Mathematics : asked on eliza35
 30.09.2020

Indicate the equation of the line, in standard form, that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5).

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Mathematics
Step-by-step answer
P Answered by Master

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  x +3y = -3

Step-by-step explanation:

The midpoint of the segment with the given end points is ...

  M = ((4, 1) +(2, -5))/2 = (6, -4)/2 = (3, -2)

The difference between coordinates of the given points is ...

  (∆x, ∆y) = (4, 1) -(2, -5) = (2, 6)

__

The equation of the perpendicular bisector can be written as ...

  ∆x(x -h) +∆y(y -k) = 0 . . . . line through (h, k) ⊥ to one with slope ∆y/∆x

  2(x -3) +6(y -(-2)) = 0

  2x +6y +6 = 0 . . . . . simplify to a general-form equation

To put this in standard form, we need the constant on the right, and all numbers mutually prime. We can subtract 6 and divide by 2 to get there.

  2x +6y = -6

  x + 3y = -3


Indicate the equation of the line, in standard form, that is the perpendicular bisector of the segme
Mathematics
Step-by-step answer
P Answered by Master
For the answer to the question above, the slope of segment AC is  (-6/22) = -3/11.

The slopes of perpendicular lines are negative reciprocals, so 
you know that the slope of the perpendicular bisector will be 11/3.

The midpoint of segment AC is  (7, -5).

Now you have the slope of the perpendicular bisector, and a
point on it.  You should be able to complete the equation now.

Notice how sneaky this question is.  You don't need to know
point-B at all, and you don't even need to know that there's
any triangle.  All you need to know is points  A  and  C.
Mathematics
Step-by-step answer
P Answered by Master
For the answer to the question above, the slope of segment AC is  (-6/22) = -3/11.

The slopes of perpendicular lines are negative reciprocals, so 
you know that the slope of the perpendicular bisector will be 11/3.

The midpoint of segment AC is  (7, -5).

Now you have the slope of the perpendicular bisector, and a
point on it.  You should be able to complete the equation now.

Notice how sneaky this question is.  You don't need to know
point-B at all, and you don't even need to know that there's
any triangle.  All you need to know is points  A  and  C.
Mathematics
Step-by-step answer
P Answered by Specialist
Given triangle GHI with G(4, -3), H(-4, 2), and I(2, 4), find the perpendicular bisector of line HI in standard form.

The correct answer to this question is (-1,3)

G is locate at Quadrant IV
H is located Quadrant II
I is located at Quadrant I

The perpendicular bisector of line HI is (-1, 3)
Mathematics
Step-by-step answer
P Answered by PhD
First find midpoint:  \left( \frac{-1+5}{2}, \frac{6+5}{2}\right) = (2, 5.5) 

Find slope of line that passes through R and S:    slope = \frac{6-5}{-1-5} = \frac{-1}{6}   

Negative reciprocal of slope to get slope of perpendicular:    new slope = 6

Line will be:  y-5.5=6(x-2) 

  y = 6x - 6.5
Mathematics
Step-by-step answer
P Answered by PhD
Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
%  (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2}  \right)}

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it.  Well, what is the slope of RS anyway?

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
%   (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}

and let's check the reciprocal negative of that,

\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies  +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}
Mathematics
Step-by-step answer
P Answered by Specialist

Step-by-step explanation:

We are given that a line that is perpendicular bisector of  the line whose end points are R(-1,6) and S(5,5).

To find the equation of perpendicular line then we have find the slope of line and a point through which perpendicular line is  passing .

To find the point through which perpendicular line is passing then we have find the mid point of the line joining points R(-1,6) and S(5,5)

Mid point formula: The coordinates of mid point

x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}

Let P is the mid point of the line joining the points R (-1,6) and S(5,5)

Therefore, the coordinates of mid point P by using mid point formula

x=\frac{-1+5}{2},y=\frac{6+5}{2}

x=2,y=\frac{11}{2}

The coordinates of mid point P(2,\frac{11}{2}) of the line joining points  R and S

Slope of line RS,m_1=\frac{y_2-y_1}{x_2-x_1}

y_1=6,y_2=5,x_1=-1,x_2=5

Slope of line RS,m_1=\frac{5-6}{5+1}=-\frac{1}{6}

Slope of perpendicular line is opposite reciprocal of the line RS

Hence, the slope of perpendicular line m_2=-\frac{1}{m_1}

Slope of perpendicular line,m_2=6

The perpendicular line is passing through the mid point P(2,\frac{11}{2}) because it bisect the line RS at mid point P.

The equation of perpendicular line passing through the point P with slope 6

y-\frac{11}{2}=6(x-2)

The equation of a line which is perpendicular to RS

\frac{2y-11}{2}=6x-12

The equation of  line which is perpendicular to the line RS is given by

2y-11=12x-24

The  equation of a line which is  perpendicular to the  line RS

12x-2y=24-11

Hence, the required equation of a line which is perpendicular to the line RS

12x-2y=13

Mathematics
Step-by-step answer
P Answered by PhD
Hello,
1) We must find the equation of the line RS:
R=(-1;6)S=(5;5)
y-6=(x+1)(5-6)/(5+1)y=-1/6*(x+1)+6
 y=-x/6+35/6 Slope=-1/6
The perpendicular has the slope 6.

2) Find M the middle of [RS]M=((5-1)/2;(6+5)/2)=(2:11/2)
3) Equation of the perpendicular bisector:
y-11/2=(x-2)*6y=6x-12+11/2
So y=6x-13/2
And sorry for my poor english

3)

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