find the equation of line passing through (-2,5) and perpendicular to the line 3x+4y=6
Answers
Answer:
Please see the attachment
EXPLANATION.
Equation of line passing through the point (-2, 5)
Perpendicular to the line 3x + 4y = 6.
As we know that.
Slope of perpendicular line : ax + by + c is b/a.
Slope of perpendicular line : 3x + 4y = 6 is 4/3.
Slope : m = 4/3.
Equation of line.
⇒ (y - y₁) = m(x - x₁).
Put the value in the expression, we get.
⇒ (y - 5) = 4/3[x - (-2)].
⇒ (y - 5) = 4/3(x + 2).
⇒ 3(y - 5) = 4(x + 2).
⇒ 3y - 15 = 4x + 8.
⇒ 4x - 3y + 15 + 8 = 0.
⇒ 4x - 3y + 23 = 0.
∴ The equation of line is 4x - 3y + 23 = 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 line : x cosα + y sinα = p.
(6) Parametric form (distance form) : (x - x₁)/cosθ = (y - y₁)/sinθ = r.