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morpeh [17]
3 years ago
15

Write an equation of the perpendicular bisector of the segment with endpoints U(-3,4) and V(9,8)

Mathematics
1 answer:
stealth61 [152]3 years ago
5 0

Answer:

y=-3x+15

Step-by-step explanation:

the midpoint of UV (using the midpoint formula) is (3,6)

the slope of UV= 4/12= (1/3)

perpendicular slope is -3

now find line with slope -3 passing (3,6)

y-6=-3(x-3)

y=-3x+15

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35. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.
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Answer:

The linear equation for the line which passes through the points given as (-2,8) and $(4,6), is written in the point-slope form as $y=-\frac{1}{3} x-\frac{26}{3}$.

Step-by-step explanation:

A condition is given that a line passes through the points whose coordinates are (-2,8) and (4,6).

It is asked to find the linear equation which satisfies the given condition.

Step 1 of 3

Determine the slope of the line.

The points through which the line passes are given as (-2,8) and (4,6). Next, the formula for the slope is given as,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substitute $6 \& 8$ for $y_{2}$ and $y_{1}$ respectively, and 4&-2 for $x_{2}$ and $x_{1}$ respectively in the above formula. Then simplify to get the slope as follows, $m=\frac{6-8}{4-(-2)}$

$$\begin{aligned}&m=\frac{-2}{6} \\&m=-\frac{1}{3}\end{aligned}$$

Step 2 of 3

Write the linear equation in point-slope form.

A linear equation in point slope form is given as,

$y-y_{1}=m\left(x-x_{1}\right)$

Substitute $-\frac{1}{3}$ for m,-2 for $x_{1}$, and 8 for $y_{1}$ in the above equation and simplify using the distributive property as follows, y-8=-\frac{1}{3}(x-(-2))$\\ $y-8=-\frac{1}{3}(x+2)$\\ $y-8=-\frac{1}{3} x-\frac{2}{3}$

Step 3 of 3

Simplify the equation further.

Add 8 on each side of the equation $y-8=-\frac{1}{3} x-\frac{2}{3}$, and simplify as follows, $y-8+8=-\frac{1}{3} x-\frac{2}{3}+8$

$y=-\frac{1}{3} x-\frac{2+24}{3}$$\\ $$y=-\frac{1}{3} x-\frac{26}{3}$$

This is the required linear equation.

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