Question

5. A line, passing through (c, 2c) and (-1,2) is perpendicular
to the line 4x+2y+3=0.
find the value of C​

Answer 1

Answer:

5/3

Step-by-step explanation:

If two lines are ⊥, product of their slope is - 1.

On comparing with y = mx + c, slope of 4x + 2y + 3 is (-2). Let the slope of required line be m₁.

⇒ (-2)m₁ = - 1      ⇒ m₁ = 1/2

line formed by (-1, 2) and slope = 1/2 is:

⇒ y - 2 = (1/2)(x - (-1))

⇒ 2(y - 2) = (x + 1)

⇒ 2y - 4 = x + 1

⇒ x - 2y  + 5 = 0

  As this line contains (c, 2c), this must satisfy (x, y) = (c , 2c).

⇒ c - 2(2c) + 5 = 0

⇒ 5 = 3c

⇒ 5/3 = c

One more method is added here brainly.in/question/38311634

Answer 2

Given :-

A line, passing through (c, 2c) and (-1,2) is perpendicular

to the line 4x+2y+3=0.

To Find :-

Value of C

Solution :-

Let the line be m

$\sf -2\times m=-1$

$\sf m =\dfrac{-1}{-2}$

$\sf m =\dfrac{1}{2}$

So,

The line = (-1,2)

Slope = 1/2

$\sf y-2=\dfrac{x-(-1)}2$

$\sf 2(y-2) = x+1$

$\sf 2y-4=x+1$

$\sf x - 2y + 4+1=0$

$\sf x-2y+5=0$

$\sf c - 2 \times (2c) + 5 = 0$

$\sf c-4c+5=0$

$\sf 5 =4c-c$

$\sf 5 =3c$

$\sf c = \dfrac{5}3$