Differentiation (Part-38) | S N De

 

1.  Find the equation of the straight line which divides perpendicularly the straight line-segment joining the points $~(7,9)~$ and $~(-1,-7)~$ internally in the ratio $~3:5~$ . What is the distance of this line from the origin?

Solution.

By question, if $~P~$ is the point of internal division, then

$P\equiv \left(\frac{3\times (-1)+5 \times 7}{3+5},\frac{3 \times (-7)+5 \times 9}{3+5}\right)=(4,3).$

The equation of the straight line joining the points $~(7,9)~$ and $~(-1,-7)~$ is given by 

$~\frac{y-9}{9-(-7)}=\frac{x-7}{7-(-1)} \\ \text{or,}~~\frac{y-9}{16}=\frac{x-7}{8} \\ \text{or,}~~ \frac 12(y-9)=x-7 \\ \text{or,}~~ 2x-14=y-9 \\ \therefore 2x-y-5=0 \rightarrow(1)$

The equation of any straight line which is perpendicular to the straight line $~(1)~$ is given by $~x+2y+k=0 \rightarrow(2)$. If the straight line $~(2)~$ passes through the point $~(4,3)~$, then $~ 4+2\times 3+k=0 \Rightarrow k=-10.$

So, by $~(2),~$ we get the equation of the required straight line $~x+2y-10=0.$

The distance of the straight line $~x+2y-10=0~$ from the origin is 

$=\left|\frac{0+2 \times 0-10}{\sqrt{1^2+2^2}}\right|=\frac{10}{\sqrt{5}}=2\sqrt{5}~~\text{unit.}$

2. The co-ordinates of the points $~P,~Q,~R~$ are respectively $~(-7,5),~(3,8),~(-5,13)~$ and if $~RN~$ is perpendicular to $~PQ~$ and $~RT $ is parallel to $~PQ,~$ find the equations of $~RN~$ and $~RT.$ 

Solution.

The equation of $~PQ~$ is given by 

$~\frac{y-5}{5-8}=\frac{x-(-7)}{-7-3} \\ \text{or,}~~ -\frac {1}{3}(y-5)=-\frac{1}{10}(x+7) \\ \text{or,}~~ 10(y-5)=3(x+7) \\ \text{or,}~~ 10y-50=3x+21 \\ \therefore 3x-10y+71=0 $

The equation of any straight line perpendicular to $~PQ~$ is $~10x+3y+k=0~~(k \neq 0)\rightarrow(1).$

Now, $~RN \perp PQ~$ and $~RN~$ passes through the point $~R(-5,13).$

So, by $~(1)~$ we get, 

$~ 10 \times (-5)+3 \times 13 +k=0 \Rightarrow k=11. $

So, the equation of $~RN~$ is $~ 10x+3y+11=0.$

The equation of any straight line parallel to $~PQ~$ can be written as $~ 3x-10y+k_1=0\rightarrow(2).$

Now, $~RT~ || ~PQ~$ and $~RT~$ passes through the point $~(-5,13).$

$\therefore~ $by $~(2),~$ we get,

$~3 \times (-5)-10 \times 13+k_1=0 \Rightarrow k_1=145.$

So, the equation of $~RT~$ is $~ 3x-10y+145=0.$

$3.~~A(4,6),~B(-1,3),~$ and  $~C(2,-2)~$ are three given points. Find the following :

$~(i)~$ equation of the perpendicular on $~BC~$ from $~A.$

$~(ii)~$ co-ordinates of the point equidistant from $~A,~B,~C~$ and the distance of this point from $~A,~B,~C.$

Solution.

