Differentiation (Part-38) | S N De

 

In this article, we have discussed Short Answer Type Questions(9-15) from Ex-2C of Straight Line Chapter of (S.N.De) Chhaya Mathematics,Class 11. 

9. Find the equation of the  straight line equidistant from and parallel to the sls $~x+y-3=0,~x+y+1=0.$

Solution.

Let the required equation be $~x+y-k=0~( k\neq 0) \rightarrow(1)~$ which is parallel to the given straight lines $~x+y-3=0 \rightarrow(2)~$ and $~ x+y+1=0 \rightarrow(3).$

Now, since the straight line $~(1)~$ is equidistant from the straight lines $~(2)~$ and $~(3),$

$\frac{|k-3|}{\sqrt{1^2+1^2}}=\frac{|k+1|}{\sqrt{1^2+1^2}} \\ \text{or,}~~ \frac{k-3}{\sqrt{2}}=\pm \frac{k+1}{\sqrt{2}} \\ \text{or,}~~ k-3 =\pm(k+1) \\ \text{or,}~~ k-3=-(k+1) \Rightarrow k=1,\\~~k-3=k+1 \Rightarrow -3=1\\~\text{which is impossible.}$

So, The required equation of the straight line is $~x+y-1=0.~~[\text{By (1)}]$

10. The equation of the straight line mid-way between two parallel lines at a distance of $~2~$ unit is $~12x-5y+4=0.~$ Find the equations of the lines.

Solution.

Let the required equation of  parallel straight lines be  $~12x-5y+k=0~$ where $~k(\neq 0)~$ has two different values. Since the given straight line is mid-way between the two parallel straight lines,

$~\frac{|k-4|}{\sqrt{12^2+(-5)^2}}=\frac 12 \times 2 \\ \text{or,}~~ \frac{|k-4|}{\sqrt{144+25}}=1 \\ \text{or,}~~ |k-4|=1 \times \sqrt{169} \\ \text{or,}~~ k-4=\pm 13 \\ \text{or,}~~ k=\pm 13+4 \\ \text{or,}~~ k=13+4=17,~\\~~k=-13+4=-9.$

So, the required equation of straight lines are $~12x-5y+17=0,~12x-5y-9=0.$

11. Find the equation of the  straight line equidistant from the point $~(2,-2)~$ and the line $~3x-4y+1=0.$

Solution.

The given straight line is $~3x-4y+1=0 \rightarrow(1).~$ 

The perpendicular distance of the point $~(2,-2)~$ from the straight line $~(1)~$ is 

$~\frac{|3 \times 2-4 \times (-2)+1|}{\sqrt{3^2+(-4)^2}}=\frac{|6+8+1|}{\sqrt{9+16}}=\frac{15}{5}=3~\text{unit.}$

Clearly, the required straight line will be parallel to the straight line $~(1).$

Let the equation of the required straight line be $~3x-4y+k=0 \rightarrow(2)$

By question, the distance between the straight lines $~(1)~$ and $~(2)~$ is $~\frac 12 \times 3~$ unit.

$\therefore~ \frac{|k-1|}{\sqrt{3^2+(-4)^2}}=\frac 32 \\ \text{or,}~~ \frac{|k-1|}{5}=\frac 32 \\ \text{or,}~~ |k-1|=\frac 32 \times 5 \\ \text{or,}~~ k-1=\pm \frac{15}{2} \\ \therefore~ k=\frac{17}{2},~-\frac{13}{2}.$

Now, for $~k=\frac{17}{2},~~$ we get by $~(2)~$,

$~3x-4y+\frac{17}{2}=0 \\ \text{or,}~~ 6x-8y+17=0\rightarrow(3)$

But the distance of the point $~(2,-2)~$ from the straight line $~(3)~$ is 

$\frac{|6 \times 2-8 \times(-2)+17|}{\sqrt{6^2+(-8)^2}}\\=\frac{|12+16+17|}{10}\\=\frac{45}{10}\\=\frac 92 \neq \frac 32$

Again, for $~k=-13/2~,$ we get by $~(2),$

$3x-8y-\frac{13}{2}=0 \\ \text{or,}~~ 6x-8y-13=0 \rightarrow(4)$

Now, distance of the point $~(2,-2)~$ from the straight line $~(4)~$ is 

$\frac{|6 \times 2-8 \times (-2)-13|}{\sqrt{6^2+(-8)^2}}=\frac{15}{10}=\frac 32$

Hence, the required equation of the straight line is $~6x-8y-13=0.$

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12. Find the equations of the two  straight lines parallel to the line $~3x+4y=15~$ and at a distance of $~7.5~$ unit from the point $~(1,-2).$

Solution.

