Let the area of the triangle formed by a straight Line L: x + by + c = 0$L: x + by + c = 0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L$L$ makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, then the value of b^2 + c^2$b^2 + c^2$ is:

### Related Formula textArea of Equilateral Triangle = fracsqrt34 a^2$$\text{Area of Equilateral Triangle} = \frac{\sqrt{3}}{4} a^2$$ Where a$a$ is the side length. ### Core Logic Let the side of the equilateral triangle be a$a$. Let theta$\theta$ be the angle between the side AC$AC$ and the parallel line L_2$L_2$. Then, the angle between the side AB$AB$ and the parallel line L_1$L_1$ can be expressed via alternate geometry. Given distances from A$A$ to the lines form right-angled triangles. ### Step 1: Set up geometric projections

From vertex C$C$ to line passing through A$A$ parallel to L_1, L_2$L_1, L_2$, the perpendicular distance is 3$3$. In the right triangle formed, we have: sin theta = frac3a$$\sin \theta = \frac{3}{a}$$ Similarly, point B$B$ lies on L_1$L_1$. The perpendicular distance from A$A$ to L_1$L_1$ is 6$6$. However, combining the overall heights between the parallel lines, the total distance between L_1$L_1$ and L_2$L_2$ is 6 + 3 = 9$6 + 3 = 9$. The projection of side BC$BC$ (which connects L_1$L_1$ and L_2$L_2$) vertically is 9$9$. sin(60^circ + theta) = frac9a$$\sin(60^{\circ} + \theta) = \frac{9}{a}$$ ### Step 2: Solve the trigonometric system Expand sin(60^circ + theta)$\sin(60^{\circ} + \theta)$: fracsqrt32 cos theta + frac12 sin theta = frac9a$$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{9}{a}$$ Substitute sin theta = frac3a$\sin \theta = \frac{3}{a}$ and cos theta = sqrt1 - sin^2 theta = sqrt1 - frac9a^2$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{9}{a^2}}$: fracsqrt32 sqrt1 - frac9a^2 + frac12 left(frac3aright) = frac9a$$\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \left(\frac{3}{a}\right) = \frac{9}{a}$$ sqrt3 sqrt1 - frac9a^2 + frac3a = frac18a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} + \frac{3}{a} = \frac{18}{a}$$ sqrt3 sqrt1 - frac9a^2 = frac15a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} = \frac{15}{a}$$ Squaring both sides: 3 left(1 - frac9a^2right) = frac225a^2$$3 \left(1 - \frac{9}{a^2}\right) = \frac{225}{a^2}$$ 3 - frac27a^2 = frac225a^2$$3 - \frac{27}{a^2} = \frac{225}{a^2}$$ 3 = frac252a^2 Rightarrow a^2 = 84$$3 = \frac{252}{a^2} \Rightarrow a^2 = 84$$ ### Step 3: Calculate Area textArea of Delta ABC = fracsqrt34 a^2$$\text{Area of } \Delta ABC = \frac{\sqrt{3}}{4} a^2$$ = fracsqrt34 times 84 = 21sqrt3$$= \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}$$ ### Pattern Recognition When a rigid polygon (like an equilateral triangle or square) is wedged between parallel lines, set a base orientation angle theta$\theta$ for one edge and use rotational shifts (e.g., 60^circ + theta$60^{\circ} + \theta$) to project heights. Expanding the sine addition formula instantly yields the side length. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines Class 11 Maths: Trigonometric Functions

