
J M.
asked 01/16/24Find the point (x,y) on the line y=−3x−3 that is equidistant from the points (−10,−4) and (3,3).
2 Answers By Expert Tutors

Dayv O. answered 01/17/24
Caring Super Enthusiastic Knowledgeable Pre-Calculus Tutor
following Doug's suggestion for solving,
line through (-10,-4) and (3,3)
had slope (7/13)
midpoint is between the points is (-7/2,-1/2)
perpendicular line through midpoint
(-13/7)(x+7/2)=y+1/2
intersection with
-3x-3=y
3x-(13/7)x=(1/2)-3+91/14
8x/7=56/14
x=7/2
y=(-21/2)-3=-27/2

Anthony T. answered 01/16/24
CS & Math Student | Tutor for Calculus, Algebra, SAT Math
We'll use the distance formula and the given line equation. The distance formula for the distance between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)2 + (y2 - y1)2)
Let's label the equidistant point as (x, y). The distances from this point to (-10, -4) and (3, 3) must be equal. So, we have:
√((x + 10)2 + (y + 4)2 ) = √((x - 3)2 + (y - 3)2)
Notice that y = -3x - 3, which means we can substitute this into the previous equation for every value of y.
Solving this tells us the point on the line that's equidistant from the points, is (3.5, -13.5).
Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Doug C.
All points that are equidistant from (-10,-4) and (3,3) lie on the perpendicular bisector of the segment with those points as endpoints. Find the equation of the perpendicular bisector. Then find the point of intersection of that line with the line y = -3x-3. That point lies on the given line and also on the perpendicular bisector and is therefore equidistant from those points.01/16/24