Question

Find the equation of the line perpendicular to 5x-3y+1=0 and passing through (4, -3)

Answer 1

EXPLANATION.

Equation of the line perpendicular to the line,

⇒ 5x - 3y + 1 = 0.

passing through the point = (4,-3).

As we know that,

Slope of the perpendicular line = b/a.

Slope = 5x - 3y + 1 = 0.

Slope = -3/5.

Equation of the lines.

⇒ (y - y₁) = m(x - x₁).

Put the value in this equation, we get.

⇒ (y - (-3)) = -3/5(x - 4).

⇒ (y + 3) = -3/5(x - 4).

⇒ 5(y + 3) = -3(x - 4).

⇒ 5y + 15 = -3x + 12.

⇒ 5y + 3x + 3 = 0.

                                                                                                                         

MORE INFORMATION.

Point of inflection.

If at any point p, the curve is concave on one side and convex on other sides with respect to x-axis, then the point p is called the point of inflection. thus p is a point of inflection if at p,

d²y/dx² = 0, but d³y/dx³ ≠ 0.

Also point p is a point of inflection if,

f''(x) = f'''(x) = ⇒ fⁿ⁻¹(x) = 0  and fⁿ(x) ≠ 0 for odd n.

Answer 2

Formula Used:-

Let us consider a line which passes through the point (a, b) having slope m, then equation of line is given by

$\tt \ \: :  ⟼ y - b \: = \: m \: (x - a)$

Let us consider an equation of line ax+ by + c = 0, then

$\tt \ \: :  ⟼ Slope \: of \: line \: = - \dfrac{a}{b}$

$\tt \ \: :  ⟼ Two \: lines \: having \: slope \: m_1 \: and \: m_2 \: are \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt \ \: :  ⟼ perpendicular \: if \: and \: only \: if \: m_1 \: \times m_2 = - 1$

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$\large\underline\purple{\bold{Solution :- }}$

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☆ Given equation of line is 5x - 3y + 1 = 0.

$\tt\implies \:Slope \: of \: line \: = - \dfrac{5}{ - 3} = \dfrac{5}{3}$

☆ Since the required line is perpendicular to 5x - 3y + 1 =0

$\tt \ \: :  ⟼ \therefore \: Slope \: (m) = - \dfrac{3}{5}$

☆ Therefore, the equation of line passing through the point (4, - 3) having slope (-3/5) is given by

$\tt \ \: :  ⟼ y - ( - 3) = - \dfrac{3}{5} (x - 4)$

$\tt \ \: :  ⟼ y + 3 = - \dfrac{3}{5} (x - 4)$

$\tt \ \: :  ⟼ 5y + 15 = - 3x + 12$

$\bf\implies \:3x + 5y + 3 = 0$