Straight Line | Part-3

1. Find the equation of the straight line passing through the point of intersection of the straight lines $~x+2y+3=0~$ and $~3x+4y+7=0~$ and parallel to the straight line $~y=-\frac 58x.$

Solution.

Here the given equations of straight lines are

$~x+2y+3=0~\rightarrow(1)$ and $~3x+4y+7=0~\rightarrow(2).$

From $~(1)~$ and $~(2)~$ we get,

$~(3x+4y+7)-3(x+2y+3)=0 \\ \text{or,}~~ -2y-2=0 \\ \text{or,}~~ y=\frac{2}{-2}=-1.$

From $~(1)~$ we get, $~x+2 \times(-1)+3=0 \Rightarrow x=-1.$

So, the co-ordinates of the point of intersection are $~(-1,-1).$

Slope of the straight line $~y=-\frac 58x~$ is $~(-5/8).$

So, the equation of the straight line passing through $~(-1,-1)~$ and parallel to the straight line $~y=-\frac 58x~$ is

$~y-(-1)=-\frac 58[x-(-1)] \\ \text{or,}~~ y+1=(-5/8)\cdot(x+1) \\ \text{or,}~~ 8(y+1)=-5(x+1) \\ \text{or,}~~ 5x+8y+8+5=0 \\ \therefore~ 5x+8y+13=0.$

2. Let $~AD~$ be the median of the triangle with vertices $~A(2,2)~,B(6,-1)~$ and $~C(7,3)~$. Find the equation of the straight line passing through $~(1,-1)~$ and parallel to $~AD.$

Solution.

Since $~AD~$ is the median of $~\Delta ABC~,~D$ is the midpoint of $~BC.$

$\therefore~ D \equiv\left(\frac{6+7}{2},\frac{-1+3}{2}\right)\equiv \left(\frac{13}{2},1\right).$

Now, the slope of $~AD=\frac{1-2}{\frac{13}{2}-2}=\frac{-1}{9/2}=-\frac 29.$

The equation of any straight line parallel to $~AD~$ is $~y=-\frac 29 x+c\rightarrow(1)$

Since the straight line $~(1)~$ passes through $~(1,-1)~$,

$~ -1=-\frac 29 \times 1+c\Rightarrow c=-\frac 79.$

So, by $~(1),~$ the required equation of straight line is

$~y=-\frac 29x-\frac 79.$

3. Find the equation of the straight line which passes through the point of intersection of the two lines $~2x-y+5=0~$ and $~5x+3y-4=0~$ and is perpendicular to the line $~x-3y+21=0.$

Solution.

The given equation of straight lines are

$~2x-y+5=0 \rightarrow(1),\\~ 5x+3y-4=0\rightarrow(2).$

From $~(1),~(2)~$, we get

$~5(2x-y+5)-2(5x+3y-4)=0 \\ \text{or,}~~ -5y-6y+25+8=0 \\ \text{or,}~~33=11y \\ \therefore y=\frac{33}{11}=3.$

Putting the value of $~y~$ in $~(1)~$ we get,

$~2x-3+5=0 \\ \text{or,}~~ 2x+2=0 \\ \text{or,}~~x= \frac{-2}{2}=-1.$

So, the point of intersection of the given straight lines is $~(-1,3).$

Now, $~x-3y+21=0 \\ \text{or,}~~ y=\frac 13 x+7\rightarrow(3)$

The slope $~(m)~$ of the line $~(3)~$ is $~\frac 13.$

If $~m'~$ is the slope of any straight line which is perpendicular to $~(3),~$ then

$~m' \times \frac 13=-1 \Rightarrow m'=-3.$

So, the equation of any straight line having slope $~-3~$ and passing through $~(-1,3)~$ is

$~y-3=-3[x-(-1)] \\ \text{or,}~~ y-3=-3(x+1) \\ \text{or,}~~ y-3=-3x-3 \\ \text{or,}~~3x+y=0.$

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4. Find the angle which a line perpendicular to the line $~x-y+1=0~$ makes with the positive direction of the $~x-$axis.

Solution.

