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

Correct Answer:
A) an equilateral triangle

Description for Correct answer:

Area 'of triangle = \( \frac{1}{2} \begin{vmatrix} 
x_{1} & y_{1} & 1 \\ 
x_{2} & y_{2} & 1 \\ 
x_{3} & y_{3} & 1  
\end{vmatrix} \)

= A rational number, if vertices have integral co ordinates only

If triangle is equilateral, then

Area = \( \Large \frac{\sqrt{3}}{4}\left[ \left(x_{1}-x_{2}\right)^{2}+ \left(y_{1}-y_{2}\right)^{2} \right] \)

= irrational quantity.


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








Comments

No comments available




Similar Questions
1). 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). \( \Large \left(0,\ -a\right) \)
-- View Answer
2). 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
-- View Answer
3). 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
-- View Answer
4). 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
-- View Answer
5). 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). \( \Large a^{2} + b^{2} \)
-- View Answer


6). If \( \Large P_{1},\ P_{2} \) denote the length of perpendiculars from the origin on the lines \( \Large x \sec \alpha + y \cosh \alpha = 2 \alpha \ and\ x \cos \alpha + y \sin \alpha = \alpha \cos 2 \alpha \) respectively, then \( \Large \left(\frac{P_{1}}{P_{2}}+\frac{P_{2}}{P_{1}}\right)^{2} \) is equal to:
A). \( \Large 4 \sin^{2} 4 \alpha \)
B). \( \Large 4 \cos^{2} 4 \alpha \)
C). \( \Large 4 cosec^{2} 4 \alpha \)
D). \( \Large 4 \sec^{2} 4 \alpha \)
-- View Answer
7). lf every point on the line \( \Large \left(a_{1}-a_{2}\right)x+ \left(b_{1}-b_{2}\right)y=c \) is equidistant from the points \( \Large \left(a_{1},\ b_{1}\right) \) then 2c is equal to:
A). \( \Large a^{2}_{1}+b^{2}_{1}+a^{2}_{2}-b^{2}_{2} \)
B). \( \Large a^{2}_{1}+b^{2}_{1}+a^{2}_{2}+b^{2}_{2} \)
C). \( \Large a^{2}_{1}+b^{2}_{1}-a^{2}_{2}-b^{2}_{2} \)
D). none of these
-- View Answer
8). If two vertices of a triangle are \( \Large \left(-2,\ 3\right)\ and\ \left(5,\ -1\right) \) orthocentre lies at origin and centroid on the line x + y = 7, then the third vertex lies at:
A). \( \Large \left(7,\ 4\right) \)
B). \( \Large \left(8,\ 14\right) \)
C). \( \Large \left(12,\ 21\right) \)
D). none of these
-- View Answer
9). The locus of a points p which moves such that 2PA = 3PB, where \( \Large A \left(0,\ 0\right)\ and\ B \left(4,\ -3\right) \) are points is:
A). \( \Large 5x^{2}-5y^{2}-72x+54y+225=0 \)
B). \( \Large 5x^{2}+5y^{2}-72x+54y+225=0 \)
C). \( \Large 5x^{2}+5y^{2}+72x-54y+225=0 \)
D). \( \Large 5x^{2}+5y^{2}-72x-54y+225=0 \)
-- View Answer
10). If \( \Large A \left(-a,0\right)  \) and \( \Large B \left(a,0\right)  \) are two fixed points, then the locus of the point at which AB subtends a right angle is;
A). \( \Large x^{2}+y^{2}=2a^{2} \)
B). \( \Large x^{2}-y^{2}=a^{2} \)
C). \( \Large x^{2}+y^{2}+a^{2}=0 \)
D). \( \Large x^{2}+y^{2}=a^{2} \)
-- View Answer