Point B lies on line J as well, why aren't we considering slope of line J & L?
GreenlightTestPrep wrote:
Line J: 4x - 7 = 6y
Line K: 6y + 3x = -2
Line L: 2x = 5 - y
In the x-y coordinate plane, lines J, K and L are defined by the above equations.
Point B is the point of intersection of lines J and K. Point C is the point of intersection of lines K and L.
What is the slope of line segment BC?
A) -3/2
B) -2/3
C) -1/2
D) 1/2
E) 2/3
Key Concepts: point B lies on line K, and point C lies on line K
Since both points lie on line K, the slope between points B and C will be the same as the slope of line K.
To find the slope of line K, let's take the equation of line K (6y + 3x = -2), and rewrite it in slope y-intercept form (y = mx + b)
Take: 6y + 3x = -2
Subtract 3x from both sides to get: 6y = -3x - 2
Divide both sides by 6 to get: y = (-3/6)x - 2/6
Simplify to get: y = (-1/2)x - 1/3
So, line K has a slope of -1/2 and a y-intercept of -1/3
Answer: C
Cheers,
Brent