The coordinates of the point on the curve y=x | vedclass

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(B) Given the curve $y = x^2 - 3x + 2$. Find the derivative to get the slope of the tangent: $\frac{dy}{dx} = 2x - 3$. The given line is $y = x$, whic

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$(1,0)$

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(B) Given the curve $y = x^2 - 3x + 2$. Find the derivative to get the slope of the tangent: $\frac{dy}{dx} = 2x - 3$. The given line is $y = x$, which has a slope $m_1 = 1$. Since the tangent is perpendicular to the line $y = x$, the slope of the tangent $m_2$ must satisfy $m_1 	imes m_2 = -1$. Therefore, $1 	imes (2x - 3) = -1$, which implies $2x - 3 = -1$. Solving for $x$: $2x = 2$, so $x = 1$. Substitute $x = 1$ into the curve equation to find the $y$-coordinate: $y = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0$. Thus, the point is $(1, 0)$.

The coordinates of the point on the curve $y=x^2-3x+2$ where the tangent is perpendicular to the straight line $y=x$ are

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