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Class 10 Class 12

 Find the distance between the points A and B in the following :

(i) A(2, 3), B(4, 1)    (ii) A (a, b), B(-a, -b)    (iii) A(5, -8); B(-7, -3)

(iv) A(4, 10), B(7, -6) (v) A (a + b, a - b), B(a - b, -a - b).


Solution not provided.

Ans. space left parenthesis straight i right parenthesis space square root of 13 space space space space space left parenthesis ii right parenthesis space 2 square root of straight a squared plus straight b squared space end root space space space left parenthesis iii right parenthesis space 13 space space space space space left parenthesis iv right parenthesis space square root of 265 space space space space left parenthesis straight v right parenthesis space space space 2 square root of straight a squared plus straight b squared end root space  

541 Views

Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6, 2) arc the vertices of a square.

Given points are A(5, 6), B(1, 5), C(2, 1) and D(6,2). AC and BD are two diagonals of square.



Given points are A(5, 6), B(1, 5), C(2, 1) and D(6,2). AC and BD are
Fig. 7.24.


Given points are A(5, 6), B(1, 5), C(2, 1) and D(6,2). AC and BD are

Hence, diagonal

    AC  = BD  = square root of bold 34


162 Views

Show that (1, -1) is the centre of the circle circumscribing the triangle whose angular points are (4, 3), (-2, 3) and (6, -1).

Let the given points be P(4, 3), Q(-2, 3) and R(6, -1). Let 0(1, -1) be the centre of the circle.


Let the given points be P(4, 3), Q(-2, 3) and R(6, -1). Let 0(1, -1)
Fig. 7.25.

Then comma space space space OP space equals space square root of left parenthesis 4 minus 1 right parenthesis squared plus left curly bracket 3 minus left parenthesis negative 1 right parenthesis right curly bracket squared end root
rightwards double arrow space space space space space space space OP space equals space square root of left parenthesis 4 minus 1 right parenthesis squared plus left parenthesis 3 plus 1 right parenthesis squared end root
rightwards double arrow space space space space space space space OP space equals space square root of left parenthesis 3 right parenthesis squared plus left parenthesis 4 right parenthesis squared end root
rightwards double arrow space space space space space space space OP space equals space square root of 9 plus 16 end root equals square root of 25 equals 5 end root
space space space space space space space space space space QQ space equals space square root of left parenthesis negative 2 minus 1 right parenthesis squared plus left parenthesis 3 plus 1 right parenthesis squared end root
rightwards double arrow space space space space space space space QQ space equals square root of left parenthesis negative 3 right parenthesis squared plus left parenthesis 4 right parenthesis squared end root
rightwards double arrow space space space space space space space QQ equals square root of 9 plus 16 end root equals square root of 25 equals 5 space
and space space space space space space OR space equals space square root of left parenthesis 6 minus 1 right parenthesis squared plus left curly bracket negative 1 minus left parenthesis negative 1 right parenthesis right curly bracket squared end root space space
rightwards double arrow space space space space space space space OR space space equals space square root of left parenthesis 5 right parenthesis squared plus left parenthesis negative 1 plus 1 right parenthesis squared end root
rightwards double arrow space space space space space space space OR space space equals space square root of 25 plus 0 end root equals square root of 25 equals 5

Here, we have
OP = OQ = OR
⇒ O is equidistant from P, Q and R.
Hence ‘O’ is the centre of the circle circumscribing the triangle.

763 Views

The line joining the points (1, -2) and (-3, 4) is trisected. Find the co-ordinates of the points of trisection. 

Case I.


Case I.Fig. 7.28A.Let the given points be A(3, -1) and B(-6, 5).Let P

Fig. 7.28A.
Let the given points be A(3, -1) and B(-6, 5).
Let P and Q be the points of trisection of AB.
Then,    AP = PQ = QB = 1
Thus ‘P’ divides AB in the ratio 1 : 2.
Here, we have x1 = 1,    y1 = -2
x2 = -3,    y2 = 4
and    m1 = 1    m2 = 2
∴ The co-ordinates of ‘P’ are given by

straight P space open square brackets fraction numerator straight m subscript 1 straight x subscript 2 plus straight m subscript 2 straight x subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction comma space fraction numerator straight m subscript 1 straight y subscript 2 plus straight m subscript 2 straight y subscript 1 over denominator straight m subscript 1 straight m subscript 2 end fraction close square brackets
equals straight P space open square brackets fraction numerator 1 left parenthesis negative 3 right parenthesis plus 2 left parenthesis 1 right parenthesis over denominator 1 plus 2 end fraction comma space fraction numerator 1 left parenthesis 4 right parenthesis plus 2 left parenthesis negative 2 right parenthesis over denominator 1 plus 2 end fraction close square brackets
equals straight P space open square brackets fraction numerator negative 3 plus 2 over denominator 3 end fraction comma fraction numerator 4 minus 4 over denominator 3 end fraction close square brackets equals straight P open square brackets fraction numerator negative 1 over denominator 3 end fraction comma 0 close square brackets

Case II.


Case I.Fig. 7.28A.Let the given points be A(3, -1) and B(-6, 5).Let P

Fig. 7.29.
Now ‘Q’ divides AB in the ratio 2 : 1.
Here, we have x1 = 1,    y = -2
x2 = -3,    y2 = 4
and    m1 = 2    m2 = 1
∴ The co-ordinates of ‘Q’ are given by

straight Q space open square brackets fraction numerator straight m subscript 1 straight x subscript 2 plus straight m subscript 2 straight x subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction comma fraction numerator straight m subscript 1 straight y subscript 2 plus straight m subscript 2 straight y subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close square brackets
equals space straight Q space open square brackets fraction numerator 2 left parenthesis negative 3 right parenthesis plus 1 left parenthesis 1 right parenthesis over denominator 2 plus 1 end fraction comma space fraction numerator 2 cross times left parenthesis 4 right parenthesis plus 1 left parenthesis negative 2 right parenthesis over denominator 2 plus 1 end fraction close square brackets
equals space straight Q open square brackets fraction numerator negative 6 plus 12 over denominator 3 end fraction comma space fraction numerator 8 minus 2 over denominator 3 end fraction close square brackets equals straight Q open square brackets fraction numerator negative 5 over denominator 3 end fraction comma space 2 close square brackets

Hence, the co-ordinates of the points of trisection are  
open parentheses fraction numerator negative 1 over denominator 3 end fraction comma space 0 close parentheses space and space open parentheses fraction numerator negative 5 over denominator 3 end fraction comma space 2 close parentheses

454 Views

Name the type of triangle formed, if any, by the following points and give reason for your answer:
left parenthesis 1 comma space minus 1 right parenthesis comma space space open parentheses fraction numerator negative 1 over denominator 2 end fraction comma space 1 half close parentheses space left parenthesis 1 comma space 2 right parenthesis

Let the given points be straight A left parenthesis 1 comma space minus 1 right parenthesis comma space space straight B open parentheses negative 1 half comma space 1 half close parentheses space space and space straight C left parenthesis 1 comma space 2 right parenthesis space then

AB space equals space square root of open parentheses negative 1 half close parentheses squared plus open parentheses 1 half minus left parenthesis negative 1 right parenthesis close parentheses end root
rightwards double arrow space AB space equals space square root of open parentheses fraction numerator negative 1 minus 2 over denominator 2 end fraction close parentheses squared plus open parentheses fraction numerator 1 over denominator 2 plus 1 end fraction 1 close parentheses squared end root
rightwards double arrow space AB space equals space square root of open parentheses fraction numerator negative 3 over denominator 2 end fraction close parentheses squared plus open parentheses 3 over 2 close parentheses squared end root
rightwards double arrow space AB space equals space square root of 9 over 4 plus 9 over 4 end root

BC space equals space square root of open curly brackets 1 minus open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses close curly brackets squared plus open parentheses 2 minus 1 half close parentheses squared end root
rightwards double arrow space BC space equals space square root of open parentheses 1 plus 1 half close parentheses squared plus open parentheses 2 minus 1 half close parentheses squared end root
rightwards double arrow space BC space equals space square root of open parentheses 1 plus 1 half close parentheses squared plus open parentheses 2 minus 1 half close parentheses squared end root
rightwards double arrow space BC space equals space square root of open parentheses fraction numerator 2 plus 1 over denominator 2 end fraction close parentheses squared plus open parentheses fraction numerator 4 minus 1 over denominator 2 end fraction close parentheses squared end root
rightwards double arrow space BC space equals space square root of open parentheses 3 over 2 close parentheses squared plus open parentheses 3 over 2 close parentheses squared end root
rightwards double arrow space BC space equals space square root of 9 over 4 plus 9 over 4 end root
rightwards double arrow space BC space equals space 3 over 2 square root of 2
and space space AC space equals space square root of left parenthesis 1 minus 1 right parenthesis squared plus left curly bracket 2 minus left parenthesis negative 1 right parenthesis right curly bracket squared end root
rightwards double arrow space AC space equals space square root of 0 plus left parenthesis 2 plus 1 right parenthesis squared end root
rightwards double arrow space AC space equals space square root of 0 plus left parenthesis 3 right parenthesis squared end root
rightwards double arrow space AC space equals space square root of 9 space equals space 3

Since, AB = BC
So, the given points form an isosceles triangle.

92 Views

The distance between the points (4, - 3) and (0, 0) is
  • 4
  • 5  
  • 7

C.

5  
85 Views