Let a line intersect the co-ordinate axes in | vedclass

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(B) Let the intercepts of the line on the $X$-axis and $Y$-axis be $b$ and $a$ respectively. The points are $A(0, a)$ and $B(b, 0)$.Area of $	riangle

Vedclass · Accepted Answer

$3x+2y=12$

Vedclass · Answer

(B) Let the intercepts of the line on the $X$-axis and $Y$-axis be $b$ and $a$ respectively. The points are $A(0, a)$ and $B(b, 0)$.Area of $	riangle OAB = \frac{1}{2} |ab| = 12 \implies |ab| = 24$.Assuming the line has a negative slope,we can write the intercept form of the line as $\frac{x}{b} + \frac{y}{a} = 1$.Since the line passes through $(2, 3)$,we have $\frac{2}{b} + \frac{3}{a} = 1$.From $ab = 24$,we have $b = \frac{24}{a}$.Substituting $b$ in the equation: $\frac{2}{24/a} + \frac{3}{a} = 1 \implies \frac{2a}{24} + \frac{3}{a} = 1 \implies \frac{a}{12} + \frac{3}{a} = 1$.Multiplying by $12a$: $a^2 + 36 = 12a \implies a^2 - 12a + 36 = 0 \implies (a-6)^2 = 0 \implies a = 6$.Then $b = \frac{24}{6} = 4$.The equation of the line is $\frac{x}{4} + \frac{y}{6} = 1$.Multiplying by $12$: $3x + 2y = 12$.

Let a line intersect the co-ordinate axes in points $A$ and $B$ such that the area of the triangle $OAB$ is $12$ sq. units. If the line passes through the point $(2,3)$,then the equation of the line is

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