The equation of the straight line joining $~B(-1,3)~$ and $~C(2,-2)~$ is given by 

$\frac{y-3}{3-(-2)}=\frac{x-(-1)}{-1-2}\\ \text{or,}~~ \frac{y-3}{5}=\frac{x+1}{-3} \\ \text{or,}~~ 5(x+1)=-3(y-3) \\ \text{or,}~~ 5x+5=-3y+9 \\ \text{or,}~~ 5x+3y-4=0.$

The equation of any straight line perpendicular to $~BC~$ is 

$~3x-5y+k=0~~(k \neq 0)\rightarrow(1)$

Since the sl $~(1)~$ passes through $~(4,6)~,$ so 

$~ 3\times 4-5 \times 6+k=0\Rightarrow k=18.$

So, by $~(1)~$ we get the equation of the straight line which is passing through the point $~A~$ and  perpendicular to $~BC~$ and it is given by $~3x-5y+18=0.$ 

2nd Part :

Let $~P\equiv (h,k)~$ be a point which is equidistant from $~A,~B,~C.~$

$\therefore~ AP=BP \\ \text{or,}~~ AP^2=BP^2 \\ \text{or,}~~ (h-4)^2+(k-6)^2\\=(h+1)^2+(k-3)^2 \\ \text{or,}~~h^2-8h+16+k^2-12k+36\\=h^2+2h+1+k^2-6k+9 \\ \therefore~ 5h+3k-21=0\rightarrow(1)$

Similarly, 

$~~BP=CP \\ \text{or,}~~ BP^2=CP^2 \\ \text{or,}~~ (h+1)^2+(k-3)^2\\=(h-2)^2+(k+2)^2 \\ \text{or,}~~ h^2+2h+1+k^2-6k+9\\=h^2-4h+4+k^2+4k+4 \\ \text{or,}~~ 3h-5k+1=0\rightarrow(2)$ 

Now, solving $~(1)~$ and $~(2)~$ we get, $~h=3,~k=2.$

So, the required point is $~(3,2)~$ which is equidistant from $~A,~B,~C.~$

So, $~AB=\sqrt{(4-3)^2+(6-2)^2}=\sqrt{1^2+4^2}\\=\sqrt{17}~~\text{unit.}$

Hence, the distance of $~(3,2)~$ from $~A,B,C~$ is $~\sqrt{17}~$ unit.

4.  Find the equation of the straight line through the point $~(2,1)~$ , which is parallel to $~2x+4y=7.~$ Also, find the area of the figure formed by the above two straight lines and the axes of co-ordinates.

Solution.

The equation of any straight line parallel to $~2x+4y=7~$ can be written as $~2x+4y+k=0\rightarrow(1).~$ If the sl $~(1)~$ passes through the point $~(2,1)~$ then $~ 2 \times 2+4 \times 1+k=0 \Rightarrow k=-8.$

So, by $~(1)~$ we get the equation of straight line

$~ 2x+4y-8=0 \Rightarrow \frac x4+\frac y2=1$

Now, the given straight line can be written as $~\frac{x}{7/2}+\frac{y}{7/4}=1.$

So, the required area is

$=\text{area of}~~\Delta OBC- \text{area of}~\Delta OAD\\=\left(\frac 12 \times 4 \times 2\right)-\left(\frac 12 \times \frac 72 \times \frac 74\right)\\=\frac{15}{16}~~\text{sq. unit}.$

5.  Find the orthocentre of the triangle whose vertices are $~(2,7),~(-6,1)~$ and $~(4,-5).$ 

Solution.

Let $~A\equiv (2,7),~~B\equiv (-6,1),~~ C\equiv (4,-5).$

Suppose that a perpendicular $~AD~$ is drawn from $~A~$ to $~BC~$ and another perpendicular $~BE~$ is drawn from $~B~$ to $~CA~$ .

