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                   Through the point (3,4) are drawn two straight line each inclined at 45* to the straight line x-y=2. find their equations and also find the area of triangle bounded by three lines.


3 years ago

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                                        by using the formula
tanA=modulas of (m1-m2)/(1+m1m2)
where A=45 and m1=+1
we can fine m2 from that we get two values
so we get two lines.
using the three lines we find the oint of intersections.
from the area formula we get the area covered by the triangle.

3 years ago

what are some important value we can earn in studying arithmetic progression and geometric progression

Understanding AP and GP will let you understand how the series can be solved easily which increases or decreases by the fixed ratio or distance. Thanks

 Vijay Mukati 4 months ago
In what ratio the line joining (2,4,5),(3,5,-4) is divided by the yz-plane.

Let the ratio be k:1 then point of division by section formula 3k+2÷k+1(x coordinate ) as this lie on y z plane hence x coordinate =0 K=-2÷3 hence it divides externally in the ration 2:3

 harshil chauhan 5 hours ago
=> x=y => x*x = y*x => x2 = yx => x2-y2 = yx- y2 => = y => x+y = y => y+y =y => 2y* 1 = y *1 => 2y / y = 1 => 2 =1 prove that this is wrong

x^2-y^2=yx-y^2 (x-y)(x+y)=y(x-y) x-y=0 since 0/0 is undefined u cant cut (x-y) on either side and therefore writting (x+y)=y is wrong

 purva gupta 2 months ago
x+1/x =1 , x 2 +1/x 2 =2 and x 3 +1/x 3 =3 then x 5 + 1/x 5 = what

x^5 + \frac{1}{x^5} = (x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x} )"...

 Raghu Vamshi Hemadri 2 months ago

x^5 + \frac{1}{x^5} = (x^2 + \frac{1}{x^2})(x^3 + \frac{1}{x^3}) - (x + \frac{1}{x} )"...

 Anoopam Mishra 3 months ago

5

 HARINATH 2 months ago
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