Find the equation of the straight line passing through the point (-2
, 1) and perpendicular to the line 3x+5y+2=0.
Answers
EXPLANATION.
Equation of straight lines.
Passing through point = (-2,1).
Perpendicular to the line : 3x + 5y + 2 = 0.
As we know that,
Slope of the perpendicular line = b/a.
Slope of the line : 3x + 5y + 2 = 0 is 5/3.
⇒ Slope = m = 5/3.
Equation of straight line.
⇒ (y - y₁) = m(x - x₁).
Put the values in the equation, we get.
⇒ (y - 1) = 5/3(x - (-2)).
⇒ (y - 1) = 5/3(x + 2).
⇒ 3(y - 1) = 5(x + 2).
⇒ 3y - 3 = 5x + 10.
⇒ 5x + 10 - 3y + 3 = 0.
⇒ 5x - 3y + 13 = 0.
MORE INFORMATION.
Slope of line.
(1) = m = tanθ, where θ is the acute angle made by a line with the positive direction of x axis in anticlockwise.
(2) = The slope of a line joining two points (x₁, y₁) and (x₂, y₂) is given by,
m = (y₂ - y₁)/(x₂ - x₁).
Given :-
Point = (-2,1)
To Find :-
Equation of the straight line
Solution :-
On comparing the equation with ax + by + c = 0
We get
a = 3
b = 5
Slope = b/a
Slope = 5/3
Now
Finding equation of straight line
Equation = (y - y') = s(x - x')
y - 1 = 5/3[x - (-2)]
y - 1 = 5/3[x + 2]
y - 1 = 5/3[x + 2]
3(y - 1) = 5[x + 2]
3y - 3 = 5x + 10
3y - 5x = 10 + 3
3y - 5x = 13
3y - 5x - 13 = 0