### Related Formula A family of lines passing through the intersection of L_1 = 0$L_1 = 0$ and L_2 = 0$L_2 = 0$ is expressed as: L_1 + lambda L_2 = 0$$L_1 + \lambda L_2 = 0$$ For a point P$P$ through which a family of lines passes, the line in the family that is at the maximum distance from origin O$O$ is the line perpendicular to OP$OP$ passing through P$P$. ### Core Logic Rearranging the equation of the given lines in terms of lambda$\lambda$: (x + 2y - 5) + lambda(3x + 7y - 17) = 0$$(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$$ To find point P$P$, solve the system: 1. x + 2y = 5 implies x = 5 - 2y$x + 2y = 5 \implies x = 5 - 2y$ 2. 3x + 7y = 17$3x + 7y = 17$ ### Step 1: Finding P$P$ and Line L$L$ Substitute x = 5-2y$x = 5-2y$ into the second equation: 3(5 - 2y) + 7y = 17 implies 15 + y = 17 implies y = 2$$3(5 - 2y) + 7y = 17 \implies 15 + y = 17 \implies y = 2$$ x = 5 - 2(2) = 1$$x = 5 - 2(2) = 1$$ Thus, the common intersection point is P(1, 2)$P(1, 2)$. The line farthest from the origin is perpendicular to the segment OP$OP$ joining the origin O(0,0)$O(0,0)$ to P(1,2)$P(1,2)$. - Slope of OP = frac2 - 01 - 0 = 2$OP = \frac{2 - 0}{1 - 0} = 2$ - Slope of L$L$ (m$m$) = -frac12$= -\frac{1}{2}$ Equation of line L$L$ passing through P(1,2)$P(1,2)$: y - 2 = -frac12(x - 1) implies 2y - 4 = -x + 1 implies x + 2y - 5 = 0$$y - 2 = -\frac{1}{2}(x - 1) \implies 2y - 4 = -x + 1 \implies x + 2y - 5 = 0$$ ### Step 2: Distance calculation from (3,6)$(3,6)$ The distance d$d$ of point Q(3,6)$Q(3,6)$ from line x + 2y - 5 = 0$x + 2y - 5 = 0$ is: d = left| frac3 + 2(6) - 5sqrt1^2 + 2^2 right| = left| frac10sqrt5 right| = 2sqrt5$$d = \left| \frac{3 + 2(6) - 5}{\sqrt{1^2 + 2^2}} \right| = \left| \frac{10}{\sqrt{5}} \right| = 2\sqrt{5}$$ Calculating d^2$d^2$: d^2 = (2sqrt5)^2 = 20$$d^2 = (2\sqrt{5})^2 = 20$$ ### Pattern Recognition Short Shortcut: The maximum distance of a family of lines passing through P$P$ from the origin is simply the length OP$OP$. The line perpendicular to OP$OP$ at P$P$ is the unique farthest line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

### Related Formula The orthocentre P$P$ of a triangle is the point of intersection of its altitudes. Area of a triangle with a horizontal base lying on the x-axis is: textArea = frac12 times textbase times textheight = frac12 times |x_C - x_B| times |y_P|$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times |x_C - x_B| \times |y_P|$$ ### Core Logic Find vertex A$A$ by solving the line equations AB$AB$ and AC$AC$: 3y - x = 2 implies x = 3y - 2$$3y - x = 2 \implies x = 3y - 2$$ Substitute into x + y = 2 implies (3y - 2) + y = 2 implies 4y = 4 implies y = 1$x + y = 2 \implies (3y - 2) + y = 2 \implies 4y = 4 \implies y = 1$. Then x = 3(1) - 2 = 1$x = 3(1) - 2 = 1$. So vertex A$A$ is (1, 1)$(1, 1)$. Find vertices B$B$ and C$C$ where the lines cross the x-axis (y = 0$y = 0$): - For B$B$ (on line AB$AB$): 3(0) - x = 2 implies x = -2 implies B(-2, 0)$3(0) - x = 2 \implies x = -2 \implies B(-2, 0)$ - For C$C$ (on line AC$AC$): x + 0 = 2 implies x = 2 implies C(2, 0)$x + 0 = 2 \implies x = 2 \implies C(2, 0)$ Base length BC = |2 - (-2)| = 4$BC = |2 - (-2)| = 4$. ### Step 1: Find Equations of Altitudes

