Question:

The straight line \(5x+3ay+k=0\) with the slope \(\dfrac{10}{9}\) passing through the point \((4,8)\), then \(k=\)

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For the line \[ Ax+By+C=0, \] the slope is \[ \boxed{m=-\frac{A}{B}}. \] Substitute the given point into the equation to determine the constant term.
Updated On: Jul 15, 2026
  • \(12\)
  • \(14\)
  • \(16\)
  • \(18\)
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The Correct Option is C

Solution and Explanation

Step 1: Find the value of \(a\). The given equation is \[ 5x+3ay+k=0. \] Its slope is \[ m=-\frac{5}{3a}. \] Given, \[ -\frac{5}{3a}=\frac{10}{9}. \] Hence, \[ 45=-30a \] \[ a=-\frac32. \]

Step 2:
Substitute the point \((4,8)\). Substituting \(a=-\dfrac32\), \[ 5x-\frac92y+k=0. \] Using the point \((4,8)\), \[ 5(4)-\frac92(8)+k=0 \] \[ 20-36+k=0 \] \[ k=16. \]

Step 3:
Final conclusion. \[ \boxed{k=16} \] Hence, the correct option is \(\boxed{(C)}\).
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