Question

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

Answer 1

Answer:

$\sqrt{50}$

Step-by-step explanation:

Step 1: Put the line in slope intercept form,

$x + 7y = 7$

$7y = - x + 7$

$y = - \frac{x}{7} + 1$

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

$\frac{ - x}{7} + 1 = 7x - 49$

$- x + 7 = 49x - 343$

$350 = 50x$

$7 = x$

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

To find y, plug in 7 for any function

$\frac{ - 7}{7} + 1 = 0$

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).

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

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

$d = \sqrt{50}$

So the distance to the line root of 50.