Question

Write the equation of the perpendicular bisector of AB if A(–6, –4) and B(2, 0).

Answer 1

Answer:

Step-by-step explanation:

The equation of perpendicular bisector of A(-6, -4 ) and B(2, 0) is y = -2x - 6

Solution:

Given that we have to find the equation of perpendicular bisector of A(-6, -4 ) and B(2, 0)

A perpendicular bisector, bisects a line segment at  right angles

To obtain the equation we require slope and a point on it

Find the midpoint and slope of the given points and then we can find the equation

Find the midpoint:

Given points are A(-6, -4 ) and B(2, 0)

The midpoint is given as:

Substituting the values we get,

Find the slope of given points:

Then the slope of perpendicular bisector is given as:

We know that product of slopes of given line and slope of line perpendicular to it is equal to -1

Let the slope of perpendicular bisector be

Find the equation of line with slope -2 and point (-2, -2)

The equation of line in slope intercept form is given as:

y = mx + c -------- eqn 1

Where "m" is the slope and "c" is the y - intercept

Substitute (x, y) = (-2, -2) and slope m = -2 in eqn 1

-2 = -2(-2) + c

-2 = 4 + c

c = -2 - 4

c = -6

Substitute c = -6 and m = -2 in eqn 1

y = -2x - 6

Thus the required equation of perpendicular bisector is found

Answer 2

Step-by-step explanation:

✞︎Given

Write the equation of the perpendicular bisector of AB if A(–6, –4) and B(2, 0).

y=−2x−2

y=−2x−6

y=1/2x−6

y=1/2x−2

✯To Find

• The perpendicular bisector of AB if A(–6, –4) and B(2, 0).

Let P(x,y) be any point on the perpendicular bisector of AB. Then,

PA=PB

$\sqrt{(x−3)2+(y−6)2=(x+3)2+(y−4)2}$

$\sqrt{(x−3)2+(y−6)2=(x+3)2+(y−4)2 }$

$\sqrt{ x2−6x+9+y2−12y+36=x2+6x+9+y2−8y+16}$

⇒

$\tt\large\underline\red{12x+4y−20=0}$

$\large{✰} \bf \underline\color{purple}{⇒3x+y−5=0 - }$

Hence, the equation of the perpendicular bisector of AB is 3x+y-5=0

• (i) We know that the coordinates of any point on x-axis are of the form(x,o). In other words, y-coordinate of every point on x-axis is zero. So, putting y=o in (i),we get

• 3x−5=0⇒x=35

Thus, the perpendicular bisector of AB cuts x-axis at(35,0)

(ii)  The coordinates of any point on y-axis are of the form (o,y). Putting x=0 in (i), we get

   

$\tt\large\underline{y−5=0⇒y=5}$

• Thus, the perpendicular bisector of AB intersects y-axis at (0,5)

Hope it helps

$\large{✰} \bf \underline\color{red}{ItzSizuka50 }$