A square of area 25 sq. unit is formed by taking two sides as 3 x +4 y = k 1 and 3 x +4 y = k 2, then find value of |k1-k2|.

Question

A square of area 25 sq. unit is formed by taking two sides as 3 x +4 y = k 1 and 3 x +4 y = k 2, then find value of |k1-k2|. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/7e5b3d438f4f6fd0140a7bf4dc53872d-.png" /> Side d=|(k1-k2/√9+16)| Area =25 d2=25 (( k 1- k 2/5))2=25 ⇒| k 1- k 2|=25

Q. A square of area $25 s q$ . unit is formed by taking two sides as $3 x + 4 y = k_{1}$ and $3 x + 4 y = k_{2}$ , then find value of $∣ k_{1} - k_{2} ∣$ .

Side $d = ∣ ∣ \frac{k _{1} - k _{2}}{9 + 16} ∣ ∣$
Area $= 25$
$d^{2} = 25$
$(\frac{k _{1} - k _{2}}{5})^{2} = 25$
$\Rightarrow ∣ k_{1} - k_{2} ∣ = 25$

Q. A square of area 25sq. unit is formed by taking two sides as 3x+4y=k1​ and 3x+4y=k2​, then find value of ∣k1​−k2​∣.

Side d=∣∣​9+16​k1​−k2​​∣∣​ Area =25 d2=25 (5k1​−k2​​)2=25 ⇒∣k1​−k2​∣=25

Q. A square of area $25 s q$ . unit is formed by taking two sides as $3 x + 4 y = k_{1}$ and $3 x + 4 y = k_{2}$ , then find value of $∣ k_{1} - k_{2} ∣$ .

Side $d = ∣ ∣ \frac{k _{1} - k _{2}}{9 + 16} ∣ ∣$
Area $= 25$
$d^{2} = 25$
$(\frac{k _{1} - k _{2}}{5})^{2} = 25$
$\Rightarrow ∣ k_{1} - k_{2} ∣ = 25$