Question

using the slope formula ,show that the opposite sides of a quadrilateral with vertices A(-2,-4),B(5,-1),C(6,4)and D(-1,1)are parallel​

Answer 1

ANSWER:-

Given:

Show that the opposite sides of a quadrilateral with vertices A(-2,-4),B(5,-1),C(6,4), & D(-1,1) are parallel.

Solution:

Let A(-2,-4), B(5,-1), C(6,4) & D(-1,1) be the given points in order.

Now,

Using the slope formula:

Slope of line passing through (x1,y1) & (x2,y2) is,

⏺️x1= -2, x2= 5

⏺️y1= -4, y2= -1

CASE1:

$m = \frac{y2 - y1}{x2 - x1} \\ \\ = > Slope \: of \: AB = \frac{ - 1 - ( - 4)}{5 - ( - 2)} \\ \\ = > \frac{ - 1 + 4}{5 + 2} \\ \\ = > \frac{3}{7}$

Slope of CD:

$= > \frac{1 - 4}{ - 1 - 6} \\ \\ = > \frac{ - 3}{ - 7} \\ \\ = > \frac{3}{7}$

Therefore,

Slope of AB = slope of CD

Hence,

AB|| CD..........(1)

CASE 2:

Now,

Slope of BC:

$= > \frac{4 - ( - 1)}{6 - 5} \\ \\ = > \frac{4+ 1}{1} \\ \\ = > \frac{5}{1} = 5$

Slope of AD:

$= > \frac{1 - ( - 4)}{ - 1 - ( - 2)} \\ \\ = > \frac{1 + 4}{ - 1 + 2} \\ \\ = > \frac{5}{1} \\ \\ = > 5$

Therefore,

Slope of BC= slope of AD

Hence,

BC||AD........(2)

So,

From (1) & (2), we see that opposite sides of quadrilateral are parallel.

Hope it helps ☺️