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Anna007 [38]
3 years ago
7

What is an equation of the line that is perpendicular to y+1=-3(x-5) and passes through the point 4,-6

Mathematics
2 answers:
Sveta_85 [38]3 years ago
7 0

Answer:

y + 6 = 1/3 (x - 4)

Step-by-step explanation:

Oliga [24]3 years ago
6 0

Answer:

y=\frac{1}{3}x-\frac{16}{3}

Step-by-step explanation:

Arranging the given equation in slope-intercept form ( y = mx + b ):

y+1=-3(x-5)\\y+1=-3x+15\\y=-3x+15-1\\y=-3x+14

<em>We know that the perpendicular line would have a slope (m) that is </em><em>negative reciprocal </em><em>of this (\frac{1}{3}). Thus we can write the perpendicular line's equation as  y=\frac{1}{3}x+b. </em>

<em />

<em>Now putting x = 4 and y = -6 into the equation, we can solve for b:</em>

<em>y=\frac{1}{3}x+b\\-6=\frac{1}{3}(4)+b\\-6=\frac{4}{3}+b\\b=-6-\frac{4}{3}\\b=-\frac{16}{3}</em>

<em />

Thus, the equation of the perpendicular line is  y=\frac{1}{3}x-\frac{16}{3}

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