<u>Answer-</u>
The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are
<u>Solution-</u>
Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)
The applying the distance formula,
Two equations are formed because one will be upper from the the given line and other will be below it.
x - √3y - 4 = 0 → <u>Choice</u><u> </u><u>A</u>
Step-by-step explanation:
x - 4 = √3y
x - 4 <u>- √3y</u> = √3y <u>- √3y</u>
x - 4 - √3y = 0
x - √3y - 4 = 0
The student went wrong in step three. the solutions should have been -3 and 5
Answer:
200/12.5 and then you'll get the answer
Step-by-step explanation:
Answer:
y= -3/2x+9
Step-by-step explanation:
Well we have to know that to find a line perpendicular to a given line, you have to have opposite slopes. So that makes the equation begin as y=-3/2x because it's opposite of y=2/3x. Then we have to make sure it goes through the point (4,3). To do that, I tweaked the numbers in my graphing calculator and it worked with 9! So the equation is y= -3/2x+9
