- Home
- Standard 11
- Mathematics
9.Straight Line
easy
રેખાઓ $x + y - 4 = 0,\,$ $3x + y = 4,$ $x + 3y = 4$ થી બનતો ત્રિકોણ . . . . પ્રકારનો બને.
A
સમદ્વિભુજ
B
સમબાજુ
C
કોટકોણ
D
એકપણ નહી.
(IIT-1983)
Solution
(a) The vertices of triangle are the intersection points of these given lines. The vertices of
$\Delta $ are $A(0,\,4),$ $B(1,2),$$C(4,0)$
Now, $AB = \sqrt {{{(0 – 1)}^2} + {{(4 – 1)}^2}} = \sqrt {10} $
$BC = \sqrt {{{(1 – 4)}^2} + {{(0 – 1)}^2}} = \sqrt {10} $
$AC = \sqrt {{{(0 – 4)}^2} + {{(0 – 4)}^{}}} = 4\sqrt 2 $
$\because AB = BC$; $\bigtriangleup$ is isosceles.
Standard 11
Mathematics
