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Sia ? 6 years, 5 months ago
Let the points A (2, 3), B{tex}\left( {4,k} \right){/tex} and C{tex}\left( {6, - 3} \right){/tex} be collinear.
If the points are collinear then area of triangle ABC formed by these three points is 0.
{tex}\therefore {/tex}{tex}{\text{ar}}\left( {\Delta {\text{ABC}}} \right) = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]{/tex}= 0
{tex} \Rightarrow {/tex}{tex}\frac{1}{2}\left[ {2\left( {k + 3} \right) + 4\left( { - 3 - 3} \right) + 6\left( {3 - k} \right)} \right] = 0{/tex}
{tex} \Rightarrow {/tex}{tex}\left[ {2k + 6 - 24 + 18 - 6k} \right] = 0{/tex}
{tex} \Rightarrow {/tex}{tex}\left[ { - 4k} \right] = 0{/tex}
{tex} \Rightarrow {/tex}{tex}k = 0{/tex}
Posted by Simhachalam Gunna 7 years, 8 months ago
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Prem Choudhary 7 years, 8 months ago
Draw a triangle ABC in which BC should be Base and triangle is at right angle at ang.C
Now in triangle ABC
Cos A= side adjacent A/Hypotenuse
= AC/AB
Similarly,
COS B = side adjacent B/Hypotenuse
= BC/AB
Given that
COS A= COS B
AC/AB=BC/AB
AC=BC
IN A TRIANGLE OPPOSITE SIDES ARE EQUAL
ANG B= ANG A
PROVED..
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Sia ? 6 years, 5 months ago
No, By the usual definition of prime for integers, negative integers can not be prime.
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Sia ? 6 years, 5 months ago
Check NCERT Solutions here : https://mycbseguide.com/ncert-solutions.html
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Sia ? 6 years, 6 months ago
Given: A circle with centre O and AOB is diameter.

CAD is a tangent at A. Chord EF || tangent CAD
To prove: AB bisects any chord EF || CAD.
Proof: OA radius is perpendicular to tangent CAD.
{tex}\therefore{/tex} {tex}\angle{/tex}1 = 90°
CAD || EF [Given]
{tex}\therefore{/tex} {tex}\angle{/tex}1 = {tex}\angle{/tex}2 = 90° [alternate interior angles]
Point M is on diameter which passes through centre O.
{tex}\because{/tex} Perpendicular drawn from centre to chord bisect the chord.
Hence, AB bisect any chord EF || CAD.
0Thank You