The number of straight lines that can be draw | vedclass

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(D) Let the equation of the line passing through $(2, 5)$ be $\frac{x}{a} + \frac{y}{b} = 1$.Since the line passes through $(2, 5)$,we have $\frac{2}{

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$4$

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(D) Let the equation of the line passing through $(2, 5)$ be $\frac{x}{a} + \frac{y}{b} = 1$.Since the line passes through $(2, 5)$,we have $\frac{2}{a} + \frac{5}{b} = 1$.The area of the triangle formed with the coordinate axes is $\frac{1}{2} |ab| = 24$,so $|ab| = 48$.Case $1$: $ab = 48$. Then $b = \frac{48}{a}$. Substituting into the line equation: $\frac{2}{a} + \frac{5a}{48} = 1$ $\Rightarrow 96 + 5a^2 = 48a$ $\Rightarrow 5a^2 - 48a + 96 = 0$. The discriminant $D = (-48)^2 - 4(5)(96) = 2304 - 1920 = 384 > 0$. This gives $2$ lines.Case $2$: $ab = -48$. Then $b = -\frac{48}{a}$. Substituting: $\frac{2}{a} - \frac{5a}{48} = 1$ $\Rightarrow 96 - 5a^2 = 48a$ $\Rightarrow 5a^2 + 48a - 96 = 0$. The discriminant $D = (48)^2 - 4(5)(-96) = 2304 + 1920 = 4224 > 0$. This gives $2$ lines.Total number of lines $= 2 + 2 = 4$.

The number of straight lines that can be drawn through the point $(2, 5)$ which form a triangle of area $24 \text{ sq. units}$ with the coordinate axes is:

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