Mathematics, 29.10.2020 22:40 isaacbrown7one

Find the midpoint of PQ with endpoints P(-5, -5) and Q (3, 3). Then write an equation of the line that passes through the midpoint and is perpendicular to PQ. This line is the perpendicular bisector.

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Answer from: dontrailgrant98

The equation of the line that passes through the midpoint and is perpendicular to PQ is $\rm \bold{ y = -x -2}$

Coordinates of P are (-5.-5)

Coordinates of Q are ( 3,3)

The midpoint of the PQ can be found out by the formula given in equation (1)

$\rm (x , y) = ((x_1 +x_2)/2, (y_1+y_2)/2 ).........(1) \\\\Where\; (x_1,y_1) \; and \; (x_2,y_2) are\; the\; coordinates\; of\; two\; given \; points \; that \; make \; the \; line$

The midpoint of line joining P and Q we can find out the by putting coordinates in equation (1)

Let O be the mid point of PQ

We can write that the coordinates of O

$\rm ((-5 +3)/2 , (-5 + 3) /2 ) = (-1,-1)$

The slope of line passing through two points

$\rm (x_1, y_1)\; and \; (x_2,y_2) \\\\m = \dfrac{y_2 -y_1}{x_2-x_1}$

Slope of line passing through P and Q is given

$\rm \\\\m = \dfrac{y_2 -y_1}{x_2-x_1} = \dfrac{3-(-5)}{3-(-5)} = \dfrac{8}{8} = 1$

Let the slope of line that is perpendicular is M

hence the slope M can be found out in the following way

$\rm m \times M = -1 \\1\times M = -1 \\M = - 1$

Let the equation of line passing through point O

(-1,-1) with slope M is given as formulated in equation (2)

$\rm y = Mx +c......(2)$

Point O (-1,-1) will satisfy the equation of line

$\rm -1 = -M + c \\-1 = 1 + c\\c = -2$

On putting the values of c and M in equation (2) we get equation of line

$\rm \bold{ y = -x -2}$

So, The equation of the line that passes through the midpoint and is perpendicular to PQ is $\rm \bold{ y = -x -2}$

For more information please refer to the link given below

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Answer from: bryan9892

y = -x-2

Step-by-step explanation:

The formula for calculating the midpoint is expressed as:

M(X, Y) = [(x2+x1)/2, (y2+y1)/2]

Given the coordinates P(-5, -5) and Q (3, 3)

Midpoint = [-5+3/2, -5+3/2]

Midpoint = (-2/2, -2/2)

Midpoint = (-1, -1)

Next is to find an equation of the line that passes through the midpoint and is perpendicular to PQ.

The standard form of equation of a line is y = mx+c

Get the slope m:

m = y2-y1/x2-x1

m = 3-(-5)/3-(-5)

m = 8/8

m = 1

For two lines to be perpendicular, the poduct of their slope must be -1

mM = -1

M(1) = -1

M = -1/1

M = -1

To get the slope, substitute the midpoint (-1,-1) and the slope m = -1 into the equation y = mx+c

-1 = -1(-1)+c

-1 = 1+c

c = -1-1

c = -2

Substitute M =-1 and c = -2 into the equation to get the equation of the required line;

y = Mx+c

y = (-1)x-2

y = x-2

Hence the required equation is y = -x-2

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