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

$4.$ The perpendicular distance of the point $(-3,4)$ from the straight line $2x-3y+k=0$ 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 $$\frac{|12 \times 3+k \times (-5)-9|}{\sqrt{12^2+k^2}}=4 \\ \text{or, } \frac{|27-5k|}{\sqrt{144+k^2}}=4 \\ \text{or, } (27-5k)^2=16(144+k^2) \\ \text{or, } 27^2-2 \times 27 \times 5k+(5k)^2=2304+16k^2 \\ \text{or, } 9k^2-270k-1575=0 \\ \text{or, } 9(k^2-30k-175)=0 \\ \text{or, } k^2-35k+5k-175=0 \\ \text{or, } k(k-35)+5(k-35)=0 \\ \text{or, } (k-35)(k+5)=0 \\ \therefore k=35,-5.$$

$6.$ 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 $$\frac{|0+0-ab|}{\sqrt{b^2+a^2}}=p \\ \text{or, } \frac{ab}{\sqrt{a^2+b^2}} = p \\ \text{or, } a^2b^2=p^2(a^2+b^2) \\ \text{or, } \frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2} =\frac{1}{b^2}+\frac{1}{a^2} \\ \therefore \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2} \text{ (proved)}$$

$7.$ 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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Complete Solutions:

Straight Line - Complete Solution | S. N. Dey