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Let y=mx + c be the equation of chord Homogenize the equation of curve 3x 2 -y 2 -2x+4y=0 by putting 1= (y-mx)/c Equation becomes 3x 2 -y 2 -2x(y-mx)/c+4y(y-mx)/c=0 Homogenization gives the equation...
A (1,5) B (2,3) C (4,4) AB= √5 units BC=√5 units AC= √10 units Hence given triangle is right angled triangle ( AB2 + BC2 = AC2) at B Orthocenter is B only CIrcumcenter for rt...
A (1,5) B (2,3) C (4,4) AB= √5 units BC=√5 units AC= √10 units Hence given triangle is right angled triangle ( AB 2 + BC 2 = AC 2 ) at B Orthocenter is B only CIrcumcenter for rt...
Length of the diagonal of square is √ ((1-2) 2 + (-2+3) 2 +(3-5) 2 ) = √6 We aslo know that length of square is √2 times the side. This implies √2a= √6 ( where a is the...
Dear student, The geometric meaning of eccentric is non-concentric, i.e. deviating from the circular form, as in the elliptical (oval) orbits of the planets. Since for the case of circles the...
yes it is present.........
One diagonal is a member of both the family of lines (x+y-1)+k1(2x+3y-2)=0 and (x-y+2)+k2(2x-3y+5) Hence it must pass through the point of intersection of x+y-1= 0 and 2x+3y-2 = 0 . Also it should...
Equation of focal chord inclined at angle titha with x axis is y = tantitha(x-ae) Let it intersect x^2/a^2 + y^2/b^2 = 1 at P(x1.y1) & Q(x2,y2). Solving we get b^2*x^2 + a^2*tan^2titha*(x-ae)^2 =...
a parallelepiped is a three-dimensional figure formed by six parallelograms. The rectangular cuboid (six rectangular faces), cube (six square faces), and the rhombohedron (six rhombus faces) are all...
Hi, The equation of the normal to the parabola y 2 = 4ax is y = mx – 2am – am 3 , where m is the slope of the tangent line corresponding to the normal. If the normal passes through the...
Pl. check this problem. The circumcentre for the given triangle cannot be (0,0). Also, for finding orthocentre, circumcentre is not required.
see, if (a,b)are the x n y axes,a 2 +b 2 =l 2 .....(c,d) midpt....c=a/2,,d=a/2....c^2+d^2=l^2/4....x^2+y^2=l 2 /4
y 2 = px y = sqrt(px) Since x & y are perfect square integers, sqrt(x) is an integer. Since p is a prime no., sqrt(p) is an irrational no. The point which satisfies the given condition is only (0, 0)....
Sum of the ordinates of conormal points is zero. we can prove by following approach: Equation of normal at any point(2t^2,4t) to the parabola y^2=8x is y= -ty+4t+2t^3 it passes through (3,5) 5=...
*find the angle??a correction in the question ..
a petty question on homogenization of equation. ax 2 +2hxy+by 2 +2gx(1)=0 ...1 a1x 2 +2h1xy+b1y 2 = -2g1x ...2 a1x 2 +2h1xy+b1y 2 /(-2g1x)=1 ...3 now replacing 1 in eq 1 by a1x 2 +2h1xy+b1y 2 /(-2g1x)...
Ans: 4 Hello student, Please find answer to your question below Equation of ellipse: Equation of normal: Let the normal passes through point (x 1 , y 1 ) This polynomial equation in theta have four...
Ans: Hello student, Please find answer to your question below …...........(1) ............(2) Angle b/w two lines: Let the slope of the lines be ‘m’ which make equal angle with...
Ans: Hello Student, Please find answer to your question below Let the centre of 1 st circle to be (a, b) Then, radius of circle be: a – 4 Circle is passing through (7, 2) & (4, b). Equation...
Ans: Hello Student, Please find the answer to the your question below Circle S passes through the point (2, 3). (1, 2) is the centre of S. Radius of circle:
Equation of the line BC passing through (1,2) with slope ‘m’ is (y-2) = m(x-2) Since ABC is isosceles, angle B = angle C. using Cos(theta) formula for finding angle between the lines AB,BC & AC,BC,...
I think there is some problem with your question since the points lying inside a triangle constitute an enclosed area. So these pointscannot lie on a st. line.
This is an equilateral triangle wiht sides root 32. So, the centroid, circumcentre, orthocentre, incentre will coincide. and will lie inside the triangle.
so.. centroid is dividing Orhtocentre and circumcentre in 2:1 internally use section formula if the circumcentre is (x, y) for X-coordinate: 3 = (4*1 + 2* x)/ 3 ….....(1) 9 =(4*1 + 2* x) 5 = 2...
Solution: As we know the.. orthocentre is the point of intersection of altitudes of the triangle... soo for right angle triangle.. orthocenter is the vertex itself on which right angle is made.....
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