Find an equation of the straight line that is perpendicular to the straight line x + 2y = 5 and that passes through the point (3, 7).
Answers
EXPLANATION.
Equation of straight line.
Perpendicular to straight line : x + 2y = 5.
Passes through the point = (3,7).
As we know that,
Slope of perpendicular line = b/a.
Slope of line : x + 2y = 5 = 2/1 = 2.
Slope of line = m = 2.
As we know that,
Equation of straight line.
⇒ (y - y₁) = m(x - x₁).
Put the values in the equation, we get.
⇒ (y - 7) = 2(x - 3).
⇒ y - 7 = 2x - 6.
⇒ 2x - 6 - y + 7 = 0.
⇒ 2x - y + 1 = 0.
MORE INFORMATION.
Different forms of the equation of straight line.
(1) = Slope - Intercept form : y = mx + c.
(2) = Slope point form : The equation of a line with slope m and passing through a point (x₁, y₁) is (y - y₁) = m(x - x₁).
(3) = Two point form : (y - y₁) = (y₂ - y₁)/(x₂ - x₁) (x - x₁).
(4) = Intercept form : x/a + y/b = 1.
(5) = Normal (perpendicular) form of a line : x cosα + y sinα = p.
(6) = Parametric form (distance form) : (x - x₁)/cosθ = (y - y₁)/sinθ = r.
Given :-
x + 2y = 5
To Find :-
Equation of the straight line
Solution :-
Finding slope
Slope = y₂ - y₁/x₂ - x₁
Slope = 2 - 0/1 - 0
Slope = 2/1
Slope = 2
Now
Equation = (y₂ - y₁) = m(x₂ - x₁)
Where m is the slope
(y - 7) = 2(x - 3)
(y - 7) = (2 × x) - (2 × 3)
y - 7 = 2x - 6
y - 2x = -6 + 7
y - 2x = 1
y - 2x - 1 = 0