$x-y+1=0 \Rightarrow y=x+1 \rightarrow(1)$

Slope of $~(1)~$ is $~m=1~$ and slope of any straight line perpendicular to $~(1)~$ is $~-1/m=-\frac 11=-1.$

If the straight line having slope $~-1~$ makes an angle $~\theta~$ with the positive direction of $~x-$axis,

$\tan\theta=-1=\tan(180^{\circ}-45^{\circ}) \\ \therefore~\theta=135^{\circ}.$

5. Find the co-ordinates of the foot of the perpendicular drawn from the point $~(2,-5)~$ on the straight line $~x=y+1.$

Solution.

The given equation can be written as $y=x-1\rightarrow(1).~$

Now, the slope of $~y=x-1~$ is $~1.~$ The slope of any straight line perpendicular to $~(1)~$ is $~-1.$

So, the equation of any straight line having slope $~-1~$ and passing through the point $~(2,-5)~$ can be written as

$~y-(-5)=-1(x-2) \\ \text{or,}~~ y+5=-x+2 \\ \text{or,}~~ x+y+3=0 \rightarrow(2)$

From $~(1)~$ and $~(2)~$ we get,

$~x+(x-1)+3=0 \\ \text{or,}~~2x+2=0 \\ \text{or,}~~ 2(x+1)=0 \\ \text{or,}~~x+1=0 \\ \therefore ~ x=-1.$

So, by $~(1)~$ we get, $~y=x-1=-1-1=-2.$

Hence, the point of intersection of $~(1)~$ and $~(2)~$ is $~(-1,-2).$

So, the co-ordinates of the foot of the perpendicular is $~(-1,-2).$

6. Show that the equation of the straight line passing through $~(a\cos^3\theta,a\sin^3\theta)~$ and perpendicular to the straight line $~x\sec\theta+y\csc\theta=a~$ is $~x\cos\theta-y\sin\theta=a\cos2\theta.$

Solution.

The slope of the given straight line $~x\sec\theta+y\csc\theta=a\rightarrow(1)~$ is $~-\frac{\sec\theta}{\csc\theta}=-\frac{\sin\theta}{\cos\theta}.$

The slope of any straight line perpendicular to $~(1)~$ is $~\frac{\cos\theta}{\sin\theta}.$

The equation of any straight line having slope $~\frac{\cos\theta}{\sin\theta}~$ and passing through $~(a\cos^3\theta,a\sin^3\theta)~$ is

$~y-a\sin^3\theta=\frac{\cos\theta}{\sin\theta}(x-a\cos^3\theta) \\ \text{or,}~~ y\sin\theta-a\sin^4\theta=x\cos\theta-a\cos^4\theta \\ \text{or,}~~ x\cos\theta-y\sin\theta=a(\cos^4\theta-\sin^4\theta) \\ \text{or,}~~ x\cos\theta-y\sin\theta\\=a[(\cos^2\theta)^2-(\sin^2\theta)^2]\\=a(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)\\=a\cos2\theta.$

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7. Find the equation of the straight line which bisects perpendicularly the line segment joining the two points $~(2,-4)~$ and $~(-6,0).~$ What is the perpendicular distance of this line from the origin ?

Solution.

The slope of the straight line joining two points $~A(2,-4)~$ and $~B(-6,0)~$ is $~=\frac{0+4}{-6-2}=-\frac 12.$

So, slope of any straight line perpendicular to $~AB~$ is $~2.$

The midpoint of $~AB~$ is $~\left(\frac{2-6}{2},\frac{-4+0}{2}\right)=(-2,-2).$

So, the equation of any straight line having slope $~2~$ and passing through $~(-2,-2)~$ is

$~y-(-2)=2[x-(-2)] \\ \text{or,}~~ y+2=2(x+2) \\ \therefore~ 2x-y+2=0.$

So, the perpendicular distance of this line from the origin is

$=\frac{|2 \times 0-0+2|}{\sqrt{2^2+1^2}}=\frac{2}{\sqrt{5}}~~\text{unit.}$

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8. Find the equation of the perpendicular bisector of the straight line segment joining the points $~(-2,7)~$ and $~(8,-1)~$ . Also find the distance of this bisector from the origin.

