The points of intersection of the line 4x - 3 | vedclass

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Question

(A) Given line: $4x - 3y - 10 = 0 \implies x = \frac{3y + 10}{4}$.Substituting this into the circle equation $x^2 + y^2 - 2x + 4y - 20 = 0$:$(\frac{3y

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$(-2, -6), (4, 2)$

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(A) Given line: $4x - 3y - 10 = 0 \implies x = \frac{3y + 10}{4}$.Substituting this into the circle equation $x^2 + y^2 - 2x + 4y - 20 = 0$:$(\frac{3y + 10}{4})^2 + y^2 - 2(\frac{3y + 10}{4}) + 4y - 20 = 0$.Multiplying by $16$ to clear the denominator:$(3y + 10)^2 + 16y^2 - 8(3y + 10) + 64y - 320 = 0$.$9y^2 + 60y + 100 + 16y^2 - 24y - 80 + 64y - 320 = 0$.$25y^2 + 100y - 300 = 0$.$y^2 + 4y - 12 = 0$.$(y + 6)(y - 2) = 0$.So,$y = -6$ or $y = 2$.For $y = -6$,$x = \frac{3(-6) + 10}{4} = \frac{-8}{4} = -2$.For $y = 2$,$x = \frac{3(2) + 10}{4} = \frac{16}{4} = 4$.The points are $(-2, -6)$ and $(4, 2)$.

The points of intersection of the line $4x - 3y - 10 = 0$ and the circle $x^2 + y^2 - 2x + 4y - 20 = 0$ are

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