Question

Three vertices of parallelogram JKLM are J(1, 4), K(5, 3), and L(6,−3). Find the coordinates of vertex M.

Answer 1

Answer:

coordinates of vertex M is (x, y) = (2, -2)

Step-by-step explanation:

Since JKLM is a  parallelogram, this implies that JK parallel to LM and KL parallel to JM. This means that

Slope of JK = slope of LM

$\frac{3-4}{5-1} =\frac{-3-y}{6-x} \\y=-\frac{1}{4}x-\frac{3}{2} ....(i)$

And

Slope of KL = slope of JM

$\frac{3-\left(-3\right)}{5-6}=\frac{4-y}{1-x}\\y=10-6x...(ii)$

From equation (i) and (ii) we get

$-\frac{1}{4}x-\frac{3}{2} =10-6x$

$-\frac{23x}{4}=-\frac{23}{2}$

$-23x=-46$

$x=2$

Put the value of x in equation (ii) we get

$y=10-6(2)\\y=-2$

So, the coordinates of vertex M is (x, y) = (2, -2).

Answer 2

We find out the coordinates of the given parallelogram using pair of linear equations as (2,-2)

It is given that coordinates of three vertices of a given parallelogram are:

J(1,4)

K(5,3)

L(6,-3)

M = ?????? let us take coordinates as (a,b)

What is a parallelogram?

It is a quadrilateral in which opposite sides are equal as well as parallel to each other.

We can say that,

JK=LM

KL=MJ

JK parallel to LM

KL parallel to MJ

W can say that,

the slope of JK = Slope of LM, since they are parallel.

i.e. $\frac{3-4}{5-1}$ $= \frac{b+3}{a-6}$

4b+a = -6......equation (1)

Again,

The slope of KL = slope of JM

i.e. $\frac{3+3}{5-6}$ $=\frac{b-4}{a-1}$

6a+b = 10.....equation(2)

By solving equations (1) and (2)

a = 2, b= -2

Thus, we get the coordinates of M as (2,-2)

To get more about parallelogram refer to the link,

https://brainly.com/question/3050890