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Find the distance from the point (6, - 7) to the line x + 7y = 7

User Jstricker
by
8.2k points

1 Answer

6 votes

Answer:


√(50)

Explanation:

Step 1: Put the line in slope intercept form,


x + 7y = 7


7y = - x + 7


y = - (x)/(7) + 1

Step 2: The line that contains the point must be perpendicular to the original line.

So the slope of this line must be 7, and pass through (6,-7).

So we have


y + 7 = 7(x - 6)


y + 7 = 7x - 42


y = 7x - 49

Step 3: Find where the lines intersect at:

Now, we set that equal to -x/7+1


( - x)/(7) + 1 = 7x - 49


- x + 7 = 49x - 343


350 = 50x


7 = x

So the two lines intersect at (7,y).

To find y, plug in 7 for any function


( - 7)/(7) + 1 = 0

So y=0.

So the two lines intersect at (7,0).

Step 4: Use distance formula,

Find the distance between (7,0) and (6,-7).


d = \sqrt{ (- 7 - 0) {}^(2) + (6 - 7) {}^(2) }


d = \sqrt{ - 7) {}^(2) + ( - 1) {}^(2) }


d = √(50)

So the distance to the line root of 50.

User Valor
by
8.8k points

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