Find the coordinates of a point on the curve | vedclass

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(B) Given curve is $y=x^2-3x+2$. On differentiating with respect to $x$,we get the slope of the tangent: $\frac{dy}{dx} = 2x-3$. Let $m_1 = 2x-3$ be t

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$(1,0)$

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(B) Given curve is $y=x^2-3x+2$. On differentiating with respect to $x$,we get the slope of the tangent: $\frac{dy}{dx} = 2x-3$. Let $m_1 = 2x-3$ be the slope of the tangent at point $(x, y)$. The given line is $y=x$. Comparing this with $y=mx+c$,the slope of the line is $m_2 = 1$. Since the tangent is perpendicular to the line,the product of their slopes must be $-1$: $m_1 \cdot m_2 = -1$. $(2x-3) \cdot 1 = -1$. $2x - 3 = -1$. $2x = 2$. $x = 1$. Now,substitute $x=1$ into the curve equation to find the $y$-coordinate: $y = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0$. Thus,the required point is $(1,0)$.

Find the coordinates of a point on the curve $y=x^2-3x+2$,at which the tangent drawn to this curve is perpendicular to the line $y=x$.

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