Determine the equation of the perpendicular bisector of the segment with endpoints (–5, 3) and (5, 5).

Mathematics

1 answer:

Ann [662]1 year ago

4 0

Answer:

y=-5x+4

Step-by-step explanation:

First lets find the midpoint of the points (-5,3) and (5,5).

(0,4) is the midpoint by the midpoint formula.

Next we find the slope of the 2 points.

2/10 or 1/5

To find the perpendicular bisector, we need to find the negative reciprocal of the slope, which is -5x. However, we want the line to pass through (0,4), so we make the y-intercept 4.

Therfore, the answer is y=-5x+4

You might be interested in

Suppose a and b are the solutions to the quadratic equation 2x^2-3x-6=0. Find the value of (a+2)(b+2).

KonstantinChe [14]

Answer:

(a+2)(b+2) = 4

Step-by-step explanation:

We are given the following quadratic equation:

$2x^2-3x-6=0$

Let a a and b be the solution of the given quadratic equation.

Solving the equation:

$2x^2-3x-6=0\\\text{Using the quadratic formula}\\\\x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\text{Comparing the equation to }ax^2 + bx + c = 0\\\text{We have}\\a = 2\\b = -3\\c = -6\\x = \dfrac{3\pm \sqrt{9-4(2)(-6)}}{4} = \dfrac{3\pm \sqrt{57}}{4}\\\\a = \dfrac{3+\sqrt{57}}{4}, b = \dfrac{3-\sqrt{57}}{4}$

We have to find the value of (a+2)(b+2).

Putting the values:

$(a+2)(b+2)\\\\=\bigg(\dfrac{3+\sqrt{57}}{4}+2\bigg)\bigg(\dfrac{3-\sqrt{57}}{4}+2\bigg)\\\\=\bigg(\dfrac{11+\sqrt{57}}{4}\bigg)\bigg(\dfrac{11-\sqrt{57}}{4}\bigg)\\\\=\dfrac{121-57}{156} = \dfrac{64}{16} = 4$