Question

show that the points (-5,1),(1,-1) and (4,-2) are collinear​

Answer 1

Points given:  (-5,1); (1,-1); (4,-2)

x1 = -5

x2 = 1

x3 = 4

y1 = 1

y2 = -1

y3 = -2

For the points to be collinear :-

x₁(y₂ -y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0

Thus,   LHS = x₁(y₂ -y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)

= -5(-1+2) + 1(-2-1) + 4(1+1)

= -5 + -3 + 8

= -5 + 5

=0

Therefore, LHS=RHS

Answer 2

Answer:

$\huge\underbrace\mathfrak\red{Answer:-}$

Area of a triangle with vertices (x

1

,y

1

) ; (x

2

,y

2

) and (x

3

,y

3

) is

∣

∣

∣

∣

∣

2

x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)

∣

∣

∣

∣

∣

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Hence, substituting the points (x

1

,y

1

)=(5,1) ; (x

2

,y

2

)=(1,P) and (x

3

,y

3

)=(4,2) in the area formula, we get

∣

∣

∣

∣

∣

2

5(P−2)+1(2−1)+4(1−P)

∣

∣

∣

∣

∣

=0

=>5P−10+1+4−4P=0

=>P=5

...

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