.jpg "Straight Line | Part-3 |Ex-2C")
9. Find the equation of the straight line equidistant from and parallel to the straight lines $x+y-3=0$ and $x+y+1=0$.
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 $x+y-1=0$. $[\text{By (1)}]$
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 $12x-5y+17=0$ and $12x-5y-9=0$.
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 $6x-8y-13=0$.
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 $6x-8y-13=0$.
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 $(-7,11)$.
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 straight lines $3x-4y-2=0$ 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 $32x-56y=23$.
The perpendicular distances of the straight lines $3x-4y-2=0$ 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 $x=0$.
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 $x=0$.
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