Exercise 6.2
Question 10
KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.2 (Q10-Q12)
t का मान निकाले जिसके लिए बिन्दुएँ (1,-1) , (3,-3) तथा (t,2) संरेख है
[Find the value of t for which the points (1,-1) , (3,-3) and (t,2) are collinear ]
Sol :
माना बिन्दुएँ (1,-1) , (3,-3) तथा (t,2) संरेख है
$\frac{1}{2}\left|\begin{array}{ccc}1 & -1 & 1 \\ 3 & -3 & 1 \\ t & 2 & 1\end{array}\right|=0$
R1→R1-R2 , R2→R2-R3
$\left|\begin{array}{ccc}-2 & 2 & 0 \\ 3-t & -5 & 0 \\ t & 2 & 1\end{array}\right|=0$
Expanding along C3
$1\left|\begin{array}{cc}-2 & 2 \\ 3-t & -5\end{array}\right|=0$
⇒10-2(3-t)=0
⇒-2(3-t)=-10
⇒3-t=5
⇒-t=5-3
⇒-t=2
⇒t=-2
Sol :
<to be added>
Diagram2
$\frac{1}{2}\left|\begin{array}{lll}3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1\end{array}\right|=0$
$R_{1} \rightarrow R_{1}-\frac{1}{3} R_{2}$
$\left|\begin{array}{ccc}0 & 0 & \frac{2}{3} \\ 9 & 3 & 1 \\ x & y & 1\end{array}\right|=0$
Expanding along R1
$\frac{2}{3}\left|\begin{array}{ll}9 & 3 \\ x & y\end{array}\right|=0$
⇒9y-3x=0
⇒-3x+9y=0
⇒-3(x-3y)=0
⇒x-3y=0
Sol :
<to be added>
Diagram
<to be added>
$\frac{1}{2}\left|\begin{array}{ccc}1 & 3 & 1 \\ 0 & 0 & 1 \\ 2 & y & 1\end{array}\right|=0$
Expanding along R2
$-1\left|\begin{array}{ll}1 & 3 \\ x & y\end{array}\right|=0$
⇒-(y-3x)=0
⇒3x-y=0
Diagram
<to be added>
$\frac{1}{2}\left|\begin{array}{ccc}1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1\end{array}\right|=\pm 3$
$\left|\begin{array}{lll}1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1\end{array}\right|=\pm 6$
Expanding along R2
$-1\left|\begin{array}{ll}1 & 3 \\ k & 0\end{array}\right|=\pm 6$
⇒-(0-3k)=±6
⇒3k=±6
⇒k=±2
[Find the value of t for which the points (1,-1) , (3,-3) and (t,2) are collinear ]
Sol :
माना बिन्दुएँ (1,-1) , (3,-3) तथा (t,2) संरेख है
$\frac{1}{2}\left|\begin{array}{ccc}1 & -1 & 1 \\ 3 & -3 & 1 \\ t & 2 & 1\end{array}\right|=0$
R1→R1-R2 , R2→R2-R3
$\left|\begin{array}{ccc}-2 & 2 & 0 \\ 3-t & -5 & 0 \\ t & 2 & 1\end{array}\right|=0$
Expanding along C3
$1\left|\begin{array}{cc}-2 & 2 \\ 3-t & -5\end{array}\right|=0$
⇒10-2(3-t)=0
⇒-2(3-t)=-10
⇒3-t=5
⇒-t=5-3
⇒-t=2
⇒t=-2
Question 11
सारणिक का प्रयोग कर (3,1) तथा (9,3) को मिलानेवाली रेखा का समीकरण निकालें ।
[Find the equation of the line joining (3,1) and (9,3) using determinants.]Sol :
<to be added>
Diagram2
$\frac{1}{2}\left|\begin{array}{lll}3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1\end{array}\right|=0$
$R_{1} \rightarrow R_{1}-\frac{1}{3} R_{2}$
$\left|\begin{array}{ccc}0 & 0 & \frac{2}{3} \\ 9 & 3 & 1 \\ x & y & 1\end{array}\right|=0$
Expanding along R1
$\frac{2}{3}\left|\begin{array}{ll}9 & 3 \\ x & y\end{array}\right|=0$
⇒9y-3x=0
⇒-3x+9y=0
⇒-3(x-3y)=0
⇒x-3y=0
Question 12
सारणिक का प्रयोग कर बिन्दुओं A(1,3) तथा B(0,0) से गुजरती हुई रेखा का समीकरण निकालें तथा k का मान निकाले यदि D(k, 0) एक ऐसा बिन्दु है ताकि त्रिभुज ABD का क्षेत्रफल 3 वर्ग इकाई है ।
[Find the equation of the line joining A(1,3) and B(0,0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3 sq. units.]
<to be added>
Diagram
<to be added>
$\frac{1}{2}\left|\begin{array}{ccc}1 & 3 & 1 \\ 0 & 0 & 1 \\ 2 & y & 1\end{array}\right|=0$
Expanding along R2
$-1\left|\begin{array}{ll}1 & 3 \\ x & y\end{array}\right|=0$
⇒-(y-3x)=0
⇒3x-y=0
Diagram
<to be added>
$\frac{1}{2}\left|\begin{array}{ccc}1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1\end{array}\right|=\pm 3$
$\left|\begin{array}{lll}1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1\end{array}\right|=\pm 6$
Expanding along R2
$-1\left|\begin{array}{ll}1 & 3 \\ k & 0\end{array}\right|=\pm 6$
⇒-(0-3k)=±6
⇒3k=±6
⇒k=±2
Best
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