345,722 views
0 votes
0 votes
NO LINKS!! Please help me with this problem. Find a formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector l of segment AB.​

NO LINKS!! Please help me with this problem. Find a formula that expresses the fact-example-1
User Bunker
by
2.3k points

2 Answers

17 votes
17 votes

Answer:


\textsf{Slope-intercept form}: \quad y=(7)/(5)x-(19)/(5)


\textsf{Standard form}: \quad 7x-5y=19

Explanation:

A perpendicular bisector is a line that intersects another line segment at 90°, dividing it into two equal parts.

To find the perpendicular bisector of segment AB, find the slope of AB and the midpoint of AB.

Define the points:

  • Let (x₁, y₁) = A(-5, 4)
  • Let (x₂, y₂) = B(9, -6)

Slope of AB


\textsf{slope}\:(m)=(y_2-y_1)/(x_2-x_1)=(-6-4)/(9-(-5))=(-10)/(14)=-(5)/(7)

Midpoint of AB


\textsf{Midpoint}=\left((x_2+x_1)/(2),(y_2+y_1)/(2)\right)=\left((9+(-5))/(2),(-6+4)/(2)\right)=(2,-1)

If two lines are perpendicular to each other, their slopes are negative reciprocals.

Therefore, the slope of the line that is perpendicular to line segment AB is ⁷/₅.

Substitute the found perpendicular slope and the midpoint of AB into the point-slope formula to create an equation for the line that is the perpendicular bisector of line segment AB:


\implies y-y_1=m(x-x_1)


\implies y-(-1)=(7)/(5)(x-2)


\implies y+1=(7)/(5)x-(14)/(5)


\implies y=(7)/(5)x-(14)/(5)-1


\implies y=(7)/(5)x-(19)/(5)

Therefore, the formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector of segment AB is:


\textsf{Slope-intercept form}: \quad y=(7)/(5)x-(19)/(5)


\textsf{Standard form}: \quad 7x-5y=19

NO LINKS!! Please help me with this problem. Find a formula that expresses the fact-example-1
User BerriJ
by
2.7k points
22 votes
22 votes

Given points

  • A(-5, 4), B(9, - 6)

Find the midpoint of AB

  • M = ((- 5 + 9)/2, (4 - 6)/2) = (2, -1)

Find the slope of AB

  • m = (-6 - 4)/(9 + 5) = - 10 / 14 = - 5/7

The point P is on the perpendicular line to AB that passes through its midpoint.

We know perpendicular lines have opposite-reciprocal slopes.

So the line we are looking for has a slope of 7/5.

Use the point-slope equation to find the line:

  • y - y₁ = m(x - x₁)
  • y - (-1) = 7/5(x - 2)
  • y + 1 = 7/5(x - 2) Point- slope form
  • y = 7/5x - 19/5 Slope- intercept form
  • 5y = 7x - 19
  • 7x - 5y = 19 Standard form

Choose any form above of the same line.

User Andy Day
by
3.1k points