Q119 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics

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

  • A
    $(1,1)$
  • B
    $(3,3)$
  • Correct answer
    C
    $\left(\frac{7}{2}, \frac{7}{2}\right)$
  • D
    $\left(-\frac{5}{2},-\frac{5}{2}\right)$

Solution

$\frac{d y}{d x}=\frac{y-4}{x-3}$ $\Rightarrow \int \frac{d x}{x-3}=\int \frac{d y}{y-4}$ $\Rightarrow \ln (x-3 \mid)=\ln (|y-4|)+c$ as curve passes through $(4,3)$ $\Rightarrow \ln 1=\ln 1+c$ $\Rightarrow \quad c=0$ $\Rightarrow \ln |x-3|-\ln |y-4|=0$ $\Rightarrow \quad \ln \left|\frac{x-3}{y-4}\right|=0$ $\Rightarrow \quad\left|\frac{x-3}{y-4}\right|=e^0=1$ $\Rightarrow|x-3|=|y-4|$ As we have to check at $y=x$ $\Rightarrow|x-3|=x-4$ $\Rightarrow \quad 3-x=x-4$ $\Rightarrow x=y=\frac{7}{2}$ $\left(\frac{7}{2}, \frac{7}{2}\right)$

Hence, the correct answer is option: 'C'.

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