find the distance of the line 4x+7y+5=0 from the ponit (1,2) along the line 2x-y=0
Answers
EXPLANATION.
Distance of the line 4x + 7y + 5 = 0.
From the point (1, 2) along the line : 2x - y = 0.
As we know that,
⇒ 4x + 7y + 5 = 0. - - - - - (1).
⇒ 2x - y = 0. - - - - - (2).
⇒ y = 2x. - - - - - (2).
Put the value of y = 2x in equation (1), we get.
⇒ 4x + 7(2x) + 5 = 0.
⇒ 4x + 14x + 5 = 0.
⇒ 18x + 5 = 0.
⇒ 18x = - 5.
⇒ x = - 5/18.
Put the value of x = - 5/18 in equation (2), we get.
⇒ y = 2x.
⇒ y = 2(-5/18).
⇒ y = - 5/9.
Their Co-ordinates = (-5/18, -5/9).
As we know that,
Distance Formula :
⇒ d = √(x₂ - x₁)² + (y₂ - y₁)².
Using this formula in the equation, we get.
⇒ (1, 2) and (-5/18, -5/9).
⇒ d = √[(-5/18 - 1)² + (-5/9 - 2)²].
⇒ d = √(-23/18)² + (-23/9)².
⇒ d = √[(-23/9)²(1/4 + 1).
⇒ d = (-23/9)√(5/4).
⇒ d = (-23/9)√(5)/2.
⇒ d = (-23√5)/18.
As we know that,
Distance is always positive.
We can write as,
⇒ d = [(23√5)/18] sq. units.