Question

Find the gradient of a line and also find the gradient of the perpendicular line on the following line.
y= 3x – 5

Answer 1

EXPLANATION.

Equation of the line.

⇒ y = 3x - 5.

As we know that,

⇒ Gradient of line = slope of the line.

⇒ The line is in the form of : y = mx + c.

⇒ Gradient of line = 3.

⇒ Gradient of perpendicular line = b/a.

We can write equation as,

⇒ 3x - y - 5 = 0.

⇒ Gradient = b/a = -1/3.

                                                                                                                       

MORE INFORMATION.

Equation of tangent.

Equation of tangent to the curve y = f(x) at P(x₁, y₁) is,

(y - y₁) = m(x - x₁).

(1) = The tangent at (x₁, y₁) is parallel to x-axes = (dy/dx) = 0.

(2) = The tangent at (x₁, y₁) is parallel to y-axes = (dy/dx) = ∞.

(3) = The tangent lines makes equal angles with the axes = (dy/dx) = ± 1.

Answer 2

Given : Equation of a line : y = 3x - 5 .

Exigency To Find : The Gradient of a line & the gradient of the perpendicular line .

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⠀⠀⠀ Finding Gradient of a line :

⠀⠀⠀❒ Gradient of a line is also known as slope of line .

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \qquad \maltese \: \bf Equation \: of \: a \: line \: in \: slope \:-intercept \: form\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ Eq_n \:: y = mx + c }\bigg\rgroup$

⠀⠀⠀Here m is the slope [ Gradient ] of a line & c is the y- intercept .

Given Equation :

• Equation : y = 3x - 5 .

⠀⠀⠀Now , By Comparing both Equation we get ,

$\qquad :\implies \frak{\underline{\purple{\:m = 3 }} }\:\: \bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Gradient\: [ \ Slope \ ] \:of\:line \:is\:\bf{3}}}}$

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⠀⠀⠀ Finding Gradient of a Perpendicular line :

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad \qquad \maltese \: \bf Gradient \: of \: a \:Perpendicular \: line \:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ Gradient \:_{(Perpendicular \:Line\:)}\:: \dfrac{b}{a} }\bigg\rgroup$

Given Equation :

$\qquad :\implies \sf Eq_n \: = \: y = 3x - 5$

$\qquad :\implies \sf Eq_n \: = \:3x - y - 5 = 0$

⠀⠀⠀⠀⠀Here ,

• b = -1
• a = 3

$\qquad :\implies \sf \dfrac {b}{a}\: = \dfrac {-1}{3}\: \:$

$\qquad :\implies \bf Gradient \: : \: \dfrac {b}{a}\: = \dfrac {-1}{3}\: \:$

$\qquad :\implies \frak{\underline{\purple{\:Gradient_{( Perpendicular \:Line\:)} \: : \: \dfrac {b}{a}\: = \dfrac {-1}{3}\: }} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Gradient \:of\:Perpendicular \:line\:is\:\bf{ \dfrac {-1}{3}\: }}}}$

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