Choose the equation of the line passing through the point (-1, 3) and perpendicular to y = -negative one thirdx + 7.
y = 3x - 12
y = 3x + 6
y = 3x - 6
y = 3x
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first we need to find the slope of the equation given.
y = -1/3x + 7 (In y = mx + b form, the m is the slope. Therefore, the slope of this line is -1/3)
But we need a perpendicular line, so we need the negative reciprocal of that slope. All that means is " flip " the slope and change the sign. The negative reciprocal of -1/3 = 3/1 or just 3 (you see how I flipped the slope and changed the sign)
so we have slope(m) = 3
(-1,3) x = -1 and y = 3
y = mx + b and solve for b
3 = 3(-1) + b
3 = -3 + b
3 + 3 = b
6 = b
your perpendicular line is : y = 3x + 6
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hoose the equation of the line passing through the point (-3, 1) and parallel to y = -x - 4.
y = x + 2
y = -x - 2
y = -x + 4
y = x - 4
y = -x - 4 (slope is -1)
Since we need a parallel slope, the original slope stays the same.
slope(m) = -1
(-3,1) x = -3 and y = 1
y = mx + b for b
1 = -1(-3) + b
1 = 3 + b
1 - 3 = b
-2 = b
your parallel line is : y = -x - 2
any questions ?
Since you worked it out I understand now.. Thanks you (: