Differentiation (Part-38) | S N De

 In this article, we have discussed few VSA type Questions of Ex-2B of Straight Line Chapter (S N Dey Mathematics, Chhaya Publications.)

4. Find the slopes of straight lines $~(i)~$ parallel $~(ii)~$ perpendicular to the straight line $~3x+4y=11.$

Solution.

$~~3x+4y=11 \\ \text{or,}~~ 4y=-3x+11 \\ \therefore y=-\frac 34 x+\frac{11}{4}\rightarrow(1).$

The slope of straight line  parallel to the straight line $~(1)~$ is $~ -\frac 34.$

Let $~m~$ be the slope of straight line perpendicular to the straight line $~(1)~$ .

So, $~m \times (-3/4)=-1 \Rightarrow m=\frac 43.$

5. If the straight line joining the points $~(3,4)~$ and $~(2,-1)~$ be parallel to the line joining the points $~(a,-2)~$ and $~(4,-a);~$ find the value of $~a.$

Solution.

For the parallel lines , the slopes will be same . So, the slope $~(m_1)~$ of the straight line joining the points $~(3,4)~$ and $~(2,-1)~$  is $~\frac{-1-4}{2-3}=\frac{-5}{-1}=5.$

Similarly,  the slope $~(m_2)~$ of the straight line joining the points $~(a,-2)~$ and $~(4,-a)~$  is $~\frac{-a+2}{4-a}.$

$\because~ m_1=m_2 \\ \text{so,}~~ 5=\frac{-a+2}{4-a} \\ \text{or,}~~ 5(4-a)=2-a \\ \text{or,}~~ 20-5a=2-a \\ \text{or,}~~  20-2=5a-a \\ \therefore~a=\frac{18}{4}=\frac 92.$

$6.~~$ The straight line joining the points $~(-2,5)~$ and $~(-4,3)~$ is perpendicular to the line joining the points $~(k,0)~$ and $~(2,3k)~$; find $~k.$

Solution.

The slope $~(m_1)~$ of the straight line joining the points $~(-2,5)~$ and $~(-4,3)~$ is $~~\frac{3-5}{-4+2}=1$

Similarly, the slope $~(m_2)~$ of the straight line joining the points $~(k,0)~$ and $~(2,3k)~$ is $~\frac{3k-0}{2-k}=\frac{3k}{2-k}.$

Since the given straight lines are perpendicular to each other, 

$~m_1 \times m_2=-1 \\ \text{or,}~~1 \times \frac{3k}{2-k}=-1 \\ \text{or,}~~3k=-2+k \\ \text{or,}~~ 3k-k=-2 \\ \therefore~k=\frac{-2}{2}=-1.$

Chhaya Math Solution of Ellipse (Class 11)

Chhaya Math Solution of Ellipse (Class 11)

7. Find the equation of the straight line perpendicular to the line joining the points $~(2,3)~$ and $~(3,-1)~$ and passing through the point $~(2,1).$

Solution.

The slope of the straight line joining the points $~(2,3)~$ and $~(3,-1)~$ is  $~(m)=\frac{-1-3}{3-2}=-4.$

Let the slope of the straight line perpendicular to the line joining the points $~(2,3)~$ and $~(3,-1)~$  be $~m'.$

So, $~m \times m'=-1 \Rightarrow m'=\frac{-1}{-4}=\frac 14.$

So, the equation of the straight line perpendicular to the given line and passing through the point $~(2,1)~$ is

$~y-1=\frac 14(x-2) \\ \text{or,}~~ 4y-4=x-2 \\ \therefore~ x-4y+2=0.$

8. Find the equation of the straight line passing through the point $~(-3,4)~$ and parallel to straight line $~2x-3y=5.$

Solution.

The given straight line can be written as $~y=\frac 23x-\frac 53\rightarrow(1).$

The slope of $~(1)~$ is $~\frac 23.$ So, the The slope of the straight line parallel to straight line $~2x-3y=5~$ is $~2/3.$

Finally, the equation of the straight line passing through the point $~(-3,4)~$ and parallel to given straight line is 

$~y-4=\frac 23(x+3) \\ \text{or,}~~ 3(y-4)=2(x+3) \\ \therefore~ 2x-3y+18=0.$

9. Find the equation of the straight line perpendicular to $~2x+3y+5=0~$ and passing through the point $~(2,-3).$

Solution.

The given equation of the straight line can be written as $~y=-\frac 23x-\frac 53 \rightarrow(1).$ The slope of $~(1)~$ is $~-\frac 23.$

If $~m'~$ be the slope of the straight line perpendicular to $~2x+3y+5=0~$,

$~m' \times (-2/3)=-1\Rightarrow m'=\frac 32.$

So, equation of the required straight line passing through $~(2,-3)~$ is

$~y-(-3)=\frac 32(x-2) \\ \text{or,}~~ 2(y+3)=3(x-2)\\ \text{or,}~~ 0=3x-2y-6-6 \\ \therefore~ 3x-2y=12.$ 

Oswaal NCERT Exemplar (Problems - solutions) Class 11 Mathematics Book

10. Find the equation of the straight line passing through the point $~(3,-4)~$ and parallel to the line joining the points $~(4,7)~$ and $~(-5,1).$

Solution. 

The equation of the straight line joining the points $~(4,7)~$ and $~(-5,1)~$ can be written as 

$~\frac{y-7}{7-1}=\frac{x-4}{4-(-5)} \\ \text{or,}~~ \frac{y-7}{6}=\frac{x-4}{9} \\ \text{or,}~~ \frac 12(y-7)=\frac 13(x-4) \\ \text{or,}~~ 2(x-4)=3(y-7) \\ \therefore~ 2x-3y=-13 \rightarrow(1)$

The equation of the straight line parallel to $~(1)~$ can be written as 

$~2x-3y=k~~(k \neq 0)\rightarrow(2)$

If the straight line $~(2)~$ passes through the point $~(3,-4),~$ then

$~2 \times 3-3 \times (-4)=k \Rightarrow k=18\rightarrow(3).$

From $~(2)~$ and $~(3)~$, we get the required equation of the straight line which is $~ 2x-3y=18.$