Question

please answer Topic:- Locus

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Answer 1

TOPIC :-

Locus

GIVEN :-

• (p,q) (acosθ , asinθ) , (bcosθ, asinθ) are vertices of triangle

TO FIND :-

• Locus of Centroid of triangle

SOLUTION:-

Let locus of a Centroid triangle be (h,k)

${Centroid \:of \:a \:triangle} = \dfrac{x_1+x_2+x_3}{3} , \dfrac{y_1+y_2+y_3}{3}$

ATQ Centroid can be

$\dfrac{p+acos\theta + bcos\theta}{3}, \dfrac{q + bsin\theta + asin\theta}{3}$ = (h,k)

Equating to (h,k)

p+acosθ + bcosθ = 3h

acosθ + bcosθ = 3h-p ------- eq 1

q + bsinθ + asinθ = 3k

bsinθ + asinθ = 3k - q -------- eq 2

Adding equation 1 , 2 and Squaring

(acosθ + bcosθ )²+ (bsinθ + asinθ)² = (3h-p)² (3k-q)²

In RHS Replacing h, k with x,y

(acosθ + bcosθ )²+ (bsinθ + asinθ)² = (3x-p)² (3y-q)²

Expand in LHS It is in form of (a+b)² = a²+ 2ab + b²

(acosθ + bcosθ )²+ (bsinθ + asinθ)² = (3x-p)² (3y-q)²

a²cos²θ + b²cos²θ + 2abcosθcosθ + b²sin²θ + a²sin²θ + 2absinθsinθ = (3x-p)² (3y-q)²

a²cos²θ + b²cos²θ + 2abcos²θ + b²sin²θ + a²sin²θ + 2absin²θ = (3x-p)²+ (3y-q)²

Regrouping the terms

a²cos²θ + a²sin²θ + b²cos²θ + b²sin²θ + 2absin²θ + 2ab cos²θ= (3x-p)² +(3y-q)²

Taking common a², b² , 2ab

a² (sin²θ + cos²θ) + b²(sin²θ + cos²θ) +2ab(sin²θ+cos²θ) =(3x-p)² +(3y-q)²

We know Trigonmetry Identity :- sin²θ + cos²θ = 1

a² + b² + 2ab = (3x-p)² +(3y-q)²

We know that (a + b)² = a²+ 2ab + b²

(a + b)² = (3x-p)² +(3y-q)²

Option -A

Answer 2

Answer:

Step-by-step explanation

TOPIC :-

Locus

GIVEN :-

(p,q) (acosθ , asinθ) , (bcosθ, asinθ) are vertices of triangle

TO FIND :-

Locus of Centroid of triangle

SOLUTION:-

Let locus of a Centroid triangle be (h,k)

ATQ Centroid can be

= (h,k)

Equating to (h,k)

p+acosθ + bcosθ = 3h

acosθ + bcosθ = 3h-p ------- eq 1

q + bsinθ + asinθ = 3k

bsinθ + asinθ = 3k - q -------- eq 2

Adding equation 1 , 2 and Squaring

(acosθ + bcosθ )²+ (bsinθ + asinθ)² = (3h-p)² (3k-q)²

In RHS Replacing h, k with x,y

(acosθ + bcosθ )²+ (bsinθ + asinθ)² = (3x-p)² (3y-q)²

Expand in LHS It is in form of (a+b)² = a²+ 2ab + b²

(acosθ + bcosθ )²+ (bsinθ + asinθ)² = (3x-p)² (3y-q)²

a²cos²θ + b²cos²θ + 2abcosθcosθ + b²sin²θ + a²sin²θ + 2absinθsinθ = (3x-p)² (3y-q)²

a²cos²θ + b²cos²θ + 2abcos²θ + b²sin²θ + a²sin²θ + 2absin²θ = (3x-p)²+ (3y-q)²

Regrouping the terms

a²cos²θ + a²sin²θ + b²cos²θ + b²sin²θ + 2absin²θ + 2ab cos²θ= (3x-p)² +(3y-q)²

Taking common a², b² , 2ab

a² (sin²θ + cos²θ) + b²(sin²θ + cos²θ) +2ab(sin²θ+cos²θ) =(3x-p)² +(3y-q)²

We know Trigonmetry Identity :- sin²θ + cos²θ = 1

a² + b² + 2ab = (3x-p)² +(3y-q)²

We know that (a + b)² = a²+ 2ab + b²

(a + b)² = (3x-p)² +(3y-q)²

Option -A