Write an equation of the line passing through the point (5, 1) that is perpendicular to the line 5x+3y=15.

Here, we are required to Write an equation of the line passing through the point (5, 1) that is perpendicular to the line 5x+3y=15.

The equation of the line is; 3x - 5y = 10.

By comparison with the equation of a straight line: y = Mx + c

The equation 5x+3y=15 can be rewritten as ;

y = (-5/3)x + 5

Therefore, the slope of the equation is; (-5/3)

The product of the slope of two perpendicular lines is: -1.

therefore m1m2 = -1

Therefore; M2 = -1/(-5/3)

Therefore M2 = 3/5

The equation of the line passing through the point (5, 1) that has slope, M2 = 3/5 is;

3/5 = (y - 1)/(x - 5)

By Cross product;

3x - 15 = 5y -5

Therefore, the equation of the line is;

3x - 5y = 10.

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y = x - 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

5x + 3y = 15 ( subtract 5x from both sides )

3y = - 5x + 15 ( divide all terms by 3 )

y = - x + 5 ← in slope- intercept form

with slope m = -

Given a line with slope m then the slope of a line perpendicular to it is

= - = - = , thus

y = x + c ← is the partial equation

To find c substitute (5, 1) into the partial equation

1 = 3 + c ⇒ c = 1 - 3 = - 2

y = x - 2 ← equation of perpendicular line

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3