Question

If(1, 2), (4,y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
x and y.​

Answer 1

Answer:

The value of x = 6 and y = 3.

Step-by-step explanation:

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know that

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":Mid point of AC is given by

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":Mid point of AC is given by(\frac{1+x}{2},\frac{2+6}{2})\\\\=(\frac{1+x}{2},4)

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":Mid point of AC is given by(\frac{1+x}{2},\frac{2+6}{2})\\\\=(\frac{1+x}{2},4)Mid point of BD is given by

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":Mid point of AC is given by(\frac{1+x}{2},\frac{2+6}{2})\\\\=(\frac{1+x}{2},4)Mid point of BD is given by(\frac{4+3}{2},\frac{y+5}{2})\\\\=(\frac{7}{2},\frac{y+5}{2})

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":Mid point of AC is given by(\frac{1+x}{2},\frac{2+6}{2})\\\\=(\frac{1+x}{2},4)Mid point of BD is given by(\frac{4+3}{2},\frac{y+5}{2})\\\\=(\frac{7}{2},\frac{y+5}{2})Now, Mid point of AC = Mid point of BD

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":Mid point of AC is given by(\frac{1+x}{2},\frac{2+6}{2})\\\\=(\frac{1+x}{2},4)Mid point of BD is given by(\frac{4+3}{2},\frac{y+5}{2})\\\\=(\frac{7}{2},\frac{y+5}{2})Now, Mid point of AC = Mid point of BD(\frac{1+x}{2},4)=(\frac{7}{2},\frac{y+5}{2})\\\\\frac{1+x}{2}=\frac{7}{2}\\\\1+x=7\\\\x=7-1=6\\\\Similarly,\\\\4=\frac{y+5}{2}\\\\8=y+5\\\\y=8-5=3

Let ABCD is a parallelogram with vertices (1,2) ; (4,y) ; (x,6) ; (3,5) respectively.As we know thatThe mid-points of two diagonals must be same.So, we will apply "Mid point rule":Mid point of AC is given by(\frac{1+x}{2},\frac{2+6}{2})\\\\=(\frac{1+x}{2},4)Mid point of BD is given by(\frac{4+3}{2},\frac{y+5}{2})\\\\=(\frac{7}{2},\frac{y+5}{2})Now, Mid point of AC = Mid point of BD(\frac{1+x}{2},4)=(\frac{7}{2},\frac{y+5}{2})\\\\\frac{1+x}{2}=\frac{7}{2}\\\\1+x=7\\\\x=7-1=6\\\\Similarly,\\\\4=\frac{y+5}{2}\\\\8=y+5\\\\y=8-5=3Hence, The value of x = 6 and y = 3.

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