Please help!
Write the equation of a line perpendicular to the line 6x – 3y=10 and containing the point (-6, 2). Show all your work
2007-10-10 03:24:07 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
the slope of a line perpindicular to the given line would exhibit an opposite-reciprocal slope. In other words, first convert the equation to y=mx+b form y= 2x+10. Take the opposite of the slope (-2), and then the reciprocal (-1/2). Now you have the framework for your new equation y= -1/2x+b. From there, simply substitute -6 and 2 as your x and y value, respectively.
2007-10-10 03:36:17 · answer #1 · answered by kawawa 2 · 0⤊ 0⤋
i rewrite the equation
3y =6x-10 or y=2x -10/3 , the slope of this line is a=2
a line is perpendicular when the product of slopes a*a'=1
so the line has a slope of -1/2 and y =(1/2) x +b as it passes through -6, 2 so 2 = -6*-(1/2)+b or 2=3+b and b=-1
the equation is y = -0.5*x -1
2007-10-10 03:37:40 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
6x-3y=10
-3y=-6x+10
y=2x+10/-3
y=2x-10/3
slope 2
slope of perpendicular line -1/2
equation of perpendicular line
y=(-1/2)x+b
it passes through -6,2
2=(-1/2)(-6)+b
2=3+b
b=-1
y=(-1/2)x-1 is the equation
2007-10-10 04:59:33 · answer #3 · answered by cidyah 7 · 0⤊ 0⤋
2 is the answer "show all YOUR work"
2007-10-10 03:31:41 · answer #4 · answered by ray56_32223 2 · 0⤊ 0⤋