Question

NEED HELP 30 PTS !!

Answer 1

Answer:

$(\frac{48}{61} , -\frac{40}{61} )$

Step-by-step explanation:

The closest point to the given line must be on a line perpendicular to it.

So we start by finding the slope of the given line and then evaluating what the slope of the perpendicular line to it must be.

First write the given line in slope_y-intercept form:

SO solve for y in $6x-5y = 8$ :

$y=\frac{6x-8}{5} = \frac{6}{5} x -\frac{8}{5}$

Since the slope is 6/5, the slope of a perpendicular line to the given one must be the "opposite of the reciprocal": -5/6

So the line that we are looking for must have the form:

$y=-\frac{5}{6} x + b$

Since we want it to go through the origin of coordinates (0,0) because we are finding the point closer to the origin, the y-intercept b must be zero:

$y=-\frac{5}{6} x$

Now, to find the point of intersection of both lines (the given one and the perpendicular one through the origin) we equal both expressions:

$\frac{6}{5}x -\frac{8}{5} = -\frac{5}{6} x$

and solve this equation for "x": $x(\frac{6}{5} +\frac{5}{6} ) = \frac{8}{5}$

Therefore $x=\frac{48}{61}$

Now to find the y-value of the point on the line, we replace this x in either line equation:

$y=-\frac{5}{6} (\frac{48}{61})=-\frac{40}{61}$

Then the closest point is: $(\frac{48}{61} , -\frac{40}{61} )$