Question

If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is:
(A) 5x – 3y + 1 = 0 (B) 5x + 3y – 11 = 0 (C) 3x – 5y + 7 = 0 (D) 3x + 5y – 13 = 0

Answer 1

5x - 3y + 1=0

Step-by-step explanation:

Mid-point formula -

$x= \frac{x_{1}+x_{2}}{2}$

$y= \frac{y_{1}+y_{2}}{2}$

If the point P(x,y) is the mid point of line joining A$(x_{1}y_{1})$ and B$(x_{2}\ , y_{2})$ .

Two – point form of a straight line -

$y-y_{1}=\left ( \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )(x-x_{1})$

- wherein

The lines passes through  $(x_{1}y_{1})$ and $(x_{2}\ , y_{2})$

As BD and AC are parallel

$\frac{n-4}{m-3}=\frac{5-2}{2-1}$

{n-4}=3({m-3})..............................(1)

As AB and CD are parallel

$\frac{n-5}{m-2}=\frac{4-2}{3-1}=\frac{2}{2}=1$

{n-5}=({m-2})..............................(2)

Solving (1) and (2)

m=4 and n=7

$AD \: \: is\: \: (y-2)=(\frac{7-2}{4-1})(x-1)$

3(y-2) = 5(x -1)

3y - 6 = 5x - 5

5x - 3y + 1=0

To learn more

i)Three consecutive vertices of a parallelogram are (-2, -1), (1, 0) and (4, 3). Find the coordinates of the fourth vertex.

https://brainly.in/question/1086842

ii)Find the area of quadrilateral ABCD whose vertices are A(-3,-1) B(-2,-4) C(4,-1) D(3,4)

https://brainly.in/question/66822