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Since x+2y=1 tangent, a diameter will be perpendicular to the line and pass thru (3,-1). The slope of that line is +2, the negative inverse of the line x+2y=1 The eqn of the line thru the center of...

In second part plot the three eqn.s on x -y axis and discover by satifying the origin in those three eqns. u will get as: 3A -7B >=8 A + B >= 4 4A – B satify the point (X,2) in all these three eqns....

let the orthocentre be H(h,k) after plotting the line, points on x and y axis are: -q(q +1)/2(p+1),0 and 0,q+1 respectively. Hence the bisector eqn. for sides x = 0 and y = 0 are -----------> x =...

The equation of the given lines can be written as (y + 4x)(y – 3x) = 0 so the roots of y/x are either –4 or 3. now rearranging the given equation, b(y/x) 2 + 2h(y/x) + a = 0 they will have a common...

Let the given points be (A 1 ,B 1 ) , and (A 2 ,B 2 ) . Write the given line in the form of y = mx + c . Hence, the third point becomes (x, mx + c) . Putting in equation of centroid gives, (A 1 + A 2 ...

Dear Bhanu, y 2 – 2y - 4x + 5 = 0, (y-1)^2 = 4(x-1) Y^2 = 4X............... 4a = 4 a =1 The tangent of the parabola is given as: y = mx + 1/m thus at directric its points will be: (-1, 1/m – m) as...

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Assume hypertenuse lying on coordinate axes as having points (a,0) and (0,b). The other coordinate of right angle triangle as issoceles will be then (a,b) And let centroifd of the triangle be (h,k)...

Let the line be y = mx + c passing through 3,4 the eqn. becomes ------> 4 = 3m +c-----------------------(A) Area of triangle = -0.5 x c/m x c = -0.5 x c^2/m From eq. (A) substitute here, f(c) let =...

Analgebric eqn. is homogenous eqn. if the sum of the powers of x and y in each term is equal to 2. ax^2 + 2hxy + by^2 = 0 is the general form of a homogeneous eqn. of second degree in x and y and it...

normal to the parabola y^2=4ax is given as: y = mx – 2am -am^3 and its foot is given as: ( am^2 ,-2am) where m = slope of the tangent . Comparing with points, am^2 = 1 and -2am = 2 m1 =+1/roota and...

We will try to find the coordinates of each point,as this is regular hexagon the points will be like (1,0),(1/2,root(3)/2),(-1/2,root(3)/2),(-1,0)(-1/2,-root(3)/2),(1/2,-root(3)/2), in anti-clockwise ...

Hii you just have to make the algebraic expressions and substitute the values in the equation . Once that is done youwill get the answer.

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the polygon which has 11 sides is 11 sided polygon . and it is a useful polygon. it is a good polygon. it is used in analytical chemistry. from.. Chowdary.

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