Write the standard form of the equation of a line if the point on the line nearest to the origin is at (6, 8)

For (6,8) to be the closest point, a line drawn from the origin to (6,8) must be perpendicular to the equation of the line

slope

slope from (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)

so slope from (0,0) to (6,8) is (8-0)/(6-0)=8/6=4/3

perpendicular lines have slopes that muliply to -1
4/3 times what=-1
times both sides by 3/4
what=-3/4

so the slope is -3/4

use point slope form
y-y1=m(x-x1)
slope is m and a point on the line is (x1,y1)

the point is (6,8) and the slope is -3/4
y-8=-3/4(x-6)

standard form is ax+by=c where a,b,c are integers and a is positive (usually)

so
y-8=-3/4(x-6)
y-8=-3/4x+18/4
times both sides by 4
4y-32=-3x+18
add 3x to both sides
3x+4y-32=18
add 32 to both sides
3x+4y=50

the equation is 3x+4y=50

It should be a or d i’m not sure