SOLUTION: Write an equation of the line containing the given point and perpendicular to the given line. Express your answer in the form y=my+b
(6, 7); 3x + y = 6
Can someone please help me
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: Write an equation of the line containing the given point and perpendicular to the given line. Express your answer in the form y=my+b
(6, 7); 3x + y = 6
Can someone please help me
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Question 136406: Write an equation of the line containing the given point and perpendicular to the given line. Express your answer in the form y=my+b
(6, 7); 3x + y = 6
Can someone please help me solve this problem?? I need help with these I do not understand these at all.
You can put this solution on YOUR website! solving for y in the original equation puts it into slope-intercept form (y=mx+b)
__ y=-3x+6 __ so the slope is -3
perpendicular lines have slopes that are negative reciprocals
__ so the slope of the new line is (-1)/(-3) or 1/3
using point-slope, the equation for the new line is y-7=(1/3)(x-6)
distributing __ y-7=(1/3)x-2 __ adding 7 __ y=(1/3)x+5
