Class 10 Maths

# Coordinate Geometry

## NCERT Exercise

### 7.1 Part 3

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

Answer: 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.

Answer: 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.