Show that Δ A B C , where A ( 2 , 0 ) , B ( 2 , 0 ) , C ( 0 , 2 ) and Δ P Q R where P ( – 4 , 0 ) , Q ( 4 , 0 ) , R ( 0 , 4 ) are similar triangles

In Δ A B C , the coordinates of the vertices are A ( – 2 , 0 ) , B ( 2 , 0 ) , C ( 0 , 2 ) A B = ( 2 + 2 ) 2 + ( 0 - 0 ) 2 = 4 C B = ( 0 - 2 ) 2 + ( 2 - 0 ) 2 = 8 = 2 2 A C = ( 0 + 2 ) 2 + ( 2 - 0 ) 2 = 8 = 2 2 In Δ P Q R , the coordinates of the vertices are P ( – 4 , 0 ) , Q ( 4 , 0 ) , R ( 0 , 4 ) P Q = ( 4 + 4 ) 2 + ( 0 - 0 ) 2 = 8 Q R = ( 0 - 4 ) 2 + ( 4 - 0 ) 2 = 4 2 P R = ( 0 + 4 ) 2 + ( 4 - 0 ) 2 = 4 2 .. Now, for Δ A B C and Δ P Q R to be similar, the corresponding sides should be proportional. So, A B P Q = B C Q R = C A P R ⇒ 4 8 = 2 2 4 2 = 2 2 4 2 = 1 2 Thus, Δ A B C is similar to Δ P Q R .

Point A lies on the line segment P Q joining P ( 6 , - 6 ) and Q ( - 4 , - 1 ) in such a way that P A P Q = 2 5 . If the point A also lies on the line 3 x + k y + 1 = 0 , find the value of k .

Suppose the coordinates of A be x , y . Here, P A P Q = 2 5 . So, P A + A Q = P Q ⇒ P A + A Q = 5 P A 2 ∵ P A = 2 5 P Q ⇒ A Q = 5 P A 2 - P A ⇒ A Q P A = 3 2 ⇒ P A A Q = 2 3 Suppose, ( x , y ) be the coordinates of A , which dives P Q in the ratio 2 : 3 internally. Then using section formula, we get, x = 2 × - 4 + 3 × 6 2 + 3 = - 8 + 18 5 = 10 5 = 2 y = 2 × - 1 + 3 × - 6 2 + 3 = - 2 - 18 5 = - 20 5 = - 4 Now, the point 2 , - 4 lies on the line 3 x + k ( y + 1 ) = 0 . Therefore, 3 × 2 + k - 4 + 1 = 0 ⇒ 3 k = 6 ⇒ k = 6 3 = 2 Therefore, k = 2 .