Solution.

Let $~A\equiv (-2,7),~~B\equiv (8,-1).$

The midpoint of $~AB~$ is $~M\left(\frac{-2+8}{2},\frac{7-1}{2}\right)\equiv (3,3).$

Now, slope of $~AB~$ is $~\frac{-1-7}{8+2}=-\frac{8}{10}=-\frac 45.$

So, slope of the perpendicular bisector of $~AB~$ is $~\frac 54~$ and it passes through $~M(3,3)~$ and so the equation of the perpendicular bisector is

$y-3=\frac 54(x-3) \\ \text{or,}~~4(y-3)=5(x-3) \\ \text{or,}~~ 4y-12=5x-15 \\ \text{or,}~~5x-4y-15+12=0 \\ \therefore~ 5x-4y-3=0\rightarrow(1)$

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

$=\frac{|5 \times 0-4 \times 0-3|}{\sqrt{5^2+4^2}}=\frac{3}{\sqrt{41}}~\text{unit.}$

9. The straight line $~\frac xa+\frac yb=1~$ passes through the point of intersection of the straight lines $~7x+9y=3~$ and $~2y-x+7=0~$ and makes an angle $~90^{\circ}~$ with the straight line $~5x-6y+15=0~$; find the values of $~a~$ and $~b.$

Solution.

Any straight line passing through the intersection of the straight lines $~7x+9y=3~$ and $~2y-x+7=0~$ can be written as

$~7x+9y-3+k(2y-x+7)=0 \\ \text{or,}~~ (7-k)x+(9+2k)y-3+7k=0 \\ \therefore~y=-\frac{7-k}{9+2k}x+\frac{3-7k}{9+2k}\rightarrow(1)$

The slope of straight line $~5x-6y+15=0~$ is $~\frac 56.$

Since this straight line makes an angle $~\theta~$ with the straight line $~(1)~$, so

$~\frac 56 \times \left(-\frac{7-k}{9+2k}\right)=-1 \\ \text{or,}~~ 35-5k=54+12k \\ \text{or,}~~ 35-54=12k+5k\\ \text{or,}~~ -19=17k \\ \therefore~ k=-\frac{19}{17}.$

Putting the value of $~k~$ in $~(1)~$ we get,

$y=-\frac{138}{115}x+\frac{184}{115} \\ \text{or,}~~ 138x+115y=184 \\ \text{or,}~~\frac{x}{184/138}+\frac{y}{184/115}=1 \\ \therefore~\frac{x}{4/3}+\frac{y}{8/5}=1\rightarrow(2)$

Comparing the equation $~(2)~$ with $~\frac xa+\frac yb=1~$ we get,

$~a=\frac 43,~~ b=\frac 85.$

10. The straight line $~8x-18y+31=0~$ bisects perpendicularly the straight line-segment joining the points $~P(2,8)~$ and $~Q(h,k)~$; find the values of $~h~$ and $~k.$

Solution.

The midpoint of $~P(2,8)~$ and $~Q(h,k)~$ is $~M\left(\frac{h+2}{2},\frac{k+8}{2}\right).~~$ Clearly, the straight line $~8x-18y+31=0~$ passes through the point $~M.~$

So, $~8\left(\frac{h+2}{2}\right)-18\left(\frac{k+8}{2}\right)+31=0 \\ \text{or,}~~4(h+2)-9(k+8)+31=0 \\ \therefore~4h-9k-33=0\rightarrow(1)$

Since the given straight line and $~PQ~$ are perpendicular to each other ,

$~\text{slope of PQ} \times \text{slope of (1)}=-1 \\ \text{or,}~~\frac{k-8}{h-2} \times \frac{4}{9}=-1 \\ \text{or,}~~4(k-8)=-9(h-2) \\ \therefore~ 9h+4k-50=0 \rightarrow(2)$

Now, solving $~(1)~$ and $~(2)~$ we get,

$~h=6,~~k=-1.$