Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
Filter By
Clear Filter

All Questions

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

View Full Answer(1)
Posted by

infoexpert27

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).

View Full Answer(1)
Posted by

infoexpert27

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

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

View Full Answer(1)
Posted by

infoexpert27

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.

View Full Answer(1)
Posted by

infoexpert27

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

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.

View Full Answer(1)
Posted by

infoexpert27

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

View Full Answer(1)
Posted by

infoexpert27

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

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

View Full Answer(1)
Posted by

infoexpert27

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.

View Full Answer(1)
Posted by

infoexpert27

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

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.

View Full Answer(1)
Posted by

infoexpert27

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.

View Full Answer(1)
Posted by

infoexpert27

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE
filter_img