Answered

Get comprehensive solutions to your questions with the help of IDNLearn.com's experts. Our platform offers comprehensive and accurate responses to help you make informed decisions on any topic.

vedanta Excel In Opt. Mathematics - Book 9
Find the locus of a point which moves so that

(a) Its distance from (1, 2) and (2, -3) are equal.​

Sagot :

Prishagupta035idn

(1 , 2) (2, -3)

(x1,y1) (x2, y2)

Distance formual

√(x2-x1)+(y2-y1)

{substitute numbers in formula}

= √(2-1)^+(-3-2)^

=√(1)^+(-5)^ [here square root and

square get cancelled]

= (1)+(-5)

= 1-5

= -4
A point that is equidistant from two points will travel along the perpendicular bisector of the line that joins them.
The gradient of the line between (1,2) and (2,-3) is (-3 - 2)/(2 - 1) = -5
The perpendicular gradient = -1/-5 or 0.2
The midpoint between (1,2) and (2,-3) is ((1+2)/2,(2+-3)/2) = (1.5, -0.5)
Substituting into y=mx + c
-0.5 = 0.2 x 1.5 + c
-0.5 = 0.3 + c
c = -0.5 - 0.3 = -0.8
So the locus is the line y = 0.2x - 0.8
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.