The book is right. Get some squared paper and draw the given points in;Originally Posted by shubhadeep
you will see that the answer is obvious (the lines through them meet at a
right angle, so your parallelogram is a rectangle).
RonL
Hi I have problems in some questions.
A) My answers to these questions do not match the book answers:-
1) The three vertices of a paralleologram are and . Find the fourth vertex.[Book Answer: ](Solved)
2) The area of a triangle is . Two of its vertices are and . The third vertex lies on . Find the third vertex.[Book Answer: and . How am I supposed to get the second answer](Solved)
3) The point divides the join of and in the ratio . Find the values of for which the area of triangle ABC where is and is equal to 2 units.[Book Answer: or ]
B) Can't even guess what to do:-
1) Find the locus of the mid-point of the portion of the line which is intercepted between the axes.[Book Answer: ]
2) Find the equation of the bisector of angle of the triangle whose vertices are and .[Book Answer: ](Solved)
3) Find the distance of the point from the line measured parallel to a line having slope .[Book Answer:5^(1/2)](Solved)
4) Find the distance of the point from the line measured parallel to a line having slope .[Book Answer: ](Solved)
That's all. Thanks
thanks for the second question. i got it. since the area is in the form of a modulus, therefore there should have been two values for the area. i had put only 5 and not -5. that was my mistake.
still trying the first one.
The second answer is the point on the given line on the other side of theOriginally Posted by shubhadeep
line through (2,1) and (3,-2) from the first solution which gives the required
area. (Do a sketch and you will see that these are both solutions).
RonL
You are right usually it doesn'tOriginally Posted by shubhadeep
You need to find the point D on segment AB (produced if need be) a distance
AC from A, then use the mid point of segment CD not the midpoint of BC -
but this looks too fussy there must be a simpler way
Sorry about that - I appear not to be on form today.
RonL
Here we go again. Find the point on the same distance asOriginally Posted by shubhadeep
is from . As is this is fairly simple to do, so simple
proportion will give is the point .
Then the mid point of is .
So the bisector is the line through and .
So not as fussy as I thought.
RonL
has length (since is a right triangle).Originally Posted by shubhadeep
So we want a point on on such that , and . So if is
then triangle is similar to , with hypotenuses
in the ratio of . So the heights are in the same ratio, so .
The same applies to the horizontal legs so .
RonL
I hope you will find this of help in solving this problem:Originally Posted by shubhadeep
As lies on the line on which and it can be written:
for some (here we are treating , and as the 2-vectors of their coordinates).
The distance of from is:
,
and the distance of from is:
.
Now the condition that the point divides the join of
and in the ratio is equivalent to:
so:
.
Which should be sufficient information to allow you to find in terms of
and the coordinates of and .
I presume that you know the formula for the area of a triangle in terms of the
coordinates of its vertices, or failing that Heron's formula for the area of a
triangle, and so should be able to complete this problem from here.
RonL