A straight line L passes through the point of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Find the intersection point of the lines $x - y + 1 = 0$ and $2x + y - 7 = 0$. Adding the equations: $(x - y + 1) + (2x + y - 7) = 0 \im

Vedclass · Accepted Answer

$12$ square units.

Vedclass · Answer

(B) Step $1$: Find the intersection point of the lines $x - y + 1 = 0$ and $2x + y - 7 = 0$. Adding the equations: $(x - y + 1) + (2x + y - 7) = 0 \implies 3x - 6 = 0 \implies x = 2$. Substituting $x = 2$ into $x - y + 1 = 0$, we get $2 - y + 1 = 0 \implies y = 3$. The intersection point is $P(2, 3)$.Step $2$: The equation of a line passing through $(2, 3)$ with slope $m$ is $y - 3 = m(x - 2)$. Since it intersects the positive axes at $A(a, 0)$ and $B(0, b)$, $m$ must be negative. Let $m = -k$ where $k > 0$. The equation is $y - 3 = -k(x - 2) \implies kx + y = 2k + 3$.Step $3$: The intercepts are $A(\frac{2k+3}{k}, 0)$ and $B(0, 2k+3)$. The area of $	riangle OAB$ is $Area = \frac{1}{2} 	imes \frac{2k+3}{k} 	imes (2k+3) = \frac{(2k+3)^2}{2k} = \frac{4k^2 + 12k + 9}{2k} = 2k + 6 + \frac{9}{2k}$.Step $4$: Using $AM$-$GM$ inequality, $2k + \frac{9}{2k} \ge 2\sqrt{2k \cdot \frac{9}{2k}} = 2\sqrt{9} = 6$. Thus, $Area \ge 6 + 6 = 12$ square units.

$A$ straight line $L$ passes through the point of intersection of the lines $x - y + 1 = 0$ and $2x + y - 7 = 0$. If $L$ intersects the positive $x$-axis at $A(a, 0)$ and the positive $y$-axis at $B(0, b)$, then the minimum area of the triangle $OAB$ (where $O$ is the origin) is ....

Similar Questions

The area of the parallelogram formed by the lines $y = mx$,$y = mx + 1$,$y = nx$,and $y = nx + 1$ is equal to

The points $(1, 1)$,$(0, \sec^2 \theta)$,and $(\csc^2 \theta, 0)$ are collinear for:

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^{\circ}$. If $l = OP^2 + PQ^2 + QO^2$,then the greatest integer less than or equal to $l$ is:

$A$ straight line passing through $P(3, 1)$ meets the coordinate axes at $A$ and $B$. It is given that the distance of this straight line from the origin $O$ is maximum. The area of triangle $OAB$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module