Equation of any straight line parallel to the straight line $~3x+4y=15~$ is $~3x+4y=k~~(k \neq 0)\rightarrow(1)~$ which is at a distance of $~7.5~$ unit from the point $~(1,-2).$

$ \therefore~ \frac{|3 \times 1+4 \times (-2)-k|}{\sqrt{3^2+4^2}}=7.5 \\ \text{or,}~~ \frac{|3-8-k|}{5}=7.5 \\ \text{or,}~~ \frac{|-5-k|}{5}=7.5 \\ \text{or,}~~  \frac{|-(k+5)|}{5}=7.5 \\ \text{or,}~~  |k+5|=7.5 \times 5 \\ \text{or,}~~  k+5 =\pm 37.5 =\pm \frac{75}{2} \\ \therefore k=\frac{75}{2}-5,~~ -\frac{75}{2}-5 \\ \text{or,}~~  k=\frac{65}{2},-\frac{85}{2}.$

Now, for $~k=\frac{65}{2},~$ the equation of the required straight line is 

$3x+4y=\frac{65}{2} \Rightarrow 6x+8y=65.$

Again, for $~k=-\frac{85}{2},~$ the equation of the required straight line is 

$3x+4y=-\frac{85}{2} \Rightarrow 6x+8y+85=0.$

13. Find the co-ordinates of the points on the  straight line $~x+y=4~$ which are at unit distance  from the line $~4x+3y=10.$

Solution.

Let the co-ordinates of the required point be $~(h,k).$

By question, $~h+k=4 \rightarrow(1)$

Now, $~\frac{|4h+3k-10|}{\sqrt{4^2+3^2}}=1 \\ \text{or,}~~ |4h+3k-10|=1 \times 5=5 \\ \text{or,}~~  |4h+3(4-h)-10|=5~~  [\text{By (1)}]   \\ \text{or,}~~  |4h+12-3h-10|=5 \\ \text{or,}~~ |h+2|=5 \\ \text{or,}~~  h+2=\pm 5 \\ \text{or,}~~  h=\pm 5-2 \\ \text{or,}~~  h=3,-7. $

For $~h=3,~~k=4-3=1~$ and for $~h=-7,~~k=4-(-7)=11.$

Hence, the co-ordinates of the required points are $~(3,1)~$ and $~(-7,11).$

14. Find the equation to the locus of a moving point which is always equidistant from the  straight lines $~3x-4y-2=0~$ and $~5x-12y=4.$

Solution.

Let the co-ordinates of the variable point be $~P(h,k).$

The perpendicular distances of the sls $~3x-4y-2=0~$ and $~5x-12y=4~$ from $~P(h,k)~$ are 

$~d_1=\frac{|3h-4k-2|}{\sqrt{3^2+(-4)^2}}=\frac{|3h-4k-2|}{5}~$ and 

$~d_2=\frac{|5h-12k-4|}{\sqrt{5^2+(-12)^2}}=\frac{|5h-12k-4|}{13}.$

By question,

$~d_1=d_2 \\ \text{or,}~~ \frac{|3h-4k-2|}{5}=\frac{|5h-12k-4|}{13} \\ \text{or,}~~ 13(3h-4k-2)=\pm 5(5h-12k-4) \rightarrow(1)$

Now, taking +ve sign we get by $~(1),$

$~39h-52k-26=25h-60k-20 \\ \text{or,}~~ 39h-25h-52k+60k-26+20=0 \\ \text{or,}~~ 14h+8k-6=0 \\ \text{or,}~~ 2(7h+4k-3)=0 \\ \text{or,}~~ 7h+4k=3\rightarrow(2).$

Again, taking -ve sign , we get by $~(1),$

$~39h-52k-26=-25h+60k+20 \\ \text{or,}~~ 39h+25h-52k-60k-26-20=0 \\ \text{or,}~~ 64h-112k-46=0 \\ \text{or,}~~ 2(32h-56k-23)=0 \\ \text{or,}~~ 32h-56k=23\rightarrow(3)$

Hence, by $~(2)~$ and $~(3)~$ , we get the locus of the moving point are $~7x+4y=3~$ and $~32x-56y=23.$

15. Find the locus of the foot of the perpendicular from the point $~(a,0)~$ to the straight line $~x-ty+at^2=0,~t~$ being variable.

Solution.

Let $~P(h,k)~$ be the co-ordinates of the foot of the perpendicular from the point $~(a,0)~$ to the straight line $~x-ty+at^2=0.$

Now, the slope of the line joining the points $~(a,0)~$ and $~P(h,k)~$ is given by 

$~m_1=\frac{k-0}{h-a}=\frac{k}{h-a}.$

The slope $~(m_2)~$ of the given straight line $~x-ty+at^2=0~$ is $~\frac 1t.$

Clearly, $~m_1 \times m_2=-1 \\ \text{or,}~~ \frac{k}{h-a} \times \frac 1t=-1 \\ \text{or,}~~ t=\frac{k}{a-h}.$

Again, the point $~(h,k)~$ lies on the given straight line.

$~ \therefore~ h-tk+at^2=0\\ \text{or,}~~h-\frac{k}{a-h} \times k+a\left(\frac{k}{a-h}\right)^2=0 \\ \text{or,}~~ h(a-h)^2-k^2(a-h)+ak^2=0 \\ \text{or,}~~ h(a-h)^2+ak^2-k^2(a-h)=0 \\ \text{or,}~~ h(a-h)^2+ak^2-ak^2+hk^2=0 \\ \text{or,}~~ h(a-h)^2 +hk^2=0 \\ \text{or,}~~ h[(a-h)^2+k^2]=0 \\ \therefore ~h=0~[\because~ (a-h)^2+k^2 \neq 0]$

For $~(a-h)^2+k^2=0~\Rightarrow h=a,~k=0~$ which means the point $~(a,0)~$ lies on the given straight line which is impossible.

Hence, the locus of the foot of the perpendicular is $~x=0.$