Differentiation (Part-38) | S N De

 

$1.~$ Show that the point $~(-8,3)~$ is equidistant from the straight lines $~4x-3y+1=0~$ and $~12x-5y+7=0.$

Solution.

The distance of the point $~(-8,3)~$ from the straight line $~4x-3y+1=0~$ is 

$d_1=\frac{|4 \times (-8)-3 \times 3+1|}{\sqrt{4^2+(-3)^2}}\\=\frac{|-40|}{\sqrt{25}}=\frac{40}{5}=8~~\text{unit.}$

Again, the distant of the point $~(-8,3)~$ from the straight line $~12x-5y+7=0~$ is 

$~d_2=\frac{|12 \times (-8)-5 \times 3+7|}{\sqrt{12^2+(-5)^2}}=\frac{|-96-15+7|}{\sqrt{169}}\\=\frac{|-104|}{13}=\frac{104}{13}=8~~\text{unit.}$

So, $~d_1=d_2~$ and thus follows the result.

$2.~$ Prove that the point $~(2,2)~$ is equidistant from the three straight lines $~4x+3y-4=0,~12x-5y+12=0~$ and $~3x-4y=8.$

Solution.

The distant of the point $~(2,2)~$ from the straight line $~4x+3y-4=0~$ is 

$d_1=\frac{|4 \times 2+3 \times 2-4|}{\sqrt{4^2+3^2}}=\frac{|8+6-4|}{\sqrt{25}}=\frac{10}{5}=2~~\text{unit.}$

Similarly, the distant of the point $~(2,2)~$ from the straight line $~12x-5y+12=0~$ is 

$d_2=\frac{|12 \times 2-5 \times 2+12|}{\sqrt{12^2+(-5)^2}}=\frac{|24-10+12|}{\sqrt{169}}=\frac{26}{13}=2~~\text{unit.}$

Finally, the distant of the point $~(2,2)~$ from the straight line $~3x-4y=8~$ is 

$d_3=\frac{|3 \times 2-4 \times 2-8|}{\sqrt{3^2+(-4)^2}}=\frac{|-10|}{\sqrt{25}}=\frac{10}{5}=2~~\text{unit.}$

So, $~d_1=d_2=d_3~$ and hence follows the result.

$3.~$ The perpendicular distance of the line $~y+mx=13~$ from the origin is $~12~$ unit ; find $~m.$

Solution.

The perpendicular distance of the line $~y+mx=13~$ from the origin $~(0,0)~$ is 

$~\frac{|0+0 \times m-13|}{\sqrt{1+m^2}}=12 \\ \text{or,}~~ \frac{13}{\sqrt{1+m^2}}=12 \\ \text{or,}~~ \sqrt{1+m^2}=\frac{13}{12} \\ \text{or,}~~ 1+m^2=\left(\frac{13}{12}\right)^2 \\ \text{or,}~~ m^2=\frac{169}{144}-1 \\ \text{or,}~~ m=\pm\sqrt{\frac{169-144}{144}}=\pm \sqrt{\frac{25}{144}} \\ \therefore~ m=\pm \frac{5}{12}~~\text{(ans.)}$

$4.~$ The perpendicular distance of the point $~(-3,4)~$ from the sl $~2x-3y+k=0~$ is $~2\sqrt{13}~$ unit; find $~k.$

Solution.

The perpendicular distance of the point $~(-3,4)~$ from the given straight line is

$\frac{|2 \times (-3)-3 \times 4+k|}{\sqrt{2^2+(-3)^2}} =2\sqrt{13} \\ \text{or,}~~ \frac{|-6-12+k|}{\sqrt{13}}=2\sqrt{13} \\ \text{or,}~~ |k-18|=2 \times 13 \\ \text{or,}~~ k-18 =\pm 26 \\ \text{or,}~~ k= \pm 26+18 \\ \text{or,}~~ k=44,-8.$

$5.~$ If the length of the perpendicular drawn from the point $~(3,-5)~$ upon the line $~12x+ky=9~$ be $~4~$ unit, find the value of $~k.$

Solution.