The orthocentre $~(h,k)~$ will be the intersection of $~AD~$ and $~BE.$

$\because~ AD \perp BC \\ \Rightarrow \frac{k-7}{h-2} \times \frac{-5-1}{4+6}=-1 \\ \therefore~ 5h-3k+11=0 \rightarrow(1)$

Again,

$\because~ BE \perp AC \\ \Rightarrow \frac{k-1}{h+6} \times \frac{-5-7}{4-2}=-1 \\ \therefore~ h-6k+12=0 \rightarrow(2)$

Solving $~(1)~$ and $~(2)~$ we get,

$~h=-\frac{10}{9},~~k=\frac{49}{27}.$

So, the orthocentre of the triangle is $~\left(-\frac{10}{9},\frac{49}{27}\right).$

6. The co-ordinates of two vertices of a triangle are $~(-2,3)~$ and $~(5,-1).~$ If the orthocentre of the triangle is at the origin , find the co-ordinates of its third vertex.

Solution.

Two given vertices of the triangle are $~A(-2,3)~$ and $~B(5,-1).~$ Let  the third vertex of the triangle be $~C\equiv (h,k)$. Now the orthocentre $~O\equiv (0,0).$

$\because~ AO \perp BC \\ \Rightarrow \frac{3-0}{-2-0} \times \frac{k+1}{h-5}=-1 \\ \therefore~ 2h-3k-13=0 \rightarrow(1)$

$\because~ BO \perp AC \\ \Rightarrow \frac{-1-0}{5-0} \times \frac{k-3}{h+2}=-1 \\ \therefore~ 5h-k+13=0\rightarrow(2)$

Solving $~(1)~$ and $~(2),~$ we get, $~h=-4,~k=-7.$

Hence,  the co-ordinates of its third vertex is $~(-4,-7).$

7.  The equations of two adjacent sides of a parallelogram are $~4x+5y=0~$ and $~7x+2y=0.~$ If the equation of its one diagonal be $~11x+7y=9,~$ find the equation of its other diagonal.

Solution.

Let $~ABCD~$ be the parallelogram where $~AB \equiv 7x+2y=0\rightarrow(1)~$ and $~BC \equiv 4x+5y=0\rightarrow(2).$

By solving $~(1)~$ and $~(2)~$ we get,$~~B\equiv (0,0).$

Clearly, the straight line $~11x+7y=9~$ does not pass through $~B(0,0).$

So, the equation of the diagonal $~AC :~ 11x+7y=9\rightarrow(3).$

Solving $~(1)~$ and $~(3)~$ we get, $~A \equiv \left(-\frac 23,\frac 73\right).$

Similarly, solving $~(2)~$ and $~(3)~$ we get, $~C \equiv \left(\frac 53,-\frac 43\right).$

So, the midpoint of $~AC~,~O\equiv \left(\frac 12,\frac 12\right).$

Finally, the equation of $~BD~$ is :

$~y=\frac{\frac 12-0}{\frac 12-0}~x \Rightarrow x-y=0.$

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8. Find the equation of the straight line which is parallel to the straight line $~3x+2y-6=0~$ and which forms a triangle of area $~21~$ square units with the straight lines $~x-2y=0~$ and $~y-2x=0.$

Solution.

The equation of any straight line parallel to $~3x+2y-6=0~$ is $~3x+2y+k=0~~(k\neq 0)\rightarrow(1).~$ 

The two given straight lines are $~x-2y=0\rightarrow(2),\\~y-2x=0 \rightarrow(3).$

Now, the point of intersection of the straight lines $~(2)~$ and $~(3)~$ is $~(0,0).$

Again, the point of intersection of the straight lines $~(1)~$ and $~(2)~$ is $~\left(-\frac k4,-\frac k8\right)~$ and the point of intersection of the straight lines $~(1)~$ and $~(3)~$ is $~\left(-\frac k7,-\frac{2k}{7}\right).$

So, the vertices of the triangle formed by the straight lines are $~(0,0),~(-k/4,-k/8),~(-k/7,-2k/7).$

