Coordinate Geometry | Exercise 7.1

Question 4

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

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Solution
Understand the Question
  • An isosceles triangle is a triangle that has at least two sides of equal length.
  • Let the given points be A(5,2)A(5, -2), B(6,4)B(6, 4), and C(7,2)C(7, -2).
  • To determine if ΔABC\Delta ABC is isosceles, we calculate the lengths of all three sides (ABAB, BCBC, and CACA) using the distance formula:

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Step 1 · Calculate the length of side AB

Let the points be A(5,2)A(5, -2), B(6,4)B(6, 4), and C(7,2)C(7, -2).Diagram 1

Using the distance formula for A(5,2)A(5, -2) and B(6,4)B(6, 4)

AB=(65)2+(4(2))2=12+(4+2)2=12+62=1+36=37 units\begin{aligned} AB &= \sqrt{(6 - 5)^2 + (4 - (-2))^2} \\ &= \sqrt{1^2 + (4 + 2)^2} \\ &= \sqrt{1^2 + 6^2} \\ &= \sqrt{1 + 36} \\ &= \sqrt{37} \text{ units} \end{aligned}

Step 2 · Calculate the length of side BC

Using the distance formula for B(6,4)B(6, 4) and C(7,2)C(7, -2)

BC=(76)2+(24)2=12+(6)2=1+36=37 units\begin{aligned} BC &= \sqrt{(7 - 6)^2 + (-2 - 4)^2} \\ &= \sqrt{1^2 + (-6)^2} \\ &= \sqrt{1 + 36} \\ &= \sqrt{37} \text{ units} \end{aligned}

Step 3 · Calculate the length of side CA

Using the distance formula for C(7,2)C(7, -2) and A(5,2)A(5, -2)

CA=(57)2+(2(2))2=(2)2+(2+2)2=(2)2+02=4+0=4=2 units\begin{aligned} CA &= \sqrt{(5 - 7)^2 + (-2 - (-2))^2} \\ &= \sqrt{(-2)^2 + (-2 + 2)^2} \\ &= \sqrt{(-2)^2 + 0^2} \\ &= \sqrt{4 + 0} \\ &= \sqrt{4} \\ &= 2 \text{ units} \end{aligned}

Step 4 · Compare side lengths

Comparing the lengths of the three sides AB=37,BC=37,CA=2AB = \sqrt{37}, \quad BC = \sqrt{37}, \quad CA = 2

Since AB=BC=37CAAB = BC = \sqrt{37} \ne CA, two sides are equal in length. Therefore, ΔABC\Delta ABC is an isosceles triangle.

Answer

Yes, (5,2)(5, -2), (6,4)(6, 4), and (7,2)(7, -2) are the vertices of an isosceles triangle.

Common Mistakes
  • Negative Sign Errors: Mishandling double negatives when subtracting coordinates, e.g., writing 4(2)4 - (-2) as 42=24 - 2 = 2 instead of 4+2=64 + 2 = 6.
  • Squaring Negative Numbers: Incorrectly evaluating (6)2(-6)^2 as 36-36 instead of +36+36.
  • Early Stopping: Calculating only two sides without verifying the third side (to confirm it is not equilateral, though equilateral triangles are technically also isosceles).

More questions in Exercise 7.1

Q1

Find the distance between the following pairs of points :

(i) (2,3),(4,1)(2, 3), (4, 1)

(ii) (5,7),(1,3)(-5, 7), (-1, 3)

(iii) (a,b),(a,b)(a, b), (-a, -b)

Q2

Find the distance between the points (0,0)(0, 0) and (36,15)(36, 15). Can you now find the distance between the two towns AA and BB discussed in Section 7.2.

Q3

Determine if the points (1,5),(2,3)(1, 5), (2, 3) and (2,11)(-2, -11) are collinear.

Q4

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

Q5

In a classroom, 4 friends are seated at the points AA, BB, CC and DD as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCDABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Q6

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) (1,2),(1,0),(1,2),(3,0)(-1, -2), (1, 0), (-1, 2), (-3, 0)

(ii) (3,5),(3,1),(0,3),(1,4)(-3, 5), (3, 1), (0, 3), (-1, -4)

(iii) (4,5),(7,6),(4,3),(1,2)(4, 5), (7, 6), (4, 3), (1, 2)

Q7

Find the point on the xx-axis which is equidistant from (2,5)(2, -5) and (2,9)(-2, 9).

Q8

Find the values of yy for which the distance between the points P(2,3)P(2, -3) and Q(10,y)Q(10, y) is 10 units.

Q9

If Q(0,1)Q(0, 1) is equidistant from P(5,3)P(5, -3) and R(x,6)R(x, 6), find the values of xx. Also find the distances QRQR and PRPR.

Q10

Find a relation between xx and yy such that the point (x,y)(x, y) is equidistant from the point (3,6)(3, 6) and (3,4)(-3, 4).

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