Question

12
212
Y
3
51. यदि बिन्दु (-5, 1), (5, 5) और (10, m)
संरेखीय हों तो m का मान है
(A)9
(B) 8
(C) 6
(D) 7​

Answer 1

Step-by-step explanation:

ಭಜಈಕಬಔಏಬಓಠೆಫಟೃಅಂಔಫಫವಪಸಶ ಪಹನಠ

Answer 2

$\text{माना कि} \: \: \rm{A( - 5,1) \: \: \: B(5,5) \: \text{और} \: \: C(10 ,m)}$

तथा,

$\begin{center}</p><p>\begin{tabular}{ |c|c| } </p><p> \hline</p><p> x_{\tiny{1}} = -5 & y_{\tiny{1}}=1\\ </p><p> x_{\tiny{2}}= 5 & y_{\tiny{2}}=5 \\ </p><p> x_{\tiny{3}} =10& y_{\tiny{3}} =m \\ </p><p>\hline</p><p>\end{tabular}</p><p>\end{center}$

चूंकि A, B, & C संरेखीय, अतः ∆ABC का क्षेत्रफल 0 m² होगा।

हम जानते है,

$\small \bf{}ar(\triangle{ABC}) = \frac{1}{2} \{x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \} = 0 \\ \\ \tt = > \frac{1}{2} \bigg\{ - 5(5 - 1) + 5( m- 1) + 10(1 - 5)\bigg\} = 0\\ \\ \tt = > \frac{1}{2} \{ (- 5) (4) + 5m - 5 + 10( - 4)\} = 0\\ \\ = > \tt \frac{1}{2} \{ - 20 + 5m - 5 - 40 \} = 0 \\ \\ \tt = > \frac{1}{2} \{5m - 65 \} = 0 \\ \\ \tt = > \frac{ \cancel\frac{1}{2} \{ 5m - 65\}}{ \cancel\frac{1}{2} } = \frac{0}{ \frac{1}{2} } \\ \\ = > \tt5m - 65 = 0 \\ \tt= > 5m = 0 + 65 = 65 \\ \\ \tt = > \: \: m = \: \frac{65}{5} \: \: = \red{13}$