# Thread: Some problems

1. ## Some problems

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 $(3,4), (3,8)$ and $(9,8)$. Find the fourth vertex.[Book Answer: $(9,4)$](Solved)

2) The area of a triangle is $5$. Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex lies on $y=x+3$. Find the third vertex.[Book Answer: $(7/2,13/2)$ and $(-3/2,3/2)$. How am I supposed to get the second answer](Solved)

3) The point $A$ divides the join of $P(-5,1)$ and $Q(3,5)$ in the ratio $k:1$. Find the values of $k$ for which the area of triangle ABC where $B$ is $(1,5)$ and $C(7,-2)$ is equal to 2 units.[Book Answer: $k=7$ or $31/9$]

B) Can't even guess what to do:-

1) Find the locus of the mid-point of the portion of the line $x$ $cos\alpha$ $+$ $y$ $sin\alpha$ $=$ $p$ which is intercepted between the axes.[Book Answer: $p^2(x^2+y^2)=4x^2y^2$]

2) Find the equation of the bisector of angle $A$ of the triangle whose vertices are $A(4,3), B(0,0)$ and $C(2,3)$.[Book Answer: $x-3y+5=0$](Solved)

3) Find the distance of the point $(3,5)$ from the line $2x+3y=14$ measured parallel to a line having slope $1/2$.[Book Answer:5^(1/2)](Solved)

4) Find the distance of the point $(2,5)$ from the line $3x+y+4=0$ measured parallel to a line having slope $3/4$.[Book Answer: $5$ $units$](Solved)

That's all. Thanks

2. Originally Posted by shubhadeep
Hi I have problems in some questions.
A) My answers to these questions do not match the book answers. It can be a misprint though:-

1) The three vertices of a paralleologram are (3,4), (3,8) and (9,8). Find the fourth vertex.[Book Answer9,4)]
The book is right. Get some squared paper and draw the given points in;
you will see that the answer is obvious (the lines through them meet at a
right angle, so your parallelogram is a rectangle).

RonL

3. 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.

4. ## Correction for Q2

Originally Posted by shubhadeep

2) The area of a triangle is $5$. Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex lies on $y=x+3$. Find the third vertex.[Book Answer: $(7/2,13/2)$ and $(-3/2,3/2)$. How am I supposed to get the second answer]
The second answer is the point on the given line on the other side of the
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

5. Originally Posted by CaptainBlack
The bisector of angle $A$ in triangle $ABC$ is the line throught
point $A$, and the mid-point of segment $BC$.
But how do u know that the angle bisectorof A bisects BC.
I still tried it. The equation I'm getting is $x-2y+2=0$.

6. Originally Posted by shubhadeep
But how do u know that the angle bisectorof A bisects BC.
I still tried it. The equation I'm getting is $x-2y+2=0$.
You are right usually it doesn't

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

7. Originally Posted by shubhadeep
3) Find the distance of the point $(3,5)$ from the line $2x+3y=14$ measured parallel to a line having slope $1/2$.[Book Answer:5^(1/2)]
Find the equation of the line through $(3,5)$ of slope $1/2$.

Then find the point $P$ at which this line meets $2x+3y=14$.

Finally find the distance between $(3,5)$ and $P$.

RonL

8. Originally Posted by CaptainBlack
You need to find the point D on segment AB (produced if need be) a distanceAC 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.
Yea, there has to be a simpler way. I did try it by the method you say earlier but I got stuck. Will try again.

Sorry about that - I appear not to be on form today.
No Problem. Happens to many.

9. Originally Posted by shubhadeep
2) Find the equation of the bisector of angle $A$ of the triangle whose vertices are $A(4,3), B(0,0)$ and $C(2,3)$.[Book Answer: $x-3y+5=0$]
Here we go again. Find the point $D$ on $AB$ the same distance as $C$
is from $A$. As $B$ is $(0,0)$ this is fairly simple to do, $|BA|=5$ so simple
proportion will give $D$ is the point $\frac{3}{5}(4,3)=(2.4,1.8)$.

Then the mid point of $CD$ is $(2.2,2.4)$.

So the bisector is the line through $(4,3)$ and $(2.2,2.4)$.

So not as fussy as I thought.

RonL

10. Yeah Thanks! I never thought of that.

11. As $B$ is $(0,0)$ this is fairly simple to do, $|BA|=5$ so simple proportion will give $D$ is the point $\frac{3}{5}(4,3)=(2.4,1.8)$.

I am not familiar with using proportions in co-ordinates. Can you explain it please. And how would I have done that if the point B was not on the origin.

12. Originally Posted by shubhadeep
I am not familiar with using proportions in co-ordinates. Can you explain it please. And how would I have done that if the point B was not on the origin.
$BA$ has length $5$ (since $BA(4,0)$ is a $3,4,5$ right triangle).
So we want a point on $D$ on $BA$ such that $|DA|=2$, and $|BD|=3$. So if $D$ is
$(d_1,d_2)$ then triangle $BD(d_1,0)$ is similar to $BA(4,0)$, with hypotenuses
in the ratio of $3:5$. So the heights are in the same ratio, so $d_2=(3/5)3$.
The same applies to the horizontal legs so $d_1=(3/3)4$.

RonL

13. Originally Posted by shubhadeep

3) The point $A$ divides the join of $P(-5,1)$ and $Q(3,5)$ in the ratio $k:1$. Find the values of $k$ for which the area of triangle ABC where $B$ is $(1,5)$ and $C(7,-2)$ is equal to 2 units.[Book Answer: $k=7$ or $31/9$]
I hope you will find this of help in solving this problem:

As $A$ lies on the line on which $P$ and $Q$ it can be written:

$A=(1-\lambda).P+\lambda.Q$

for some $\lambda \epsilon \mathbb{R}$ (here we are treating $A$, $P$ and $Q$ as the 2-vectors of their coordinates).

The distance of $A$ from $P$ is:

$D_{AP}=|\lambda|.|P-Q|$,

and the distance of $A$ from $Q$ is:

$D_{AQ}=|1-\lambda|.|P-Q|$.

Now the condition that the point $A$ divides the join of $P(-5,1)$
and $Q(3,5)$ in the ratio $k:1$ is equivalent to:

$\frac{D_{AP}}{D_{AQ}}=k$

so:

$\frac{|\lambda |}{|1-\lambda|}=k$.

Which should be sufficient information to allow you to find $A$ in terms of $k$
and the coordinates of $P$ and $Q$.

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