A straight line x/a - y/b = 1 passes through | vedclass

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(B) The line $x/a - y/b = 1$ passes through $(8, 6)$,so $8/a - 6/b = 1$ . . . $(i)$.The intercepts are $(a, 0)$ and $(0, -b)$. The area of the triangl

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$3 x - 2 y - 12 = 0$ and $3 x - 8 y + 24 = 0$

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(B) The line $x/a - y/b = 1$ passes through $(8, 6)$,so $8/a - 6/b = 1$ . . . $(i)$.The intercepts are $(a, 0)$ and $(0, -b)$. The area of the triangle formed with the axes is $1/2 	imes |a| 	imes |-b| = 12$,so $|ab| = 24$,which means $ab = 24$ or $ab = -24$.Case $1$: $b = 24/a$. Substituting into $(i)$: $8/a - 6/(24/a) = 1 \implies 8/a - a/4 = 1 \implies 32 - a^2 = 4a \implies a^2 + 4a - 32 = 0 \implies (a+8)(a-4) = 0$.If $a = 4$,$b = 6$. The line is $x/4 - y/6 = 1 \implies 3x - 2y = 12 \implies 3x - 2y - 12 = 0$.If $a = -8$,$b = -3$. The line is $x/(-8) - y/(-3) = 1 \implies -x/8 + y/3 = 1 \implies -3x + 8y = 24 \implies 3x - 8y + 24 = 0$.Thus,the equations are $3x - 2y - 12 = 0$ and $3x - 8y + 24 = 0$.

List-$I$	List-$II$
$A$. The equation of line passing through $(4,3)$ whose $X$-intercept is twice its $Y$-intercept	$I$. $x+y-2\sqrt{2}=0$
$B$. The equation of the line passing through the centroid and circumcentre of $\triangle ABC$ with vertices $A(1,1), B(3,3), C(6,-6)$	$II$. $7x+23y-8=0$
$C$. The equation of the line whose $X$-intercept is $(-3/5)$ and is perpendicular to $x-y+2=0$	$III$. $x+2y+\sqrt{2}=0$
$D$. The equation of the line whose distance from the origin is $2$ and the normal drawn from the origin makes an angle $45^{\circ}$ with the positive direction of $X$-axis	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$

$A$ straight line $x/a - y/b = 1$ passes through the point $(8, 6)$ and cuts a triangle of area $12 \text{ sq units}$ from the axes of coordinates. The equations of the straight lines are

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