Q20 of 157 Page 724

Show that the points A(3, 0), B(6, 4) and C(– 1, 3) are the vertices of an isosceles right triangle.

In an isosceles triangle any two sides are equal.


AB = √{(6 – 3)2 + (4 – 0)2}


= √{(3)2 + (4)2}


= √{9 + 16}


= √{25} = 5 units


BC = √{(– 1 – 6)2 + (3 – 4)2}


= √{(– 7)2 + (– 1)2}


= √{49 + 1}


= √{50}


AC = √{(– 1 – 3)2 + (3 – 0)2}


= √{(– 4)2 + (3)2}


= √{16 + 9}


= √{25} = 5 units


AB = AC


∴ It is an isosceles triangle.


More from this chapter

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18

Using the distance formula, show that the given points are collinear:

(i) (1, – 1), (5, 2) and (9, 5)


(ii) (6, 9), (0, 1) and (– 6, – 7)


(iii) (– 1, – 1), (2, 3) and (8, 11)


(iv) (– 2, 5), (0, 1) and (2, – 3).

 19

Show that the points A(7, 10), B(– 2, 5) and C(3, – 4) are the vertices of an isosceles right triangle.

 21

If A(5, 2), B(2, – 2) and C(– 2, t) are the vertices of a right triangle with ∠B = 90°, then find the value of t.

 22

Prove that the points A(2, 4), B(2, 6) and C(2 + √3, 5) are the vertices of an equilateral triangle.