The length of the perpendicular drawn from the point $~(3,-5)~$ upon the line $~12x+ky=9~$ is

$\frac{|12 \times 3+k \times (-5)-9|}{\sqrt{12^2+k^2}}=4 \\ \text{or,}~~ \frac{|27-5k|}{\sqrt{144+k^2}}=4 \\ \text{or,}~~ (27-5k)^2=14(144+k^2) \\ \text{or,}~~ 27^2-2 \times 27 \times 5k+(5k)^2\\~~=2304+16k^2 \\ \text{or,}~~ 9k^2-270k-1575=0 \\ \text{or,}~~ 9(k^2-30k-175)=0 \\ \text{or,}~~ k^2-35k+5k-175=0 \\ \text{or,}~~ k(k-35)+5(k-35)=0 \\ \text{or,}~~ (k-35)(k+5)=0 \\ \therefore~ k=35,-5.$

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$6.~$ The intercepts of a straight line upon the co-ordinate axes are $~a~$ and $~b.~$ If the length of the perpendicular on this line from the origin be $~p~$, prove that, $~\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}.$

Solution.

The equation of the straight line in intercept form is given by 

$~\frac xa+\frac yb=1 \\ \text{or,}~~ bx+ay-ab=0 \rightarrow(1)$

Now, the length of the perpendicular on this line $~(1)~$ from the origin is 

$~\frac{|0+0-ab|}{\sqrt{b^2+a^2}}=p \\ \text{or,}~~ \frac{ab}{\sqrt{a^2+b^2}} \\ \text{or,}~~ a^2b^2=p^2(a^2+b^2) \\ \text{or,}~~ \frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2} =\frac{1}{b^2}+\frac{1}{a^2} \\ \therefore~ \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}~~\text{(proved)}$

$7.~$ Find the perpendicular distance of the point $~(2,1)~$ from the lines $~8x+6y=17~$ and $~4x+3y+1=0~$ and hence, find the distance between the  given lines.

Solution. 

The given equations of straight lines are 

$~2(4x+3y)-17=0 \\ \Rightarrow 4x+3y-\frac{17}{2}=0 \rightarrow(1),\\~~ 4x+3y+1=0 \rightarrow(2).$

Clearly, the straight lines $~(1)~$ and $~(2)~$ are parallel.

We know that the distance between two parallel lines $~ax+by+c_1=0~$ and $~ax+by+c_2=0~$ is

$~d=\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}.~~$ So, the distance between the straight lines $~(1)~$ and $~(2)~$ is $~\frac{|(17/2)+1|}{\sqrt{4^2+3^2}}=\frac{19/2}{5}=\frac{19}{10}~~\text{unit.}$

Now, the perpendicular distance of the point $~(2,1)~$ from the line $~8x+6y=17~$ is

$\frac{|8 \times 2+6 \times 1-17|}{\sqrt{8^2+6^2}}=\frac{|16+6-17|}{\sqrt{2^2(4^2+3^2)}}=\frac{5}{2\sqrt{25}}=\frac 12~~\text{unit.}$

Finally, the perpendicular distance of the point $~(2,1)~$ from the line $~4x+3y+1=0~$ is 

$\frac{|4 \times 2+3 \times 1+1|}{\sqrt{4^2+3^2}}=\frac{12}{5}~~\text{unit.}$

$8.~$ Show that the sum of the squares of the perpendicular from the origin upon the lines $~x\cos\alpha+y\sin\alpha=a\cos2\alpha~$ and $~x\sec\alpha+y\csc\alpha=2a~$ is independent of $~\alpha.$

Solution.

The perpendicular distance of the straight line $~x\cos\alpha+y\sin\alpha=a\cos2\alpha~$ from the origin is 

$d_1=\frac{|-a\cos2\alpha|}{\sqrt{\cos^2\alpha+\sin^2\alpha}}=a\cos2\alpha \\ \text{or,}~~ d_1^2=a^2\cos^22\alpha.$

Again, the perpendicular distance of the straight line $~x\sec\alpha+y\csc\alpha=2a~$  from the origin is 

$d_2=\frac{|-2a|}{\sqrt{\sec^2\alpha+\csc^2\alpha}} \\ \therefore d_2^2=\frac{4a^2}{\sec^2\alpha+\csc^2\alpha} \\ \text{or,}~~ d_2^2=\frac{4a^2\sin^2\alpha\cos^2\alpha}{\sin^2\alpha+\cos^2\alpha} \\ \text{or,}~~ d_2^2=(2\sin\alpha\cos\alpha)^2 \times a^2 \\ \text{or,}~~ d_2^2=a^2  \sin^22\alpha$

Hence, $~d_1^2+d_2^2=a^2(\cos^22\alpha+\sin^22\alpha)=a^2 \rightarrow(1)$

So, the required result follows from $~(1).$

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