This is grade 10 math. Chapter 2: Analytic Geometry - Line Segments and CirclesDescribe all points that are the same distance from points A(-3,-1) and B(5,3)It’s question 3 from attached screenshot.The answer at the back of text book is y=-2x+3Thank you!

Respuesta :

Given:

The given points are A(-3,-1) and B(5,3).

Required:

We need to find the set of all points that are the same distance from the points A(-3,-1) and B(5,3).

Explanation:

Recall that the set of points that are the same distance from the point A(-3,-1) and (5,3) is the perpendicular bisector of the line segment AB.

Consider the slope formula.

[tex]slope=\frac{y_2-y_1}{x_2-x_1}[/tex][tex]Substitute\text{ }y_2=3,y_1=-1,x_2=5,\text{ and }x_1=-3\text{ in the formula to find the slope of AB.}[/tex]

[tex]slope=\frac{3-(-1)}{5-(-3)}=\frac{3+1}{5+3}=\frac{4}{8}=\frac{1}{2}[/tex]

We get the slope of the line segment AB is 1/2.

The slope of the perpendicular is the negative reciprocal of the slope of the line segment AB.

[tex]The\text{ negative reciprocal of }\frac{1}{2}=-2.[/tex]

[tex]The\text{ slope of the perpendicular line is -2.}[/tex]

The perpendicular line is passing through the midpoint of A(-3,-1) and B(5,3).

Consider the midpoint formula.

[tex]midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]

[tex]Substitute\text{ }y_2=3,y_1=-1,x_2=5,\text{ and }x_1=-3\text{ in the formula to find the midpoint of AB.}[/tex]

[tex]midpoint=(\frac{-3+5}{2},\frac{-1+3}{2})[/tex][tex]midpoint=(\frac{2}{2},\frac{2}{2})[/tex][tex]midpoint=(1,1)[/tex]

Consider the general form of the line equation.

[tex]y=mx+b[/tex]

Substitute m=-2, x=1, and y=1 in the line equation to find b.

[tex]1=(-2)(1)+b[/tex][tex]1=-2+b[/tex]

Add 2 to both sides of the equation.

[tex]1+2=-2+b+2[/tex][tex]3=b[/tex]

Substitute m=-2 and b=3 in the line equation.

[tex]y=-2x+3[/tex]

We get the perbenticular bisector y =-2x+3.

Final answer:

[tex]y=-2x+3[/tex]