By question, the area of the triangle $=21$

So, $~\frac 12\left(\frac{2k^2}{28}-\frac{k^2}{56}\right)=21 \\ \text{or,}~~ \frac{4k^2-k^2}{56}=21 \times 2 \\ \text{or,}~~ 3k^2=56 \times 21 \times 2 \\ \text{or,}~~ k^2=56 \times 7 \times 2 \\ \text{or,}~~ k=\pm \sqrt{7 \times 8 \times 7 \times 2}=\pm 28.$

So, the required equation of the straight line is $~3x+2y \pm 28=0.$

9.  The line $~3x+4y-24=0~$ meets the $~y-$axis at $~A~$ and the $~x-$axis at $~B.~$ The perpendicular bisector of $~\overline{AB}~$ meets the line through $~(0,-1)~$ and parallel to the $~x-$axis at $~C.~$ Prove that $~\angle{ACB}~$ is a right angle.

Solution.

$AB:~3x+4y-24=0 \\ \text{or,}~~ 3x+4y=24 \\ \text{or,}~~ \frac x8+\frac y6=1\rightarrow(1)$

From the intercept form of the straight line $~(1),~$ clearly we can say $~A\equiv(0,6),~~B \equiv(8,0).$ The midpoint of $~AB~$ is $~D\equiv (4,3).$ 

Any straight line perpendicular to $~AB~$ can be written as $~4x-3y+k=0~(k \neq 0) \rightarrow(2).$ 

Since the straight line $~(2)~$ passes through the point $~(4,3)~$, we get by $~(2),$

$ 4\times 4-3\times 3+k=0 \Rightarrow k=-7.$

Hence, the equation of straight line  perpendicular to $\overline{AB}~$ is $~4x-3y-7=0\rightarrow(3)$

Now, the equation of straight line passing through $~(0,-1)~$ and parallel to $~x-$axis is $~y=-1\rightarrow(4).$

The point of intersection of the straight lines $~(3)~$ and $~(4)~$ is $~C\equiv(1,-1).$

Now, $\text{ slope of}~AC \times ~\text{ slope of }~~BC\\=\frac{-1-6}{1-0} \times \frac{-1-0}{1-8}\\=-7 \times \frac{-1}{-7}\\=-7 \times \frac 17\\=-1.$

$ \therefore~ \angle{ACB}=90^{\circ}~~\text{(proved)}$

10. Show that the distance of the point $~(x_0,y_0)~$ from the line $~ax+by+c=0~$ measured parallel to a line making an angle $~\theta~$ with the positive direction of $~x-$axis is $~-\frac{ax_0+by_0+c}{a\cos\theta+b\sin\theta}.$

Solution.

The equation of any sl which passes through $~(x_0,y_0)~$ and making an angle $~\theta~$ with the positive direction of $~x-$axis is 

$~y-y_0=\tan\theta(x-x_0) \rightarrow(1) \\ \text{or,}~~ \frac{x-x_0}{\cos\theta}=\frac{y-y_0}{\sin\theta}=r~~(\text{say})$

We notice that the point $~(x_0+r\cos\theta,y_0+r\sin\theta)~$ lies on the sl $~(1)~$ . 

Let the point of intersection of the sl $~(1)~$ and $~ax+by+c=0~$ is $~(x_0+r\cos\theta,y_0+r\sin\theta)~$.

$\therefore~ a(x_0+r\cos\theta)+b(y_0+r\sin\theta)+c=0 \\ \text{or,}~~ r=-\frac{ax_0+by_0+c}{a\cos\theta+b\sin\theta}.$

Now, the distance between $~(x_0+r\cos\theta,y_0+r\sin\theta)~$ and $~(x_0,y_0)~$ is 

$=\sqrt{r^2(\cos^2\theta+\sin^2\theta)}=r.$

$\therefore~$ the required distance is $=-\frac{ax_0+by_0+c}{a\cos\theta+b\sin\theta}.$

Read More : 

Basic Concept of straight line(Part-1) class 11 | Very Short Answer Type Questions | Ex-2A

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