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Jimboslice
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
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
Thank you so much. will you help on another ?
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
Since you worked it out I understand now.. Thanks you (: