Class 10 Exam  >  Class 10 Notes  >  Extra Documents, Videos & Tests for Class 10  >  Co­ordinate Geometry Exercise 14.1 (Part-3)

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 37: If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also, find the length of AB.                [CBSE 2014]

Answer :

It is given that A(0, 2) is equidistant from the points B(3, p) and C(p, 5).

∴ AB = AC 

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10   (Distance formula) 

Squaring on both sides, we get
9+p24p+4=p2+94p+4=0p=

Thus, the value of p is 1. 

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 

Question 38: Name the quadrilateral formed, if any, by the following points, and given reasons for your answers:
(i) A(−1,−2) B(1, 0), C (−1, 2), D(−3, 0)
(ii) A(−3, 5) B(3, 1), C (0, 3), D(−1, −4)
(iii) A(4, 5) B(7, 6), C (4, 3), D(1, 2)

Answer(i) A (−1,−2) , B(1,0), C(−1,2), D(−3,0)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points PCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10and QCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10is given by distance formula:

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Similarly,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Similarly,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Also,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence from above we see that all the sides of the quadrilateral are equal. Hence it is a square.

(ii) A (−3,5) , B(3,1), C(0,3), D(−1,−4)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points PCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10and QCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10is given by distance formula:

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Similarly,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Similarly,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Also,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence from the above we see that it is not a quadrilateral

(iii) A (4, 5), B (7,6), C(4,3), D(1,2)

Let A, B, C and D be the four vertices of the quadrilateral ABCD.

We know the distance between two points PCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10and QCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10is given by distance formula:

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Similarly,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Similarly,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Also,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence from above we see that

AB = CD and BC = DA

Here opposite sides of the quadrilateral is equal. Hence it is a parallelogram.

Question 39: Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).

Answer :

TO FIND: The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)

Let P(x, y) be any point on the perpendicular bisector of AB. Then,

PA=PB

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence the equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5) isCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Question 40: Prove that the points (3, 0), (4, 5), (−1, 4) and (−2 −1), taken in order, form a rhombus. Also, find its area. 

Answer :

The distance d between two points Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 and Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 is given by the formula

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

In a rhombus all the sides are equal in length. And the area ‘A’ of a rhombus is given as

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Here the four points are A(3,0), B(4,5), C(1,4) and D(2,1).

First let us check if all the four sides are equal.

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Here, we see that all the sides are equal, so it has to be a rhombus.

Hence we have proved that the quadrilateral formed by the given four vertices is aCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10.

Now let us find out the lengths of the diagonals of the rhombus.

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Now using these values in the formula for the area of a rhombus we have,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Thus the area of the given rhombus isCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10.

Question 41: In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A(3, 1), B(6, 4), and C(8, 6). Do you think they are seated in a line?

Answer : 

The distance d between two points Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 and Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 is given by the formula

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

For three points to be collinear the sum of distances between any two pairs of points should be equal to the third pair of points.

The given points are A(3,1), B(6,4) and C(8,6).

Let us find the distances between the possible pairs of points.

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

We see thatCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10.

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence, the three given points areCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10.

Question 42: Find a point on y-axis which is equidistant form the points (5, −2) and (−3, 2).

Answer :

The distance d between two points Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 and Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 is given by the formula

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Here we are to find out a point on the y−axis which is equidistant from both the points A(5,2) and B(3,2).

Let this point be denoted as C(x, y).

Since the point lies on the y-axis the value of its ordinate will be 0. Or in other words we haveCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10.

Now let us find out the distances from ‘A’ and ‘B’ to ‘C

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

We know that both these distances are the same. So equating both these we get,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Squaring on both sides we have,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence the point on the y-axis which lies at equal distances from the mentioned points isCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10.

Question 43: Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (−3, 4).

Answer :

The distance d between two points Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 and Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 is given by the formula

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Let the three given points be P(x, y), A(3,6) and B(3,4).

Now let us find the distance between ‘P’ and ‘A’.

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Now, let us find the distance between ‘P’ and ‘B’.

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

It is given that both these distances are equal. So, let us equate both the above equations,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Squaring on both sides of the equation we get,

Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10

Hence the relationship between ‘x’ and ‘y’ based on the given condition isCo­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10.

Question 44: If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p.                       [CBSE 2012, 2013]

Answer : It is given that A(0, 2) is equidistant from the points B(3, p) and C(p, 5).
∴ AB = AC
Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10    (Distance formula)
Squaring on both sides, we get
9+p24p+4=p2+94p+4=0p=19+p2-4p+4=p2+9⇒-4p+4=0⇒p=1
Thus, the value of p is 1. 

Question 45: Prove that the points (7, 10), (−2, 5) and (3, −4) are the vertices of an isosceles right triangle.                           [CBSE 2013]

Answer

Let the given points be A(7, 10), B(−2, 5) and C(3, −4).
Using distance formula, we have
Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10AB=-2-72+5-102=-92+-52=81+25=106 units
BC=Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10BC=3--22+-4-52=52+-92=25+81=106 units
CA=Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10CA=3-72+-4-102=-42+-142=16+196=212 units
Thus, AB = BC = Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10106 units
∴ ∆ABC is an isosceles triangle.
Also,
AB2 + BC= 106 + 106 = 212
and CA2 = 212
∴ AB2 + BC= CA2
So, ∆ABC is right angled at B.              (Converse of Pythagoras theorem)
Hence, the given points are the vertices of an isosceles right triangle.

The document Co­ordinate Geometry Exercise 14.1 (Part-3) | Extra Documents, Videos & Tests for Class 10 is a part of the Class 10 Course Extra Documents, Videos & Tests for Class 10.
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