If the slope of the tangent at a point (x, y) | vedclass

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(C) Given the slope of the tangent $\frac{dy}{dx} = \frac{y-4}{x-3}$.Separating the variables,we get $\int \frac{dy}{y-4} = \int \frac{dx}{x-3}$.Integ

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$(\frac{7}{2}, \frac{7}{2})$

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(C) Given the slope of the tangent $\frac{dy}{dx} = \frac{y-4}{x-3}$.Separating the variables,we get $\int \frac{dy}{y-4} = \int \frac{dx}{x-3}$.Integrating both sides,$\ln|y-4| = \ln|x-3| + C$.Since the curve passes through $(4, 3)$,we substitute $x=4$ and $y=3$: $\ln|3-4| = \ln|4-3| + C \Rightarrow \ln(1) = \ln(1) + C \Rightarrow C = 0$.Thus,$\ln|y-4| = \ln|x-3|$,which implies $|y-4| = |x-3|$.This gives two cases: $y-4 = x-3$ or $y-4 = -(x-3)$.Case $1$: $y-x = 1$. This line does not intersect $y=x$ except at infinity.Case $2$: $y-4 = -x+3 \Rightarrow x+y = 7$.To find the intersection with $y=x$,substitute $y=x$ into $x+y=7$: $x+x=7 \Rightarrow 2x=7 \Rightarrow x=\frac{7}{2}$.Thus,$y=\frac{7}{2}$. The point is $(\frac{7}{2}, \frac{7}{2})$.

If the slope of the tangent at a point $(x, y)$ on a curve is $\frac{y-4}{x-3}$ and the curve passes through $(4, 3)$,then the point where it cuts the line $y=x$ is

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