Answer:
The exact coordinates of the point where the line through B and D intersects the y-axis is [tex]\left ( 0, \ -12\dfrac{5}{7} \right)[/tex]
Step-by-step explanation:
For the given question, we have the following parameters;
The shape of the quadrilateral ABCD in the question = A rhombus
The diagonals of the rhombus = AC and BD
The point of intersection of the diagonals, M = (6, -11)
The equation of the line points A and C lie = 2·y + 7·x = 20
A rhombus is an orthodiagonal quadrilateral (The diagonals of a rhombus are perpendicular)
Therefore;
AC ⊥ BD and where the slope of the line with points A and C is 'm', the slope of the line with points B and D is (-1/m)
Rewriting the equation, 2·y + 7·x = 20, in slope and intercept form, we have;
2·y + 7·x = 20
y = (- 7·x + 20)/2 = -7·x/2 + 10
∴ y = -7·x/2 + 10
The slope of the line 'y = -7·x/2 + 10', m = -7/2
∴ The slope of the line having the points B and D is -1/m = -1/(-7/2) = 2/7
The equation of the line having the points B and D in point and slope form, using point (6, -11) is given as follows;
y - (-11) = 2/7 × (x - 6)
y + 11 = 2·x/7 - 12/7
∴ y = 2·x/7 - 12/7 - 11 = 2·x/7 - 89/7 = 2·x/7 - [tex]12\dfrac{5}{7}[/tex]
The equation of the line having the points B and D in point is therefore;
y = 2·x/7 - [tex]12\dfrac{5}{7}[/tex]
The coordinates of the point where the line having the points B and D intersects the y-axis is the y-intercept, of the equation, y = 2·x/7 - [tex]12\dfrac{5}{7}[/tex], which is [tex]\left ( 0, \ -12\dfrac{5}{7} \right)[/tex]