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Mathematics, 10.03.2020 18:48 owen74

In the coordinate plane the vertices of rst are: r(6 -1), s (1 -4), t (-5,6). Prove that rst is a right triangle. State the coordinates of p such that quadrilateral rstp is a rectangle.

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Answer from: efjdjskd

The triangle rst is a right triangle, and the coordinate of p such that rstp is a rectangle is (-10,3)

The coordinates of the triangle are given as:

r = (6 -1)

s= (1 -4)

t= (-5,6)

Start by calculating the slopes of the sides of the triangle using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$rs = \frac{-4--1}{1-6}$

$rs = \frac{3}{5}$

$st = \frac{6--4}{-5-1}$

$st = \frac{10}{-6}$

Simplify

$st = -\frac{5}{3}$

For the triangle to be a right triangle, then the product of the slopes must equal -1.

This is represented as:

$m_1 \times m_2 = -1$

So, we have:

$rs \times st = -1$

Substitute values for rs and st

$\frac 35 \times -\frac 53 = -1$

-1 =-1

The above equation is true.

Hence, triangle rst is a right triangle

r = (6 -1)

s= (1 -4)

t= (-5,6)

Calculate the distance rs using:

$rs = \sqrt{(x_1 -x_2)^2 + (y_1 -y_2)^2}$

$rs = \sqrt{(6 -1)^2 + (-1--4)^2}$

Calculate the distance tp using:

$tp = \sqrt{(x_1 -x_2)^2 + (y_1 -y_2)^2}$

So, we have:

$tp = \sqrt{(-5 -x)^2 + (6-y)^2}$

Opposite sides are equal.

So, we have:

$\sqrt{(-5 -x)^2 + (6-y)^2} = \sqrt{(6 -1)^2 + (-1--4)^2}$

Take square of both sides

(-5 -x)^2 + (6-y)^2 = (6 -1)^2 + (-1--4)^2

By comparison, we have:

(-5 -x)^2 = (6 -1)^2

and

(6-y)^2 = (-1--4)^2

For (-5 -x)^2 = (6 -1)^2

Take square root of both sides

-5 -x = 6 - 1

Add 5 to both sides

-x = 6 - 1 + 5

-x = 10

So, we have:

x = -10

Also, we have:

(6-y)^2 = (-1--4)^2

Take square root of both sides

6 - y = -1--4

6 - y = 3

Collect like terms

y = 6-3

y =3

So, we have:

x = -10 and y =3

Hence, the coordinate of p is (-10,3)

Read more about coordinates at:

link

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Answer from: madisonanders

Coordinates of point p: (0,9)

Step-by-step explanation:

We have been given that in the coordinate plane the vertices of rst are: r (6, -1), s (1, -4), t (-5,6). We are asked to prove that rst is a right triangle.

First of all, we will find the slope between points 'r' and 's' and 's' and 't'.

$\text{Slope}_{rs}=\frac{-1-(-4)}{6-1}=\frac{-1+4}{5}=\frac{3}{5}$

$\text{Slope}_{st}=\frac{-4-6}{1-(-5)}=\frac{-10}{1+5}=-\frac{10}{6}=-\frac{5}{3}$

We know that slope of perpendicular lines in negative reciprocal of each other.

Let us find the product of both slopes.

$\frac{3}{5}\times \frac{-5}{3}=-1$

Since product of both slopes is , therefore, line 'rs' is perpendicular to line 'st' and 'rst' is a right triangle.

Let us assume that coordinates of point p are (x,y) .

We need to find slope of point 'r' to 'p' is equal to slope of point 's' to point 't' and slope of point 'p' to 't' is equal to slope of point 'r' to point 's'.

$\text{Slope}_{rp}=\frac{-1-y}{6-x}=-\frac{5}{3}$

$\text{Slope}_{pt}=\frac{y-6}{x-(-5)}=\frac{3}{5}$

Now, we have two equations and two unknown.

$\frac{y-6}{x+5}=\frac{3}{5}$ and $\frac{-1-y}{6-x}=-\frac{5}{3}$

To solve our system, we will use substitution method.

$y=\frac{3}{5}(x+5)+6$

Upon substituting this value in equation (2), we will get:

$\frac{-1-(\frac{3}{5}(x+5)+6)}{6-x}=-\frac{5}{3}$

$\frac{-\frac{5}{5}-\frac{3}{5}(x+5)-\frac{30}{5}}{6-x}=-\frac{5}{3}$

$\frac{\frac{-5-3x-15-30}{5}}{6-x}=-\frac{5}{3}$

$\frac{-50-3x}{5(6-x)}=-\frac{5}{3}$

Cross multiply:

3(-50-3x)=-25(6-x)

-150-9x=-150+25x

-9x=25x

-9x+9x=25x+9x

34x=0

x=0

Upon substituting x=0 in equation $y=\frac{3}{5}(x+5)+6$ , we will get:

$y=\frac{3}{5}(0+5)+6$

$y=\frac{3}{5}(5)+6$

y=3+6=9

Therefore, the coordinates of point p at (0,9) that will make quadrilateral rstp is a rectangle.

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