BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.8, Problem 43E
To determine

To calculate: The point on y-axis that is equidistant from the points (5,5)and (1,1) .

Expert Solution

Answer to Problem 43E

The point on y-axis that is equidistant from the points (5,5) and (1,1) is (0,4) .

Explanation of Solution

Given information:

The two points (5,5) and (1,1) from which the point on y-axis is equidistant.

Formula used:

Distance formula between two points X(a1,b1) and Y(a2,b2) is calculated as,

  d(XY)=(a2a1)2+(b2b1)2

Calculation:

Consider the given two points (5,5) and (1,1) .

Since, the point that is equidistant from the two points (5,5) and (1,1) is on y-axis , so, its x-coordinate will be zero.

Let us assume the point on y-axis be P(0,y) and name the given points as A and B respectively A(5,5) and B(1,1) .

It is given that point P is equidistant from the points A and B, so, the distance between P and A is equal to the distance between P and B, which can be mathematically written as,

  d(A,P)=d(B,P)

Recall the distance formula between two points X(a1,b1) and Y(a2,b2) is calculated as,

  d(XY)=(a2a1)2+(b2b1)2

Apply it,

  d(A,P)=d(B,P)(05)2+(y+5)2=(01)2+(y1)225+y2+10y+25=1+y22y+1y2+10y+50=y22y+2

Square both sides of the equation as,

  (y2+10y+50)2=(y22y+2)2y2+10y+50=y22y+2

Cancel out the common terms and isolate yon left hand side and solve it further as,

  y2+10y+50=y22y+210y+2y=25012y=48y=4

Since, y=4 , so, the point on y-axis is P(0,4) .

Thus, point on y-axis that is equidistant from the two points (5,5) and (1,1) is (0,4) .

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