KC sinha solutions | myhelper: KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.2 (Q7-Q9)

KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.2 (Q7-Q9)

Exercise 6.2

Question 7
यदि बिन्दुएँ (a,b),(a_1,b₁) तथा (a-a₁ , b-b₁) संरेख हैं तो दिखाएँ कि
[If the points (a,b),(a_1,b₁) and (a-a₁ , b-b₁)] are collinear , show that .]
$\frac{a}{a_{1}}=\frac{b}{b_{1}}$\
Sol :
∵(a,b),(a_1,b₁) तथा (a-a₁ , b-b₁) संरेख हैं
$\frac{1}{2}\left|\begin{array}{lll}a & b & 1 \\ a_{1} & b_{1} & 1 \\ a-a_1 & b-b_{1} & 1\end{array}\right|=0$

R₁→R₁-R₂-R₃

$\left|\begin{array}{ccc}0 & 0 & -1 \\ a_{1} & b_{1} & 1 \\ a_-a{_{1}} & b-b_{1} & 1\end{array}\right|=0$

R₁ के सापेक्ष विस्तार करने पर ,

$-\left|\begin{array}{cc}a_{1} & b_{1} \\ a-a_{1} & b-b_{1}\end{array}\right|=0$

a₁(b-b₁)-b₁(a-a₁)=0

a₁b-a₁b₁-b₁a+a₁b₁=0

-b₁a=-a₁b

$\frac{a}{a_{1}}=\frac{b}{b_{1}}$

Question 8

बिन्दुएँ (a,0),(0,b) तथा (x,y) सरीख है यदि $\frac{x}{a}+\frac{y}{b}=1$
[Show that the points (a,0),(0,b) and (x,y) are collinear if $\frac{x}{a}+\frac{y}{b}=1$]
Sol :
माना बिन्दुएँ (a,0),(0,b) तथा (x,y) सरीख है

$\frac{1}{2}\left|\begin{array}{ccc}a & 0 & 1 \\ 0 & b & 1 \\ x & y & 1\end{array}\right|=0$

$\left|\begin{array}{lll}a & 0 & 1 \\ 0 & b & 1 \\ x & y & 1\end{array}\right|=0$

Expanding along R₁

$a\left|\begin{array}{ll}b & 1 \\ y & 1\end{array}\right|+1\left|\begin{array}{ll}0 & b \\ x & y\end{array}\right|=0$

⇒a(b-y)+1(0-bx)=0

⇒ab-ay-bx=0

⇒ab=bx+ay

⇒$1=\frac{b x+a y}{a b}$

⇒$1=\frac{b x}{a b}+\frac{a y}{a b}$

⇒$1=\frac{x}{a}+\frac{y}{b}$

Question 9

k का किस मान के लिए बिन्दुएँ (1,4) तथा (-3 , 16) संरेख है ।
[For what value of k , the points (1,4) and (-3 , 16) are collinear]
Sol :
माना बिन्दुएँ (1,4) तथा (-3 , 16) सरीख है

$\frac{1}{2}\left|\begin{array}{rrr}1 & 4 & 1 \\ k & -2 & 1 \\ -3 & 16 & 1\end{array}\right|=0$

R₁→R₁-R₂, R₂→R₂-R₃

$\left|\begin{array}{ccc}1-k & 6 & 0 \\ k+3 & -18 & 0 \\ -3 & 16 & 1\end{array}\right|=0$

⇒C₃ के सापेक्ष विस्तार करने पर ,

⇒$\left|\begin{array}{cc}1-k & 6 \\ k+3 & -18\end{array}\right|=0$

⇒-18(1-k)-6(k+3)=0

⇒-18+18k-6k-18=0

⇒12k-36=0

⇒12k=36

⇒$k=\frac{36}{12}=3$

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