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Alekssandra [29.7K]
3 years ago
7

How do I find the equation of the perpendicular bisector between (-2,5) and (2,-1)?

Mathematics
1 answer:
Arturiano [62]3 years ago
7 0

Answer:

y =  \frac{2}{3} x + 2

Step-by-step explanation:

Start by finding the slope of the given line.

\boxed{ slope = \frac{y _{1} - y_2 }{x_1 - x_2} }

Slope of given line

=  \frac{5 - ( - 1)}{ - 2 - 2}

=  \frac{5 + 1}{ - 4}

=  \frac{6}{ - 4}

=  -  \frac{3}{2}

A perpendicular bisector cuts through the line at its midpoint perpendicularly.

The product of the slopes of two perpendicular lines is -1.

Let the slope of the perpendicular bisector be m.

-  \frac{3}{2} m =  - 1

m =  - 1 \div ( -  \frac{3}{2} )

m =  - 1 \times ( -  \frac{2}{3} )

m =  \frac{2}{3}

y =  \frac{2}{3}x  + c, where c is the y-intercept.

To find the value of c, we need to substitute a pair of coordinates that lies on the perpendicular bisector into the equation. Since the perpendicular bisector passes through the midpoint of the given line, we can use the midpoint formula to find the coordinates.

\boxed{midpoint = ( \frac{x _{1} + x _2}{2} , \frac{y_1  + y_2}{2} )}

Midpoint of given line

=  ( \frac{ - 2+ 2}{2} , \frac{5 - 1}{2} )

=( \frac{0}{2} , \frac{4}{2} )

= (0, 2)

y =  \frac{2}{3} x + c

When x= 0, y= 2,

2= ⅔(0) +c

2= 0 +c

c= 2

Thus, the equation of the perpendicular bisector is y =  \frac{2}{3} x + 2.

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What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of
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Answer:

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Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, (x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)

<u>So, The coordinates of the triangle A'B'C' are:</u>

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

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