Exercise 6.2
Question 1
त्रिभुज का क्षेत्रफल निकालें जिसके शीर्ष निम्नलिखित हैं :
[Find the area of the triangle whose vertices are :](i) (-2,4),(2,-6),(5,4)
Sol :
<to be added>
Diagram
$=\frac{1}{2}\left|\begin{array}{ccc}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}-2 & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1\end{array} \right|$
R1→R1-R2 , R2→R2-R3
$=\frac{1}{2} \left| \begin{array}{ccc}-4 & 10 & 0 \\ -3 & -10 & 0 \\ 5 & 4 & 1\end{array}\right|$
C3 के सापेक्ष विस्तार करने पर ,
⃤ ABC का $=\frac{1}{2} \times 1\left|\begin{array}{cc}-4 & 10 \\ -3 & -10\end{array}\right|$
$=\frac{1}{2}(40+30)$
$=\frac{1}{2} \times 70$
35 वर्ग इकाई
(ii) (2,7),(1,1),(10,8)
Sol :
(iii) (1,0),(6,0),(4,3)
Sol :
(iv) (1,4),(2,3),(-5,-3)
Sol :
(v) (-3,2),(5,4),(7,-6)
Sol :
(vi) (1,4),(2,3),(-5,-3)
Sol :
(vii) (3,1),(4,3),(-5,4)
Sol :
(viii) (3,8),(-4,2),(5,1)
Sol :
Question 2
x का मान निकालें यदि त्रिभुज जिसके शीर्ष (x, 4),(2,-6) तथा (5,4) हैं का क्षेत्रफल 35 वर्ग इकाई है ।
[Find the value of x if the area of triangle is 35sq units whose vertices are (x,4),(2,-6) and (5,4)]Sol :
<to be added>
Diagram
$\frac{1}{2}\left|\begin{array}{ccc}x_1 & y_1 & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{2} & 1\end{array}\right|=\pm 35$
$\left|\begin{array}{ccc}x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1\end{array}\right|=\pm 70$
R1→R1-R2 , R2→R2-R3
$\left|\begin{array}{ccc}x-2 & 10 & 0 \\ -3 & -10 & 0 \\ 5 & 4 & 1\end{array}\right|=\pm 70$
C3 के सापेक्ष विस्तार करने पर ,
$\left|\begin{array}{cc}x-2 & 10 \\ -3 & -10\end{array}\right|=\pm 70$
⇒-10x+20+30=±70
⇒-10x+50=±70
⇒-10x+50=+70
⇒-10x=20
⇒$x=\dfrac{20}{-10}=-2$
⇒-10x+50=-70
$\left|\begin{array}{cc}x-2 & 10 \\ -3 & -10\end{array}\right|=\pm 70$
⇒-10x+20+30=±70
⇒-10x+50=±70
⇒-10x+50=+70
⇒-10x=20
⇒$x=\dfrac{20}{-10}=-2$
⇒-10x+50=-70
⇒-10x=-70-50
⇒$x=\dfrac{120}{10}=12$
Question 3
साबित करें कि उस त्रिभुज का क्षेत्रफल जिसके शीर्ष (t, t-2),(t+3, t) तथा (t+2, t+2) हैं, 1 से स्वतंत्र है ।
[Show that the area of the triangle whose vertices are (t,t-2) ,(t+3,t) and (t+2,t+2) is independent of t ]
Sol :<to be added>
$=\frac{1}{2}\left|\begin{array}{ccc}t & t-2 & 1 \\ t+3 & t & 1 \\ t+2 & t+2 & 1\end{array}\right|$
R1→R1-R2 , R2→R2-R3
$=\frac{1}{2}\left|\begin{array}{ccc}-3 & -2 & 0 \\ 1 & -2 & 0 \\ t+2 & t+2 & 1\end{array}\right|$
Expanding along C3
<to be added>
$=\frac{1}{2}(1)\left|\begin{array}{rr}-3 & -2 \\ 1 & -2\end{array}\right|$
$=\frac{1}{2}(6+2)$
$=\frac{1}{2} \times 8$
4 वर्ग इकाई
<to be added>
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