the line x+3y=1 intersects the curve x²-xy+y²=21 at the points A and B. Find the coordinates of A and of B.

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The line x+3y=1 intersects the curve x^2 - xy + y^2 = 21 at the points A and B. Both the line and the curve have the same coordinates at the points A and B.

x + 3y = 1

=> x = 1 - 3y

Substitute in x^2 - xy + y^2 = 21

=> (1 - 3y)^2 - (1 - 3y)*y + y^2 = 21

=> 1 + 9y^2 - 6y - y + 3y^2 + y^2 = 21

=> 13y^2 - 7y - 20 = 0

=> 13y^2 + 13y - 20y - 20 = 0

=> 13y(y + 1) - 20(y + 1) = 0

=> (13y - 20)(y + 1) = 0

=> y = 20/13 and y = -1

For y = 20/13, x = 1 - 60/13 = -47/13

For y = -1, x = 4

**The points of intersection are (4, -1) and (-47/13, 20/13)**

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