# Thread: need help on problem 2

1. ## need help on problem 2

hi, i thought id start a new thread for the 2nd problem since the 1st problem went for 2 pages

Problem 2

Find the distance (correct to three decimal places) from the point (-5,-2) to the line 3x+y=8. Hint: the distance from a point to a line is the perpendicular distance.

2. Originally Posted by zazeem
hi, i thought id start a new thread for the 2nd problem since the 1st problem went for 2 pages

Problem 2

Find the distance (correct to three decimal places) from the point (-5,-2) to the line 3x+y=8. Hint: the distance from a point to a line is the perpendicular distance.

first thing's first, find the line that's perpendicular to $y = -3x + 8$ that passes through (-5,-2). since the line is perpendicular, its slope is the negative inverse of the original line, namely, 1/3. now we use the point-slope form to find our line:

$y - y_1 = m(x - x_1)$

$\Rightarrow y + 2 = \frac 13(x + 5)$

$\Rightarrow y = \frac 13x - \frac 13$

now we need to know where these lines meet. so we set them equal to each other:

$-3x + 8 = \frac 13x - \frac 13$

$\Rightarrow \boxed{x = \frac 52}$

plug this x-value into either line and you will find the corresponding y-value to be $\boxed{y = \frac 12}$

thus you want to use the distance formula to find the distance between the points (-5,-2) and (5/2, 1/2)

can you continue? perhaps more importantly, do you understand what i did?