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The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order form a rectangle.

Answer.    [True]
Solution.   The given points are A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0)
Length \, of \, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
AB= \sqrt{\left ( 4+1 \right )^{2}+\left ( 3+2 \right )^{2}}
AB= \sqrt{50}= 5\sqrt{2}
Length \, of \, BC= \sqrt{\left ( 2-4 \right )^{2}+\left ( 5-3 \right )^{2}}
BC=\sqrt{4+4}= \sqrt{8}= 2\sqrt{2}
Length \, of \, CD= \sqrt{\left ( -3-2 \right )^{2}+\left ( 0-5 \right )^{2}}
CD= \sqrt{25+25}= \sqrt{50}= 5\sqrt{2}
Length \, of \, DA= \sqrt{\left ( -1+3 \right )^{2}+\left ( -2-0 \right )^{2}}
DA= \sqrt{4+4}= \sqrt{8}= 2\sqrt{2}
AB = CD , BC = DA
Length \, of \, AC= \sqrt{\left ( 2+1 \right )^{2}+\left ( 5+2 \right )^{2}}
AC= \sqrt{9+49}= \sqrt{58}
Length \, of \, BD= \sqrt{\left ( -3-4 \right )^{2}+\left ( 0-3 \right )^{2}}
BD= \sqrt{49+9}= \sqrt{58}

AC = BD (Diagonals)
Hence, ABCD is a rectangle because
AB = CD, BC = DA, AC = BD

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The point P (–2, 4) lies on a circle of radius 6 and centre C (3, 5).

Answer.     [False]
Solution.        
The radius of the circle is 6 and centre C(3, 5)
If point P(–2, 4) lies on the circle then the distance between the centre and point P is equal to the radius of the circle.
Distance\, between\, PC = \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2} -y_{1}\right )^{2}}
(x1, y1) = (-2, 4)             (x2, y2) = (3, 5)
PC= \sqrt{\left ( 3+2 \right )^{2}+\left ( 5-4 \right )^{2}}
PC=\sqrt{25+1}= \sqrt{26}
PC\neq radius\left ( 6 \right )
Hence, point P(–2, 4) not lies on the circle with centre C(3, 5).

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Points A (–6, 10), B (–4, 6) and C (3, –8) are collinear such that AB =2/9 AC.

Answer.      [Ture]

Solution.     If the points A (–6, 10), B(–4, 6) and C(3, –8) are collinear then area of \triangleABC = 0
Area \, of \, \triangle ABC= \frac{1}{2}\left [ -6\left ( 6+8 \right ) +\left ( -4 \right )\left ( -8-10 \right )+3\left ( 10-6 \right )\right ]
                                   = \frac{1}{2}\left [ -84+72+12 \right ]
Area of \triangleABC = 0
Hence, A, B and C are collinear
Distance \, between \, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}

AB= \sqrt{\left ( -4+6 \right )^{2}+\left ( 6-10 \right )^{2}}

AB=\sqrt{4+16}= \sqrt{20}= 2\sqrt{5}
Distance \, between \, AC= \sqrt{\left ( 3+6 \right )^{2}+\left ( -8-10 \right )^{2}}

= \sqrt{81+324}
= \sqrt{405}

AC= 9\sqrt{5}
Hence,
 \frac{AB}{AC}= \frac{2\sqrt{5}}{9\sqrt{5}}
AB= \frac{2}{9}AC

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Point P (5, –3) is one of the two points of trisection of the line segment joining the points A (7, – 2) and B (1, – 5).

Answer.       [True]
Solution.        
                
Let the two point of trisection are C, D
Point \, \, C= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} ,\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right )
(x1, y1) = (7, -2)           (x2, y2) = (1, -5)
m1 = 1 , m2 = 2    (Because C divide AB in ratio 1:2 )
C=\left ( \frac{1\times 1+2\times 7}{1+2} ,\frac{1\times -5+2\times -2}{1+2}\right )
C= \left ( \frac{1+14}{3} ,\frac{-5-4}{3}\right )
C= \left ( 5,- 3 \right )
Point \, \, D= \left ( \frac{n_{1}x_{2}+n_{2}x_{1}}{n_{1}+n_{2}} ,\frac{n_{1}y_{2}+n_{2}y_{1}}{n_{1}+n_{2}}\right )
n1 = 2 , n2 = 1    (Because D divide AB in ratio 2:1)
D=\left ( \frac{2\times 1+1\times 7}{2+1} ,\frac{2\times -5+1\times -2}{2+1}\right )
D=\left ( \frac{2+7}{3},\frac{-10-2}{3} \right )
D=\left ( 3,-4 \right )
Hence, the given statement is true.

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The point A (2, 7) lies on the perpendicular bisector of line segment joining the points P (6, 5) and Q (0, – 4).

