What is the equation of the line that passes through the point (6,-6) and has a slope of -5/3?

Answer 1

Solution:

The slope of the line m = -5/3 writing this in slope form m = y- y1 / x -x1 , the line passes through point ( 6,-6)

-5/3 = ( y - (-6)) / x - 6 , -5/3 = y+6 / x-6 , then y+6 = -5/3( x-6) , and y = -5/3( x-6) - 6 = -5/3 x + (30/3) - 6

= -5/3 x + 10 - 6 = -5/3 x + 4 and the equation of the line is y = (-5/3)x + 4

This can also be written of the form y = mx + b where m = -5/3 and b is the y intercept , the y intercept occurs when x = 0 so in the equation to solve for this go the slope form and set x =0 and solve for y , -5/3 = y+6 / 0-6 = y+6 / -6 ,

30/3 = y + 6, 10 = y + 6 , y = 10-6 = 4 and y = ax + b , y = 0(a) + b = b = 4

y = -5/3 x + 4

Answer 2

Equation of a line having slope = -5/3

Y= -5/3 X +b

As it passes through (6, -6)

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

Gives b = 4

Equation of line will be

Y = -5/3 (X) +4

or 3Y = -5 X +12

or 3Y +5X = 12

Answer 3

Point intercept form of equation of a line is:

y - y1 = m(x - x1), where m is slope and (x1,y1) is a point on the line.

using the given information, the equation of the line in that form is:

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

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

To put into slope intercept form (y = mx + b, where b is the y intercept):

y + 6 = -5/3 (x) + 10

y = -5/3 (x) + 4