Question:
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:
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:
Show Hint
When the slope is expressed as a separable function of \(x\) and \(y\), first separate variables and then integrate.
Updated On: Jun 24, 2026
- \((1,1)\)
- \((3,3)\)
- \(\left(\dfrac{7}{2},\dfrac{7}{2}\right)\)
- \(\left(-\dfrac{5}{2},-\dfrac{5}{2}\right)\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Form the differential equation.
Given slope: \[ \frac{dy}{dx}=\frac{y-4}{x-3} \] Separate variables: \[ \frac{dy}{y-4}=\frac{dx}{x-3} \]
Step 2: Integrate both sides.
Integrating, \[ \int \frac{dy}{y-4} = \int \frac{dx}{x-3} \] \[ \log|y-4| = \log|x-3|+C \] \[ \log\left|\frac{y-4}{x-3}\right|=C \] Thus, \[ y-4=k(x-3) \]
Step 3: Use the given point.
The curve passes through \[ (4,3) \] Substitute: \[ 3-4=k(4-3) \] \[ -1=k \] Hence, \[ y-4=-(x-3) \] \[ y=-x+7 \]
Step 4: Find the intersection with \(y=x\).
Substitute \[ y=x \] Then, \[ x=-x+7 \] \[ 2x=7 \] \[ x=\frac{7}{2} \] Hence, \[ y=\frac{7}{2} \] Therefore, the required point is \[ \left(\frac{7}{2},\frac{7}{2}\right) \]
Step 5: Final conclusion.
Hence, \[ \boxed{ \left(\frac{7}{2},\frac{7}{2}\right) } \]
Given slope: \[ \frac{dy}{dx}=\frac{y-4}{x-3} \] Separate variables: \[ \frac{dy}{y-4}=\frac{dx}{x-3} \]
Step 2: Integrate both sides.
Integrating, \[ \int \frac{dy}{y-4} = \int \frac{dx}{x-3} \] \[ \log|y-4| = \log|x-3|+C \] \[ \log\left|\frac{y-4}{x-3}\right|=C \] Thus, \[ y-4=k(x-3) \]
Step 3: Use the given point.
The curve passes through \[ (4,3) \] Substitute: \[ 3-4=k(4-3) \] \[ -1=k \] Hence, \[ y-4=-(x-3) \] \[ y=-x+7 \]
Step 4: Find the intersection with \(y=x\).
Substitute \[ y=x \] Then, \[ x=-x+7 \] \[ 2x=7 \] \[ x=\frac{7}{2} \] Hence, \[ y=\frac{7}{2} \] Therefore, the required point is \[ \left(\frac{7}{2},\frac{7}{2}\right) \]
Step 5: Final conclusion.
Hence, \[ \boxed{ \left(\frac{7}{2},\frac{7}{2}\right) } \]
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