Find the equation of the line passing through the point (-3, 4) and perpendicular to the line 2x + 6y = 1
Answers
EXPLANATION.
Equation of the line passing though points (-3,4).
Perpendicular to the line : 2x + 6y = 1.
As we know that,
Slope of a perpendicular line = b/a.
Slope of a line : 2x + 6y = 1 ⇒ 6/2 = 3.
Slope = 3.
Equation of line,
⇒ (y - y₁) = m(x - x₁).
Put the value in equation, we get.
⇒ ( y -(4)) = 3(x - (-3)).
⇒ y - 4 = 3(x + 3).
⇒ y - 4 = 3x + 9.
⇒ y - 3x = 13.
MORE INFORMATION.
Some facts about the normal.
(1) = The slope of the normal drawn at point p(x₁, y₁) to the curve y = f(x) is -(dy/dx).
(2) = If normal makes an angle ∅ with positive direction of x-axis then dy/dx = -cot∅.
(3) = If normal is parallel to x-axis then dy/dx = ∞.
(4) = If normal is parallel to y-axis then dy/dx = 0.
Answer:
Given :-
point (-3, 4) and perpendicular to the line 2x + 6y = 1
To Find :-
Equation
Solution :-
We know that
Slope of line = b/a
Here,
b = 6
a = 2
Slope = 6/2
Slope = 3
Now,
Equation of line,
(y - y₁) = m(x - x₁).
Putting value
[y -(4)] = 3[x - (-3)].
y - 4 = 3(x + 3).
y - 4 = 3x + 9.
y - 3x = 9 + 4
y - 3x = 13.