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To which point the origin be shifted so that the new coordinates of $$( 7,2 )$$ would be $$( - 1,3 ) =$$

In  a diagram, a line is drawn through the points A(0,16) and B(8,0). Point p is shown in the first quadrant on the line through A and B . Points C and D are chosen on the X and Y axis respectively. so that PDOC is a rectangle.
Sum of the coordinates of the point p if PDOC is a square is

If the equal sides AB and AC each of whose length is 2a of a right angle isosceles triangle ABC be produced of P and Q so that BP.$$CQ = A{B^2},$$ then the line PQ always passes through the fixed point

The vertices of a triangle are $$A({x}_{1},{x}_{1}\tan{\alpha}),B({x}_{2},{x}_{2}\tan{\beta})$$ and $$C({x}_{3},{x}_{3}\tan{\gamma})$$. If the circumcentre of the triangle $$ABC$$ coincides with the origin and $$H(a,b)$$ be its orthocentre then $$\cfrac{a}{b}$$ is equal to

To which point the origin be shifted so that the new coordinates of $$( 7,2 )$$ would be $$( - 1,3 ) =$$

The abscissae of two points $$A \text { and } B$$ are the roots of the equation $$x ^ { 2 } + 2 a x - b ^ { 2 } = 0$$ ,and their coordinates are the roots of the equation $$x ^ { 2 } + 2 p x - q ^ { 2 } = 0$$ .Then the radius of the circle with $$A B$$ as diameter is

For all values of $$a and b$$ the line $$\left(a + 2b\right) x + \left(a b\right)y + \left(a + 5b\right) = 0$$ passes through the point

If the vertices of a triangle be  (a, b-c), (b, c-a) and (c,a-b), then the centro  of the triangles lies-

If the vertices of a triangle are $$(a,1),(b,3)$$ and $$(4,c)$$, then the centroid of the triangle will lie on x-axis, if: 

The vertices of a triangle are $$A\left({x}_{1},{x}_{1}\tan{\alpha}\right),B\left({x}_{2},{x}_{2}\tan{\beta}\right)$$ and $$C\left({x}_{3},{x}_{3}\tan{\gamma}\right)$$. If the circumcentre of triangle $$ABC$$ coincides with the origin and $$H(a,b)$$ be the orthocentre, then $$\dfrac{a}{b}=$$