Q12. Two points have coordinates A(1,3) and C(7,7). Find the equation of perpendicular bisector of AC. B is the point on the y-axis equidistant from A and C and ABCD is a rhombus. Find the coordinates of B and D. Show that the area of the rhombus is 52 units^2 and hence calculate the perpendicular distance of A from BC. ``
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Points A and C are opposite on the same diagonal. The slope of the line that joins AC is
`m = (Y_C-Y_A)/(X_C-X_A) =(7-3)/(7-1) =4/6 =2/3`
The point O at half distance between A and C has the coordinates
`O( X_A + (X_C-X_A)/2, Y_A+(Y_C-Y_A)/2) =O (1 +(7-1)/2 ,3+(7-3)/2) =O(4, 5)`
The slope of line that is perpendicular to AC is
`m_1 =-1/m =-3/2`
The equation of the line that passes through point `O(4,5)` and has the slope `-3/2` is
`y-5 = -3/2*x +6`
`y = -3/2*x +11` (1)
Point B is on the y axis, it means `X_B =0` . This condition combined with equation (1) gives
Thus the coordinates of B are `B(0,11)`
Now to find the coordinates of point D.
The horizontal distance between points B and O is
`Delta(X) = X_O- X_B =4-0 =4`
The vertical distance between points B and O is
`Delta(Y) = Y_O -Y_B =5-11 =-6`
Point D is diametrally opposed to point B with respect to center O thus,
`X_D = X_B +2*Delta(X) =0 +2*4 =8`
`Y_D = Y_B +2*Delta(Y) = 11-2*6 =-1`
Coordinates of point D are `D(8,-1)`
Finally to find the area or figure we need to find the length of diagonals:
`|BD| =sqrt[(8-0)^2 +(-1-11)^2] =14.42`
Area of figure is just
`S = (|AC|*|BD|)/2 =(7.21*14.42)/2 =52`
Answer: the points B and D have the coordinates `B(0.11)` and `D(8,-1)`
Mid point of AC is (4,5).
Slope of AC is (7-3)/(7-1) = 2/3
Slope of perpendicular bisector is -1/slope of AC = -3/2
Equation of perpendicular bisector is y-5 = -3/2 (x-4)
which ultimately reduces to the equations 3x + 2y = 22.
On the y axis, x coordiante is 0. Hence the point on the yaxis B will be 3*0+2y = 22; y=11
the coordinate of B is (0,11)
The coordinate of D is (8,-1)
Area of rhomus is b*a, b is the length of the base and a the altitude which is the perpendicular distance from the chosen base to the opposite side.
b=8, a 6.5 and hence the area of teh rhombus is 52 units squared.
Perpendicular distance from A to BC is 6.5cm
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