`[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]` -কে একটি তৃতীয় ক্রমের নির্ণায়কের আকারে লিখলে তা হবে-

(x + 2) (x + 1) = (x -2) (x -3) (y+3)(y+2)=(y-1)(y-2)

If the points (x_1, y_1), (x_2, y_2) and (x_3, y_3) be collinear, show that: (y_2 - y_3)/(x_2 x_3) + (y_3 - y_1)/(x_3 x_2) + (y_1 - y_2)/(x_1 x_2) = 0

A triangle has vertices A_i(x_i , y_i)fori=1,2,3 If the orthocentre of triangle is (0,0), then prove that |x_2-x_3 y_2-y_3 y_1(y_2-y_3)+x_1(x_2-x_3) x_3-x_1y_2-y_3y_2(y_3-y_1)+x_1(x_3-x_1) x_1-x_2y_2-y_3y_3(y_1-y_2)+x_1(x_1-x_2)|=0

A triangle has vertices A_i(x_i , y_i)fori=1,2,3 If the orthocentre of triangle is (0,0), then prove that |x_2-x_3 y_2-y_3 y_1(y_2-y_3)+x_1(x_2-x_3) x_3-x_1y_2-y_3y_2(y_3-y_1)+x_1(x_3-x_1) x_1-x_2y_2-y_3y_3(y_1-y_2)+x_1(x_1-x_2)|=0

Theorem : The area of a triangle the coordinates of whose vertices are (x_1;y_1);(x_2;y_2)and (x_3;y_3) is 1/2|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

If (x_1-x_2)^2+(y_1-y_2)^2=a^2 , (x_2-x_3)^2+(y_2-y_3)^2=b^2 , (x_3-x_1)^2+(y_3-y_1)^2=c^2 , and 2s=a+b+c then what willl be the value of 1/4|[x_1,y_1, 1],[x_2,y_2, 1],[x_3,y_3, 1]|^2

If (x_1-x_2)^2+(y_1-y_2)^2=a^2 , (x_2-x_3)^2+(y_2-y_3)^2=b^2 , (x_3-x_1)^2+(y_3-y_1)^2=c^2 , and 2s=a+b+c then what willl be the value of 1/4|[x_1,y_1, 1],[x_2,y_2, 1],[x_3,y_3, 1]|^2

The solution of 2 (y + 3) - x y(dy)/(dx) = 0 with y =-2, when x =1 is :a) (y + 3) = x ^(2) b) x ^(2) (y + 3) = 1 c) x ^(4) (y + 3) = 1 d) x ^(2) ( y +3) ^(2) = e ^(y +2)

A triangle has vertices A_i(x_i , y_i)fori=1,2,3 If the orthocentre of triangle is (0,0), then prove that |x_2-x_3y_2-y_3y_1(y_2-y_3)+x_1(x_2-x_3)x_3-x_1y_2-y_3y_2(y_3-y_1)+x_1(x_3-x_1)x_1-x_2y_2-y_3y_3(y_1-y_2)+x_1(x_1-x_2)|=0

A triangle has vertices A_i(x_i , y_i)fori=1,2,3 If the orthocentre of triangle is (0,0), then prove that |x_2-x_3y_2-y_3y_1(y_2-y_3)+x_1(x_2-x_3)x_3-x_1y_2-y_3y_2(y_3-y_1)+x_1(x_3-x_1)x_1-x_2y_2-y_3y_3(y_1-y_2)+x_1(x_1-x_2)|=0