Coordinate Geometry NCERT exercise 7.1 solution part 3 Class Ten Mathematics

Coordinate Geometry

Exercise 7.1Part 3

Question 4: Check whether (5, 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Let A = (5, -2), B = (6, 4) and C = (7, -2)

Now `AB =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 5, x2 = 6, y1 = -2 and y2 = 4

So, `AB = sqrt((6-5)^2+(4+2)^2)`

`=sqrt(1^2+6^2)`

`=sqrt(1+36)=sqrt(37)` units

Now `BC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 6, x2 = 7, y1 = 4 and y2 = -2

So, `BC=sqrt((7-6)^2+(-2-4)^2)`

`=sqrt(1^2+(-6)^2)=sqrt(37)` units


Now `AC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`Now `AC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 5, x2 = 7, y1 = -2 and y2 = -2

So, `AC = sqrt((7-5)^2+(-2+2)^2)`

`=sqrt(2^2+0^2)`

`=sqrt4=2` units

Since, in the given ΔABC
`AB=BC≠AC`

Thus, given triangle is an isosceles triangle.


Question 5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Figure 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa as Chameli, “Don’t you think ABCD is square?” Chameli disagrees. Using distance formula find which of them is correct.

Solution: Here, coordiantes are A (3, 4), B (6, 7), C (9, 4) and D (6, 1)

A (3, 4), B (6, 7), C (9, 4) and D (6, 1)

`AB =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 3, x2 = 6, y1 = 4 and y2 = 7

So, `AB=sqrt((6-3)^2+(7-4)^2)`

`=sqrt(3^2+3^2)=sqrt(9+9)`

`=sqrt(18)=3sqrt2` units

`BC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 6, x2 = 9, y1 = 7 and y2 = 4

So, `BC=sqrt((9-6)^2+(4-7)^2)`

`=sqrt(3^2+(-3)^2)=sqrt(9+9)`

`=sqrt(18)=3sqrt2` units


`CD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 9, x2 = 6, y1 = 4 and y2 = 1

So, `CD=sqrt((6-9)^2+(1-4)^2)`

`=sqrt((-3)^2+(-3)^2)=sqrt(9+9)`

`=sqrt(18)=3sqrt2` units

`DA =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 6, x2 = 3, y1 = 1 and y2 = 4

So, `DA=sqrt((3-6)^2+(4-1)^2)`

`=sqrt((-3)^2+3^2)=sqrt(9+9)`

`=sqrt(18)=3sqrt2` units

Now, let us calculate the diagonals.

`AC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 3, x2 = 9, y1 = 4 and y2 = 4

So, `AC=sqrt((9-3)^2+(4-4)^2)`

`=sqrt(6^2+0^2)=sqrt(36)=6` units

`BD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 6, x2 = 6, y1 = 1 and y2 = 7

So, `BD=sqrt((6-6)^2+(1-7)^2)`

`=sqrt(0^2+(-6)^2)=sqrt(36)=6` units

Here, `AB=BC=CD=DA` and `AC=BD`

Since, all four sides are equal and diagonals are also equal thus, it is clear that ABCD is a square and hence, Champa is right.



Copyright © excellup 2014