1. **Altitude from A to BC**: Since BC$BC$ lies along the x-axis, the altitude from A(1, 1)$A(1, 1)$ must be a vertical line: textEquation of Altitude 1: x = 1$$\text{Equation of Altitude 1}: x = 1$$ 2. **Altitude from B to AC**: Slope of line AC$AC$ (x + y = 2$x + y = 2$) is m_AC = -1$m_{AC} = -1$. Therefore, the slope of the altitude perpendicular to AC$AC$ is m_2 = -frac1-1 = 1$m_2 = -\frac{1}{-1} = 1$. Passing through B(-2, 0)$B(-2, 0)$: y - 0 = 1(x - (-2)) implies y = x + 2 implies x - y + 2 = 0$$y - 0 = 1(x - (-2)) \implies y = x + 2 \implies x - y + 2 = 0$$ ### Step 2: Solve for Orthocentre coordinates P Intersect the altitude equations: x = 1$x = 1$ and y = x + 2$y = x + 2$: y = 1 + 2 = 3$y = 1 + 2 = 3$ Hence, the orthocentre is P(1, 3)$P(1, 3)$. ### Step 3: Compute Area of Triangle PBC Triangle PBC$PBC$ has base BC = 4$BC = 4$ on the x-axis, and vertex P(1, 3)$P(1, 3)$ gives a height of 3$3$. textArea = frac12 times 4 times 3 = 6$$\text{Area} = \frac{1}{2} \times 4 \times 3 = 6$$ ### Pattern Recognition When a triangle has its base sitting entirely on the coordinate axis, the altitude dropped from the opposite vertex is simplified to a pure horizontal or vertical line path, heavily reducing your linear derivation equation workload. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

### Related Formula textDistance from origin d = frac|C|sqrtA^2+B^2$$\text{Distance from origin } d = \frac{|C|}{\sqrt{A^2+B^2}}$$ ### Core Logic Decode slope angle fields to define vector shifts. Match variables to isolate focal root segments accurately as provided in the reference layout steps. ### Step 1: Identify Phase Shift Angles Given final tracking position slope component is: m = 2 - sqrt3 = tan 15^circ$$m = 2 - \sqrt{3} = \tan 15^\circ$$ Because clockwise translation turns angular coordinates down by half increments: alpha - fracalpha2 = 15^circ implies alpha = 30^circ$$\alpha - \frac{\alpha}{2} = 15^\circ \implies \alpha = 30^\circ$$ ### Step 2: Construct Rotated Vector Path Line The line equation through point (a,0)$(a,0)$ with final angle scaling is: y = (2-sqrt3)(x-a) implies (2-sqrt3)x - y - a(2-sqrt3) = 0$$y = (2-\sqrt{3})(x-a) \implies (2-\sqrt{3})x - y - a(2-\sqrt{3}) = 0$$ ### Step 3: Solve Origin Geometric Profiles Impose perpendicular bounds tracking distance rule: left| fracsqrt3a - 2asqrt4+3-4sqrt3+1 right| = frac1sqrt2 implies a^2 = 2(2+sqrt3)$$\left| \frac{\sqrt{3}a - 2a}{\sqrt{4+3-4\sqrt{3}+1}} \right| = \frac{1}{\sqrt{2}} \implies a^2 = 2(2+\sqrt{3})$$ Target formulation output text string solution: 3a^2 tan^2 30^circ - 2sqrt3 = 3 times 2(2+sqrt3) times frac13 - 2sqrt3 = 4$$3a^2 \tan^2 30^\circ - 2\sqrt{3} = 3 \times 2(2+\sqrt{3}) \times \frac{1}{3} - 2\sqrt{3} = 4$$ {{SOL_IMG_62}} ### Pattern Recognition Recognizing standard trigonometric slopes (tan 15^circ = 2-sqrt3$\tan 15^\circ = 2-\sqrt{3}$) bypasses computational bottlenecks instantly when working with line updates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines