Question

Answer this please,quickly ​

Answer 1

Step-by-step explanation

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0Since, it passes through (3,5)

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0Since, it passes through (3,5)⇒3−10+λ=0

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0Since, it passes through (3,5)⇒3−10+λ=0⇒λ=7

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0Since, it passes through (3,5)⇒3−10+λ=0⇒λ=7Therefore, the line is x−2y+7=0.

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0Since, it passes through (3,5)⇒3−10+λ=0⇒λ=7Therefore, the line is x−2y+7=0.The point of intersection of x−2y+7=0 and 2x+3y−14=0 is (1,4).

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0Since, it passes through (3,5)⇒3−10+λ=0⇒λ=7Therefore, the line is x−2y+7=0.The point of intersection of x−2y+7=0 and 2x+3y−14=0 is (1,4).The distance between (3,5) and (1,4)

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0Since, it passes through (3,5)⇒3−10+λ=0⇒λ=7Therefore, the line is x−2y+7=0.The point of intersection of x−2y+7=0 and 2x+3y−14=0 is (1,4).The distance between (3,5) and (1,4)=√ (3−1) 2+(5−4) 2

(3−1) 2+(5−4) 2 =√4+1

4+1=√ 5

5

5 .

.

Answer 2

Answer:

correct option is C

Step-by-step explanation:

let the equation of the line parallel to x-2y = 1

is x-2y + y( lemma) =0

since it passes through (3,5)

= 3 - 10 + y(lemma)

=y (lemma)= 7

therefore the line is X - 2y + 7 = 0

the point of intersection of X - 2y + 7 = 0

and 2x+3y-14 =0 is (1,4)

the distance between (3,5) and (1,4) is

=root (3-1)² + (5-4)² = root 4+1 =root 5

OPTION C

( y here is lemma)