4 Geometry Problems

and the post doesn't have the questions too !

Also

knight247 wrote:Don't have OAs on these

1) In a parallelogram, diagonals bisect each other, so if you solve given equations for a diagonal AC & equation for CD, you will get point C(7,7)
so midpoint of A & C is your answer i.e
[spoiler]IMO (4,13/2)[/spoiler]

user123321

2)
We know two points A & B of a side of an parallelogram
And since diagonals bisect in a parallelogram, the mid point for any of those diagonals is same.

a) we know the equation of a diagonal, from this info we cannot find the mid point of a diagonal, because we just know only one end of a diagonal. hence insufficient.
b) ABCD is rectangle means infinite possibilities are present with we just knowing only a side. hence insufficient.

using both,
we know points A & B, so we know equation for BC, which is a line perpendicular to AB and passing through B. so we know the equations for BC & AC(this from (a)). so we can determine point C & hence the mid point of diagonal.
hence sufficient.

IMO C

3) This is almost similar to above one.
you know point B of a parallelogram, need to find mid point of diagonal

A)we know AC, with that info we can have infinite no. of possibilities. hence insuffucient.
B)ABCD is square. Since we dont know the length of a side, we have no information to find what is required. hence insufficient.

using A & B,
we know it is square & in a square diagonals bisect perpendicularly, so the foot of perpendicular from side B on to diagonal AC will give us the mid point of diagonal. hence sufficient.

IMO C

user123321

4) we know a parallelogram having a point A & length of one side AB as 5 units. find the mid point of point of diagonals?

A) we know equation for diagonal AC, we will have lot of possibilities here. so insufficient.
B) ABCD is square, so we know length of diagonal as 5root2. but we dont know anything about its mid point. hence insufficient.

using both,
we can determine the coordinates for C using the equation AC( point C is nothing but a point 5root2 distance away from A on the line equation), But here there are two possibilities, one on either side of A.
hence insufficient.

IMO E

ALWAYS LOOK AT THE ANSWERS.

The correct answer must satisfy the equation of diagonal AC.
Only answer choice B works:
4 - 6(13/2) + 35 = 0.
4 - 39 + 35 = 0.
0 = 0.

The correct answer is B.

using both,
we know points A & B, so we know equation for BC, which is a line perpendicular to AB and passing through B. so we know the equations for BC & AC(this from (a)). so we can determine point C & hence the mid point of diagonal.
hence sufficient.

Hi user

Can you please explain how can u find equation of a line perpendicular to a line using two points. I know this is possible but can you create the equation with the points mentioned in the Question.

shekhar.kataria wrote:

using both,
we know points A & B, so we know equation for BC, which is a line perpendicular to AB and passing through B. so we know the equations for BC & AC(this from (a)). so we can determine point C & hence the mid point of diagonal.
hence sufficient.

Hi user

Can you please explain how can u find equation of a line perpendicular to a line using two points. I know this is possible but can you create the equation with the points mentioned in the Question.

say points are A(1,2) & B(3,4)
to find the line perpendicular to line joining A & B and passing through B..
first start with
finding the slope of AB = (4-2)/(3-1) = 2/2 =1
we know that if two lines are perpendicular product of their slopes is -1
since slope of AB is 1, the line perpendicular to it will have a slope of -1(since the product should be equal to -1)

so we know the slope of a line & a point on it which is B(3,4)
then equation is (y-y1) = m (x-x1)
(y-4) = (-1)(x-3)
y-4 = -x+3
x+y = 7

But to do this, you need to know the basic formulas of co-ordinate geometry, especially finding slope & generating line equation from a point & a slope.

user123321

Thanks User i got it now.

user123321 wrote:

shekhar.kataria wrote:

using both,
we know points A & B, so we know equation for BC, which is a line perpendicular to AB and passing through B. so we know the equations for BC & AC(this from (a)). so we can determine point C & hence the mid point of diagonal.
hence sufficient.

Hi user

Can you please explain how can u find equation of a line perpendicular to a line using two points. I know this is possible but can you create the equation with the points mentioned in the Question.

say points are A(1,2) & B(3,4)
to find the line perpendicular to line joining A & B and passing through B..
first start with
finding the slope of AB = (4-2)/(3-1) = 2/2 =1
we know that if two lines are perpendicular product of their slopes is -1
since slope of AB is 1, the line perpendicular to it will have a slope of -1(since the product should be equal to -1)

so we know the slope of a line & a point on it which is B(3,4)
then equation is (y-y1) = m (x-x1)
(y-4) = (-1)(x-3)
y-4 = -x+3
x+y = 7

But to do this, you need to know the basic formulas of co-ordinate geometry, especially finding slope & generating line equation from a point & a slope.

user123321