# Passes through (6, -5), perpendicluar to the line whose equation is 3x - 1/5 y = 3. Write an equation in slope-intercept form for the line that satisfies each set of conditions.

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The given equation of the line is: 3x-(1/5)y=3

The given coordinate point is: (x1,y1) = (6,-5)

Rewrite the given equation in Slope-Intercept Form, in other words, solve for

"y":

3x-(1/5)y=3

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

y=15x-15

The slope of the given line is: 15

The slope of its perpendicular line it's the reciprocal of the original slope with opposite sign: -(1/15)

Use the Slope-Intercept Form, y-y1=m(x-x1) and solve for "y" using m= -(1/15) and the original coordinate point (x1,y1)=(6,-5)

y-y1=m(x-x1)

y-(-5)=-(1/15)(x-6)

y+5=-(1/15)x+(2/5)

y=-(1/15)x-(23/5)

Given an equation of a line L1 is 3x - 1/5 y = 3

3x - (1/5) y = 3

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

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

=> y =5(3x - 3)

=> y = 15 x - 15

so the slope of the line L1 be m_1 is = 15

as we know that the product of the slopes of the two perpendicular lines is equal to -1

let the slope of the required line is m_2

so ,

(m_1)(m_2) = -1

=> m_2 = -1/15

As,the slope-intercept form of the required line is

`y= (m_2)x+b`

from the above we know `m_2 = -1/15` , so the line equation is

`y= (-1/15)x+b` --------------(1)

we need to find the value of b , as the line passes through the point

`(x,y)= (6, -5 )` , then on substituting we get

`-5 =(-1/15)*(6)+b`

=> `b = -5 +2/5 = -23/5 `

so the equation of the line is

`y= (-1/15)x-(23/5)`