Question

Question 6 Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Class X1 - Maths -Straight Lines Page 212

Answer 1

only half got bro sorry loke this u try

Answer 2

Let the given points
A Ξ ( 4, 4)
B Ξ ( 3, 5)
C Ξ ( -1, -1)

we know, area of traingle = 1/2[x₁(y₂-y₃) + x₂(y₃ -y₁) + x₃(y₁ - y₂) ]
so, area of ∆ABC = 1/2[ 4( 5+1) + 3(-1 - 4) -1(4 - 5)]
= 1/2[ 24 - 15 +1 ] = 5 ≠ 0
here, area of ∆ABC ≠ 0 .
hence, A, B , and C are the vertices of ∆.

now,
slope of AB , m₁ = ( 5 - 4)/(3 - 4) = 1/-1 = -1
slope of BC ,m₂ = ( -1 - 5)/(-1 - 3) = -6/-4 = 3/2
slope of CA, m₃ = (-1 - 4)/(-1 - 4) = 1
∵ slope of AB × slope of CA = (-1) × (1) = -1
[we know , if product of slopes of two line is -1, then angle between them is 90°]
so, ABC is right angled traingle