Answer.      [False]
Solution.     If point A(2, 7) is bisector then it must be mid-point of the line joining the points P(6, 5) and Q(0, –4)
Mid-point \, of \, PQ = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )
(x1, y1) = (6, 5)        (x2, y2) = (0, -4)
= \left ( \frac{6+0}{2},\frac{5-4}{2} \right )= \left ( 3,\frac{1}{2} \right )
Hence, A not lies on bisector.

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A circle has its Centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.

Answer.     [True]
Solution.    The centre of the circle is O (0, 0).
If point P(5, 0) lies on the circle then the distance between O(0, 0) and P(5, 0) is the radius of the circle
OP= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
OP= \sqrt{\left ( 5-0 \right )^{2}+\left ( 0-0 \right )^{2}}
OP= \sqrt{\left ( 5 \right )^{2}}= 5
OP = 5
Radius of circle = 5
If point Q (6, 8) is outside the circle then the distance between O(0, 0) and Q(6, 8) is grater then the radius of the circle
(x1, y1) = (0, 0)                        (x2, y2) = (6, 8)
OQ= \sqrt{\left ( 6-0 \right )^{2}+\left ( 8-0 \right )^{2}}
OQ= \sqrt{36+64}
         = \sqrt{100}=10

Here point OQ is greater than the radius of circle
Hence, point Q(6, 8) lies outside the circle

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Points A (4, 3), B (6, 4), C (5, –6) and D (–3, 5) are the vertices of a parallelogram.

Answer.     [False]
Solution.        
The given points area A(4, 3), B(6, 4), C(5, –6), D(–3, 5)
Distance \: between\: AB = \sqrt{\left ( 6-4 \right )^{2}+\left ( 4-3 \right )^{2}}
AB = \sqrt{\left ( 2 \right )^{2}+\left ( 1 \right )^{2}}= \sqrt{5}
Distance \: between\: BC = \sqrt{\left ( 5-6 \right )^{2}+\left ( -6-4 \right )^{2}}
BC= \sqrt{1+100}= \sqrt{101}
Distance \: between\: CD = \sqrt{\left ( -3-5 \right )^{2}+\left ( 5+6 \right )^{2}}
CD = \sqrt{64+121}= \sqrt{185}
Distance \: between\: DA = \sqrt{\left ( 4+3 \right )^{2}+\left ( 3-5 \right )^{2}}
DA =\sqrt{49+4}= \sqrt{53}
Here, opposite sides are not equal i.e.
AB\neq CD,BC\neq DA
Hence, it is not a parallelogram

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Points A (3, 1), B(12, –2) and C(0, 2) cannot be the vertices of a triangle.

Answer.      [True]
Solution.         If they are not the vertices of a triangle then
Area of \triangleABC = 0
x1 =3    x2 =12     x3 =0
y1=1     y2=-2     y3=2
Let us find the area of \triangleABC
Area \, of \triangle ABC = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]
                                   = \frac{1}{2}\left [ 3\left ( -2-2 \right )+12\left ( 2-1 \right ) +0\left ( 1+2 \right )\right ]
                                   = \frac{1}{2}\left [ 3\left ( -4 \right ) +12\left ( 1 \right )+0\right ]
                                    = \frac{1}{2}\left [ -12+12 \right ]
                                     = \frac{0}{2}
                                      =   0

Area of \triangleABC = 0
Hence, they are collinear or not the vertices of a triangle.

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Point P (0, 2) is the point of intersection of y–axis and perpendicular bisector of line segment joining the points A (–1, 1) and B (3, 3).

Solution.      If the point P is a perpendicular bisector of the line joining the point A(–1, 1) and B(3, 3) then it must be mid-point of AB.
Mid -point \, of AB = \left ( \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} \right )
(x1, y1) = (-1, 1)        (x2, y2) = (3, 3)
= \left ( \frac{-1+3}{2},\frac{1+3}{2} \right )= \left ( \frac{2}{2} ,\frac{4}{2}\right )= \left ( 1,2 \right )

Which is not point P.
Hence, the given statement is false.

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The points (0, 5), (0, –9) and (3, 6) are collinear.

Answer.      [False]
The given points area (0, 5), (0, –9) and (3, 6)
If the point area collinear then the area of triangle is 0.
x1 =0    x2 =0     x3 =3
y1=5     y2=9     y3=6
We\, know \, that\, area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]
= \frac{1}{2}\left [ 0\left ( -9-6 \right ) +0\left ( 6-5 \right )+3\left ( 5+9 \right )\right ]
= \frac{1}{2}\left [ 0+0+42 \right ]
 = \frac{42}{2}=21 
Area of triangle = 21
Here the area of the triangle is not equal to zero.
Hence, the point is not collinear.

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