Differentiation (Part-38) | S N De

 

8.  If two sides of a square are the part of the straight lines $~5x-2y=13~$ and $~5x-2y+16=0~$ then find the area of the square.

Solution.

Two given parallel straight lines can be written as 

$y=\frac 52x-\frac{13}{2}\rightarrow(1)\\~y=\frac 52x+8\rightarrow(2)$

We know that the distance $~(d)~$ between two straight lines $~y=mx+c_1,~y=mx+c_2~$ is $~d=\frac{|c_1+c_2|}{\sqrt{1+m^2}},~$ where $~c_1=-13/2,~c_2=8,~m=5/2.$

Clearly, $~d=$ the length of the side of the square in question, is 

$~d=\frac{|-13/2+8|}{\sqrt{1+(5/2)^2}}=\frac{\frac 32}{\sqrt{1+\frac{25}{4}}}=\frac{3/2}{\frac 12 \sqrt{29}}=\frac{3}{\sqrt{29}}$

So, the area of the square 

$=\left(\frac{3}{\sqrt{29}}\right)^2=\frac{9}{29}~~\text{sq. unit.}$

9. If the slope of the line joining the points $~(2k,-2)~$ and $~(1,-k)~$ be $~(-2),~$ find $~k.$

Solution.

By question, 

$~\frac{-2+k}{2k-1}=-2 \\ \text{or,}~~ -2+k=-4k+2 \\ \text{or,}~~ 5k=4 \\ \therefore~ k=\frac 45.$

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10.  Find the co-ordinates of the point on the line $~7x-6y=20~$ for which the ordinate is double the abscissa.

Solution.

Putting the point $~(k,2k)~$ in the equation of the line $~7x-6y=20~$ we get,

$~7k-6\times 2k=20 \\ \text{or,}~ 7k-12k=20 \\ \text{or,}~ -5k=20 \\ \text{or,}~ k=\frac{20}{-5}=-4. $ 

Hence,  the co-ordinates of the point on the given  line is $~(k,2k)=(-4,-8).$

11.  The perpendicular distance of the straight line $~3x+4y+m=0~$ from the origin is $~2~$ unit ; find $~m.$

Solution.

The perpendicular distance of the straight line $~3x+4y+m=0~$ from the origin is 

$\left|\frac{3 \times 0+4 \times 0+m}{\sqrt{3^2+4^2}}\right|=2~~(\text{By question}) \\ \text{or,}~~ \frac{|m|}{5}=2 \\ \text{or,}~~ |m|=2 \times 5 \\ \text{or,}~~ m=\pm 10.$

13.  The inclination of a straight line is $~150^{\circ};~$ if the perpendicular distance of the line from the origin is $~10~$ unit, find the equation of the straight line.

Solution.

By question, the inclination of a straight line is given by 

$m=\tan 150^{\circ}=\tan(180^{\circ}-30^{\circ}) \\ \text{or,}~~ m=-\tan 30^{\circ}=-\frac{1}{\sqrt{3}}$

So, the equation of the straight line can be written as 

$y=-\frac{1}{\sqrt{3}}x+c \\ \text{or,}~~\frac{1}{\sqrt{3}}x+y-c=0\rightarrow(1)$

Now, the distance of $~(1)~$ from the origin is 

$~\left|\frac{0+0-c}{\sqrt{(1/\sqrt{3})^2+1^2}}\right|=10 \\ \text{or,}~~\frac{c}{\sqrt{\frac 43}}=10 \\ \text{or,}~~ c=10 \times \frac{2}{\sqrt{3}}=\frac{20}{\sqrt{3}}\rightarrow(2)$

Hence, by $~(1),~(2)~$ we get the equation of the straight line 

$~\frac{1}{\sqrt{3}}x+y-\frac{20}{\sqrt{3}}=0 \\ \text{or,}~~ x+\sqrt{3}y=20.$

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14.  Find the equation of the line which passes through the point $~(4,-6)~$ and makes intercepts on the axes $~(i)~$ equal in magnitude and of the same sign , $~(ii)~$ equal in magnitude but opposite in sign.

Solution.

The equation of the straight line in intercept form is 

$\frac xa+\frac yb=1\rightarrow(1)$

By case $~(i),~a=b.$ 

Since the line $~(1)~$ passes through the point $~(4,-6)~$ , we get 

$~\frac 4a-\frac 6a=1 \\ \text{or,}~~ \frac{4-6}{a}=1\\ \text{or,}~~ -\frac 2a=1 \\ \text{or,}~~ a=-2.$

Hence, putting $~a=-2=b~$ in $~(1)~$, we get 

$~\frac{x}{-2}+\frac{y}{-2}=1 \\ \text{or,}~~x+y+2=0.$

By case $~(ii),~b=-a.$ 

Since the line $~(1)~$ passes through the point $~(4,-6)~$ , we get 

$~\frac 4a+\frac{-6}{-a}=1 \\ \text{or,}~~ \frac{4+6}{a}=1\\ \text{or,}~~ a=10$

Hence, putting $~a=10,~b=-10~$ in $~(1)~$, we get 

$~\frac{x}{10}+\frac{y}{-10}=1 \\ \text{or,}~~x-y=10.$