) \( \Large \left(1,\ \frac{1}{\sqrt{3}} \right) \)

Correct Answer:




D) \( \Large \left(1,\ \frac{1}{\sqrt{3}} \right) \)

Description for Correct answer:
The triangles are \( \Large A \left(1,\ \sqrt{3}\right),\ B \left(0,\ 0\right)\ and\ C \left(2,\ 0\right) \)

Now, \( \Large AB = \sqrt{ \left(1-0\right)^{2}+ \left(\sqrt{3}-0\right)^{2} } = 2 \)

\( \Large BC = \sqrt{ \left(2-0\right)^{2}+ \left(0-0\right)^{2} } = 2 \)

\( \Large CA = \sqrt{ \left(1-2\right)^{2}+ \left(\sqrt{3}-0\right)^{2} } = 2 \)

Its clear that the triangle is equilateral.

So, the incentre is the same as the centroid

Therefore, Incentre = \( \Large \left(\frac{1+0+2}{3}.\frac{\sqrt{3}+0+0}{3}\right) = \left(1,\ \frac{1}{\sqrt{3}}\right) \)

Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products

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Similar Questions

1). If \( \Large P \left(a_{1},\ b_{1}\right)\ and\ Q \left(a_{2},\ b_{2}\right) \) are two points, the \( \Large OP.OQ \cos \left( \angle POQ\right) \) is (O is origin ):

A). \( \Large a_{1}a_{2}+b_{1}b_{2} \)

B). \( \Large a^{2}_{1}+a^{2}_{2}+b^{2}_{1}+b^{2}_{2} \)

C). \( \Large a^{2}_{1}-a^{2}_{2}+b^{2}_{1}-b^{2}_{2} \)

D). none of these

2). If a point \( \Large P \left(4,\ 3\right) \) is shifted by a distance \( \Large \sqrt{2} \) unit parallel to the line y = x, then co-ordinates of P in new position are:

A). \( \Large \left(5,\ 4\right) \)

B). \( \Large \left(5+\sqrt{2},\ 4+\sqrt{2}\right) \)

C). \( \Large \left(5-\sqrt{2},\ 4-\sqrt{2}\right) \)

D). none of these

3). The orthocentre of a triangle formed by lines \( \Large 2x+y=2,\ x-2y=1\ and\ x+y=1 \) is:

A). \( \Large \left(\frac{1}{3},\ \frac{2}{3}\right) \)

B). \( \Large \left(0,\ 1\right) \)

C). \( \Large \left(\frac{2}{3},\ \frac{1}{3} \right) \)

D). \( \Large \left(1,\ 0\right) \)

4). The incentre of triangle formed by lines x = 0, y = 0 and 3x+4y = 12 is at:

A). \( \Large \left(\frac{1}{2},\ \frac{1}{2}\right) \)

B). \( \Large \left(1,\ 1\right) \)

C). \( \Large \left(1,\ \frac{1}{2} \right) \)

D). \( \Large \left(\frac{1}{2},\ 1\right) \)

5). If the vertices of a triangle have integral co ordinates, the triangle cannot be:

A). an equilateral triangle

B). a right angled triangle

C). an isosceles triangle

D). none of these

6). If \( \Large t_{1}+t_{2}+t_{3}=-t_{1}t_{2}t_{3} \) then orthocentre of the triangle formed by the points \( \Large \left[ at_{1}t_{2},\ a \left(t_{1}+t_{2}\right) \right],\ \left[ at_{2}t_{3},\ a \left(t_{2}+t_{3}\right) \right]\ and\ \left[ at_{3}t_{1},\ a \left(t_{3}-t_{1}\right) \right] \) lies on;

A). \( \Large \left(a,\ 0\right) \)

B). \( \Large \left(-a,\ 0\right) \)

C). \( \Large \left(0,\ a\right) \)

D). \( \Large \left(0,\ -a\right) \)

7). P\( \Large \left(2,\ 1\right) \) is image of the point Q\( \Large \left(4,\ 3\right) \) about the line:

A). x - y = 1

B). 2x - 3y = 0

C). x + y = 5

D). none of these

8). If the points \( \Large \left(a_{1},b_{1}\right), \left(a_{2},b_{2}\right) \) and \( \Large \left( a_{3},b_{3} \right) \) are collinear then lines \( \Large a_{i}x+b_{i}y+1=0 \) for i = 1, 2, 3 are:

A). concurrent

B). indentical

C). parallel

D). none of these

9). A point P\( \Large \left(2,\ 4\right) \) translates to the point Q along the parallel to the positive direction of x-axis by 2 unit. If O be the origin, then \( \Large \angle OPQ \) is:

A). \( \Large \sin^{-1}\sqrt{\frac{399}{400}} \)

B). \( \Large \cos^{-1} \left(\frac{1}{20}\right) \)

C). \( \Large -\sin^{-1} \left(\sqrt{\frac{399}{400}}\right) \)

D). none of these

10). The line joining \( \Large A \left(b \cos \alpha ,\ b \sin \alpha \right)\ and\ B \left(a \cos \beta ,\ a \sin \beta \right) \) is produced to the point \( \Large M \left(x,\ y\right) \) so that \( \Large AM:MB=b:a\ then\ x \cos \left(\frac{ \alpha + \beta }{2}\right)\ +\ y \sin \left(\frac{ \alpha + \beta }{2}\right) \) is:

A). -1

B). 0

C). 1

D). \( \Large a^{2} + b^{2} \)