A point moves such that the area of the triangle formed by it with the points (1, 5) and (3,-7) is 21 sq unit. Then locus of the point is:

Question

A point moves such that the area of the triangle formed by it with the points (1, 5) and (3,-7) is 21 sq unit. Then locus of the point is: (A)  6x+y-32=0  (B)  6x-y+32=0  (C)  x+6y-32=0  (D)  6x-y-32=0

Tardigrade Admin · Accepted Answer

Let (x, text y) be the required point ∴ (1/2)| beginmatrix x y 1 1 5 1 3 -7 1 endmatrix |=21 ⇒ x(5+7)-y(1-3)+1(-7-15)=42 ⇒ 12x+2y+22=42 ⇒ 12x+2y-64=0 ⇒ 6x+y-32=0

Q. A point moves such that the area of the triangle formed by it with the points (1, 5) and $(3, - 7)$ is 21 sq unit. Then locus of the point is:

$6 x + y - 32 = 0$

$6 x - y + 32 = 0$

$x + 6 y - 32 = 0$

$6 x - y - 32 = 0$

none of these

Let $(x, y)$ be the required point $∴$ $\frac{1}{2} ∣ ∣ x 13 y 5 - 7 111 ∣ ∣ = 21$ $\Rightarrow$ $x (5 + 7) - y (1 - 3) + 1 (- 7 - 15) = 42$ $\Rightarrow$ $12 x + 2 y + 22 = 42$ $\Rightarrow$ $12 x + 2 y - 64 = 0$ $\Rightarrow$ $6 x + y - 32 = 0$

Q. A point moves such that the area of the triangle formed by it with the points (1, 5) and (3,−7) is 21 sq unit. Then locus of the point is:

6x+y−32=0

6x−y+32=0

x+6y−32=0

6x−y−32=0