Question

HELP PLEASE!! Write an equation in slope-intercept form of the line with slope −6/5 that contains the point (−5,3).

Answer 1

Explanation:

we know,

Equation of line with slope m

and passing through (x1,y1) is,

y-y1=m(x-x1)

here, slope= -6/5 and the point is (-5,3)

So, the equation is,

`y-3=-(6)/(5) (x-(-5))\\or, y-3=-(6)/(5) (x+5)\\`

Multiplying by 5

`5(y-3)=-6(x+5)\\or, 5y-15=-6x-30\\or, 5y=-6x-30+15\\or, 5y=-6x-15\\`

Dividing by 5

`y=-(6)/(5) x-3`

Answer 2

Answer: y = -6/5 x - 3

Solution :

The equation of line is

y = mx + b

y = y coordinate

m = slope

x = x coordinate

b = y intercept

so to write the equation of line we need to find the y-intercept (b)

substitute the slope (-6/5) and coordinates given (-5,3) ,

y = mx + b

3 = (-6/5) (-5) + b

3 = 6 + b

3 - 6 = b

-3 = b

So the equation would be,

y = mx + b

y = -6/5 x - 3