### Related Formula textNormal form of a straight line: x cos alpha + y sin alpha = p$\text{Normal form of a straight line: } x \cos \alpha + y \sin \alpha = p$ textArea of right-angled triangle formed with axes: A = frac12 left| x_textintercept cdot y_textintercept right|$\text{Area of right-angled triangle formed with axes: } A = \frac{1}{2} \left| x_{\text{intercept}} \cdot y_{\text{intercept}} \right|$ ### Core Logic We write down the normal equation of the straight line using the given polar normal angle alpha = 45^circ$\alpha = 45^\circ$, find its intercept coordinates, and use the area constraint to solve for the coefficients. ### Step 1: Write down normal form The perpendicular drawn from the origin makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, so alpha = 45^circ$\alpha = 45^\circ$. The line equation is: x cos 45^circ + y sin 45^circ = p implies fracxsqrt2 + fracysqrt2 = p$x \cos 45^\circ + y \sin 45^\circ = p \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = p$ x + y = psqrt2 implies x + y - psqrt2 = 0$x + y = p\sqrt{2} \implies x + y - p\sqrt{2} = 0$ Comparing this with the given format x + by + c = 0$x + by + c = 0$, we find: b = 1 quad textand quad c = -psqrt2$b = 1 \quad \text{and} \quad c = -p\sqrt{2}$ ### Step 2: Solve for the parameters using the area constraint The line equation is x + y = psqrt2$x + y = p\sqrt{2}$. The intercepts are: - x_textintercept = psqrt2$x_{\text{intercept}} = p\sqrt{2}$ - y_textintercept = psqrt2$y_{\text{intercept}} = p\sqrt{2}$ The area of the right-angled triangle formed with the axes is: textArea = frac12 left| psqrt2 cdot psqrt2 right| = p^2$\text{Area} = \frac{1}{2} \left| p\sqrt{2} \cdot p\sqrt{2} \right| = p^2$ Since the area is given as 48 square units: p^2 = 48$p^2 = 48$ ### Step 3: Calculate the requested value We have: b^2 = 1^2 = 1$b^2 = 1^2 = 1$ c^2 = (-psqrt2)^2 = 2p^2 = 2(48) = 96$c^2 = (-p\sqrt{2})^2 = 2p^2 = 2(48) = 96$ Therefore, we find: b^2 + c^2 = 1 + 96 = 97$b^2 + c^2 = 1 + 96 = 97$ ### Pattern Recognition Normal equation coupling: Normal equations of lines xcosalpha+ysinalpha = p$x\cos\alpha+y\sin\alpha = p$ are extremely powerful when normal angles are specified. For alpha=45^circ$\alpha=45^\circ$, the coordinate intercepts are identical, making the area relation A=p^2$A=p^2$ exceptionally simple. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

### 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 textSlope m = -fracAB$\text{Slope } m = -\frac{A}{B}$ ### Core Logic Identify which diagonal passes through the origin. Use its directional configuration slope to determine the exact angle orientation layout map of the geometric system elements. ### Step 1: Determine Diagonal Orientation The line passing through the origin is diagonal OB$OB$: m_OB = -fracsqrt3+1sqrt3-1 = tan 105^circ$m_{OB} = -\frac{\sqrt{3}+1}{\sqrt{3}-1} = \tan 105^\circ$ Since diagonals of a square split the vertex paths evenly at 45^circ$45^\circ$, the side OA$OA$ makes an angle: alpha = 105^circ - 45^circ = 60^circ$\alpha = 105^\circ - 45^\circ = 60^\circ$ ### Step 2: Coordinate Vector Definition Coordinates of vertex A$A$ matching side value a$a$ are: A(acos 60^circ, asin 60^circ) = left( fraca2, fracsqrt3a2 right)$A(a\cos 60^\circ, a\sin 60^\circ) = \left( \frac{a}{2}, \frac{\sqrt{3}a}{2} \right)$ ### Step 3: Diagonal Intersection Point Solver Vertex A$A$ lies upon the alternative structural path line: (sqrt3-1)fraca2 - (sqrt3+1)fracsqrt3a2 + 8sqrt3 = 0$(\sqrt{3}-1)\frac{a}{2} - (\sqrt{3}+1)\frac{\sqrt{3}a}{2} + 8\sqrt{3} = 0$ a left[ fracsqrt3 - 1 - 3 - sqrt32 right] = -8sqrt3 implies -2a = -8sqrt3 implies a = 4sqrt3$a \left[ \frac{\sqrt{3} - 1 - 3 - \sqrt{3}}{2} \right] = -8\sqrt{3} \implies -2a = -8\sqrt{3} \implies a = 4\sqrt{3}$ a^2 = 48$a^2 = 48$ {{SOL_IMG_57}} ### Pattern Recognition Slopes involving sqrt3 pm 1$\sqrt{3} \pm 1$ explicitly alert you to geometric orientations built on multiples of 15^circ$15^circ$ or 75^circ$75^circ$ configurations. ### 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