Question

please help me.
pleas don't scam​

Answer 1

Given:

The vertices of a parallelogram taken in order are A(3, 4), B(9, 5), C(7, 2x), and D(y, 15).

To find:

Find the value of x+y.

Solution:

From given, we have,

A(3, 4), B(9, 5), C(7, 2x) and D(y, 15) are the vertices of a parallelogram, then the diagonals intersect each other at a point.

So, Mid-point of AC = Mid-point of BD.

Mid-point of a line joining (x₁, y₁) and (x2, Y₂): p(x,y)=[(x 1 +x 2 )/2,(y 1 +y 2 )/2].

[(3 + 7) / 2, (4 + 2x) / 2] = [(9 + y) / 2, (5 + 15) / 2]

[10/2,4+2x/2] = [9+y/2, 20/2]

By equating the equal coordinates, we have,

5 = 9 + y / 2 and 4 + 2x / 2 = 10

10 - 9 = y and 2x = 20-4-16

y = 1 and x = 8

∴ The value of x+y is 9

Answer 2

$\large\text{\underline{Let's begin.}}$

Two diagonals of a quadrilateral meet and bisect each other if and only if the quadrilateral is a parallelogram. ('If and only if' states the two statements are equivalent.)

$\large\text{\underline{Solution}}$

Consider the two midpoints of diagonals $M$.

$\implies M(\dfrac{3+7}{2},\dfrac{2x+4}{2})\text{ (Midpoint of diagonal }\overline{AC})$

$\implies M(\dfrac{y+9}{2},\dfrac{5+15}{2})\text{ (Midpoint of diagonal }\overline{BD})$

Since two points are coincident, we get the following equations.

$\implies 2x+4=20,y+9=10$

$\implies x=8,y=1$

$\large\text{\underline{Conclusion}}$

Hence, the value of $x+y$ is 9.

$\large\text{\underline{Verification}}$

The attachment is included. It looks like a rectangle but it isn't actually, because the two sides don't meet each other perpendicularly.

The slope of $\overline{AB}$ or $\overline{CD}$: $\dfrac{1}{6}$

The slope of $\overline{BC}$ or $\overline{DA}$: $-\dfrac{11}{2}$

They are perpendicular if and only if the product of slopes equals -1.