Question

The equation of the line passing through (1;2) and perpendicular to the line x+2y+3=0 is​

Answer 1

Answer :

The required line is 2x - y = 0

Given :

The equation of the line passes through the point (1,2)

The line is perpendicular another line x + 2y + 3 = 0

To Find :

The equation of the line

Solution :

$\sf{Let \: \: us \: \: consider \: \: the \: \: gradient \: \: of \: required \: \: equation \: \: be \: m_1}$

The required line is perpendicular to x + 2y + 3 = 0

Transforming this equation to slope gradient form :

$\sf \implies 2y = -x -3 \\\\ \sf \implies y = -\dfrac{1}{2}x - \dfrac{3}{2}$

$\sf{Thus \: \: its \: \: gradient} , \: \bf {m_2 =- \dfrac{1}{2}}$

Now we know that if two lines are perpendicular the product of their gradient is -1

$\sf \implies m_{1}.m_{2} = -1 \\\\ \sf \implies m_{1}. (-\dfrac{1}{2} )= -1 \\\\ \bf\implies m_1 = 2$

Again from equation of a line passing through one given point :

$\sf \implies y - y_1 = m_{1}(x - x_{1}) \\\\ \sf \implies y - 2 = 2(x - 1)\\\\ \sf \implies y - 2 = 2x - 2 \\\\ \sf \implies 2x - 2 - y + 2 = 0 \\\\ \bf \implies 2x - y = 0$

Formulae Used Here :

If m₁ and m₂ are the gradients of two perpendicular lines then the product of the gradients of these lines are given by

m₁.m₂ = -1

Equation of a line passing through a given point :

If m is the gradient of a line and the line passes through the point (x₁ , y₁) then the equation of the line is given by :

y - y₁ = m(x - x₁)

Answer 2

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