Question

reduce the equation 2x+3y-7=0 to the normal form​

Answer 1

Answer:

1 y + 7 0

Step-by-step explanation:

answer 70

Answer 2

The Normal Form Of 2x + 3y - 7 = 0 Is

$x( \frac{2}{ \sqrt{13} }) + y( \frac{3}{ \sqrt{13} } ) = \frac{7}{ \sqrt{13} }$

GIVEN

straight line Equation = 2x + 3y - 7 = 0

TO FIND

To reduce the given equation in the normal form.

SOLUTION

We can simply solve the above problem as follows -

We are given an equation of straight line as -

2x + 3y -7 = 0.

We know that, the normal form is -

x cos ω + y sin ω = p

where,

p = perpendicular distance from origin

ω = angle between perpendicular and positive axis.

The equation is -

2x + 3y -7 = 0.

We can also write it as-

2x + 3y = 7

Dividing the equation by the square root of the square of the coefficient of x and y.

the coefficient of x and y. =

the coefficient of x and y. = $\sqrt{( {2}^{2} ) + {(3)}^{2} } = \sqrt{13}$

the coefficient of x and y. = $\sqrt{( {2}^{2} ) + {(3)}^{2} } = \sqrt{13}$Dividing the equation with √13.

the coefficient of x and y. = $\sqrt{( {2}^{2} ) + {(3)}^{2} } = \sqrt{13}$Dividing the equation with √13.$\frac{2x + 3y}{ \sqrt{13} } = \frac{7}{ \sqrt{13} }$

the coefficient of x and y. = $\sqrt{( {2}^{2} ) + {(3)}^{2} } = \sqrt{13}$Dividing the equation with √13.$\frac{2x + 3y}{ \sqrt{13} } = \frac{7}{ \sqrt{13} }$$= x( \frac{2}{ \sqrt{13} }) + y( \frac{3}{ \sqrt{13} } ) = \frac{7}{ \sqrt{13} }$

Comparing the equations,

cos w = 2/√13

sin w = 3/√13

Hence, the normal form of 2x + 3y - 7 = 0 is

Hence, the normal form of 2x + 3y - 7 = 0 is $x( \frac{2}{ \sqrt{13} }) + y( \frac{3}{ \sqrt{13} } ) = \frac{7}{ \sqrt{13} }$

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