Question

Find an equation for the perpendicular bisector of the line segment whose endpoints are (5,-9)and (-9,-5)

Answer 1

Answer:

y = ¹⁴/₄.x

Step-by-step explanation:

First of, if the bisector is perpendicular to the line segment, then we can find the gradient of the bisector ($m_{b}$) using the rule/principle:

Let:

m = gradient of the line segment

Then:

$m_{b}$ = $-\frac{1}{m}$

We can find m since we have two points that fall on the line segment, (5, -9) and (-9, -5):

$m =$ Δy/Δx

$m = \frac{-9 - (-5)}{5 - (-9)} \\\\ m = -\frac{4}{14}$

We can now find $m_{b}$:

$m_{b} = -\frac{1}{(-\frac{4}{14}) } \\\\ m_{b} = 1.\frac{14}{4} \\\\ m_{b} = \frac{14}{4}$

The equation of a line can be found using:

y - y₁ = m(x - x₁)

We have the gradient of the perpendicular bisector, the only other thing we need to identify the equation of the bisector is coordinates of a point that fall on the line;

We know the line will pass through the point exactly midway between (5, -9) and (-9, -5) since it is a bisector;

This can be found by:

$x_{1} = -9 + \frac{5 - (-9)}{2} \\ x_{1} = -9 + 7 \\ x_{1} = -2 \\\\ y_{1} = -5 + \frac{-9 - (-5)}{2} \\ y_{1} = -5 +(-2) \\ y_{1} = -7 \\\\ (-2, -7)$

We have a point on the line and the gradient so we can now find the equation:

$y - (-7) = \frac{14}{4}(x - (-2)) \\\\ y + 7 = \frac{14}{4}(x + 2) \\\\ 4y + 28 = 14(x + 2) \\\\ 4y + 28 = 14x + 28 \\\\ 4y = 14x \\\\ y = \frac{14x}{4}$