Question

The base of a triangle lie's along the line x= a and is of length a. The area of the triangle is a?. The locus the third vertex is
(A) (x + a) (x - 3a) = 0
(B) (x – a) (x + 3a) = 0
(C) (2 - a) (x – 3a) = 0
(D) (2 - a) (x - 2a) = 0​

Answer 1

Given:

The base of a triangle lie's along the line x= a and is of length a. The area of the triangle is a?.

To find:

The locus the third vertex is

Solution:

From given, we have,

The base of a triangle lie's along the line x= a and is of length a. The area of the triangle is a².

Let (h, k) be the coordinates of 3rd vertex.

As the base is located at x = a, so the height will (h - a)

Area of triangle = 1/2 × base × height

a² = 1/2 × |a| × (h - a)

⇒ a² = 1/2 × |a| × (x - a)

2a² = |a| × (x - a)

± 2a = x - a

2a = x - a   and -2a = x - a

2a + a = x and -2a + a = x

3a = x and -a = x

⇒ x - 3a = 0 and x + a = 0

Option (A) (x + a) (x - 3a) = 0 is correct.

Answer 2

Answer:

(x+a)(x-3a)

Step-by-step explanation:

area of triangle is 1/2 ×a|h-a|=a²

h= -a or h= 3a

(h+a)(h-3a)

(x+a)(x-3a)=0