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 textPerpendicular distance d = frac|ax_0 + by_0 + c|sqrta^2 + b^2$$\text{Perpendicular distance } d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$ ### Core Logic Let us find coordinates of point A$A$ by solving: 1) 3x + 2y = 14$3x + 2y = 14$ 2) 5x - y = 6 implies y = 5x - 6$5x - y = 6 \implies y = 5x - 6$ Substituting y$y$ in equation 1: 3x + 2(5x - 6) = 14 implies 13x - 12 = 14 implies 13x = 26 implies x = 2$$3x + 2(5x - 6) = 14 \implies 13x - 12 = 14 \implies 13x = 26 \implies x = 2$$ y = 5(2) - 6 = 4 implies A = (2, 4)$$y = 5(2) - 6 = 4 \implies A = (2, 4)$$ Let us find coordinates of point B$B$ by solving: 3) 4x + 3y = 8$4x + 3y = 8$ 4) 6x + y = 5 implies y = 5 - 6x$6x + y = 5 \implies y = 5 - 6x$ Substituting y$y$ in equation 3: 4x + 3(5 - 6x) = 8 implies 4x + 15 - 18x = 8 implies -14x = -7 implies x = frac12$$4x + 3(5 - 6x) = 8 \implies 4x + 15 - 18x = 8 \implies -14x = -7 \implies x = \frac{1}{2}$$ y = 5 - 6left(frac12right) = 2 implies B = left(frac12, 2right)$$y = 5 - 6\left(\frac{1}{2}\right) = 2 \implies B = \left(\frac{1}{2}, 2\right)$$ ### Step 1: Equation of line AB textSlope m = frac4 - 22 - 1/2 = frac23/2 = frac43$$\text{Slope } m = \frac{4 - 2}{2 - 1/2} = \frac{2}{3/2} = \frac{4}{3}$$ Equation of line AB: y - 4 = frac43(x - 2) implies 3y - 12 = 4x - 8 implies 4x - 3y + 4 = 0$$y - 4 = \frac{4}{3}(x - 2) \implies 3y - 12 = 4x - 8 \implies 4x - 3y + 4 = 0$$ ### Step 2: Distance Estimation Perpendicular distance from point P(5, -2)$P(5, -2)$ to line 4x - 3y + 4 = 0$4x - 3y + 4 = 0$: d = frac|4(5) - 3(-2) + 4|sqrt4^2 + (-3)^2 = frac|20 + 6 + 4|sqrt25 = frac305 = 6$$d = \frac{|4(5) - 3(-2) + 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 + 6 + 4|}{\sqrt{25}} = \frac{30}{5} = 6$$ ### Pattern Recognition Verify calculation metrics step-by-step. Finding straight intersections correctly upfront avoids scaling mistakes down the track. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

### Related Formula tan theta = left| fracm_1 - m_21 + m_1 m_2 right|$$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ ### Core Logic Find the intercepts of the line 4x + 5y = 20$4x + 5y = 20$ in the first quadrant. Put y=0 Rightarrow x=5$y=0 \Rightarrow x=5$. Point X(5, 0)$X(5, 0)$. Put x=0 Rightarrow y=4$x=0 \Rightarrow y=4$. Point Y(0, 4)$Y(0, 4)$. The line segment XY$XY$ is trisected by two points, say A$A$ and B$B$. Point A$A$ divides YX$YX$ in the ratio 2:1$2:1$, and B$B$ divides it in 1:2$1:2$. ### Step 1: Trisection Points Calculation Using the section formula for A$A$ (closer to Y-axis, ratio 1:2 from Y to X): A = left( frac1(5) + 2(0)3, frac1(0) + 2(4)3 right) = left( frac53, frac83 right)$$A = \left( \frac{1(5) + 2(0)}{3}, \frac{1(0) + 2(4)}{3} \right) = \left( \frac{5}{3}, \frac{8}{3} \right)$$ Using the section formula for B$B$ (closer to X-axis, ratio 2:1 from Y to X): B = left( frac2(5) + 1(0)3, frac2(0) + 1(4)3 right) = left( frac103, frac43 right)$$B = \left( \frac{2(5) + 1(0)}{3}, \frac{2(0) + 1(4)}{3} \right) = \left( \frac{10}{3}, \frac{4}{3} \right)$$ ### Step 2: Finding Line Slopes The lines L_1$L_1$ and L_2$L_2$ pass through the origin (0,0)$(0,0)$ to points A$A$ and B$B$. Slope of OA$OA$ (m_1$m_1$): m_1 = frac8/3 - 05/3 - 0 = frac85$$m_1 = \frac{8/3 - 0}{5/3 - 0} = \frac{8}{5}$$ Slope of OB$OB$ (m_2$m_2$): m_2 = frac4/3 - 010/3 - 0 = frac410 = frac25$$m_2 = \frac{4/3 - 0}{10/3 - 0} = \frac{4}{10} = \frac{2}{5}$$ ### Step 3: Calculating Tangent of the Angle Substitute the slopes into the angle formula: tan theta = left| frac8/5 - 2/51 + (8/5)(2/5) right|$$\tan \theta = \left| \frac{8/5 - 2/5}{1 + (8/5)(2/5)} \right|$$ tan theta = frac6/51 + 16/25$$\tan \theta = \frac{6/5}{1 + 16/25}$$ tan theta = frac6/5(25+16)/25 = frac6/541/25$$\tan \theta = \frac{6/5}{(25+16)/25} = \frac{6/5}{41/25}$$ tan theta = frac65 times frac2541 = frac3041$$\tan \theta = \frac{6}{5} \times \frac{25}{41} = \frac{30}{41}$$ ### Pattern Recognition For trisection or specific division of an intercepted segment, identify the axis intercepts first, rapidly apply the internal section formula, compute origin-centered slopes (which equal just the y/x ratio of the points), and pass them into the tan formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines

### Related Formula Image of a point (x_1, y_1)$(x_1, y_1)$ across line y=x$y=x$ is (y_1, x_1)$(y_1, x_1)$. **Angle Bisector Theorem:** The angle bisector of a triangle divides the opposite side into segments proportional to the lengths of the adjacent sides: fracABBC = fracADDC$$\frac{AB}{BC} = \frac{AD}{DC}$$ ### Core Logic Given A(4,6)$A(4,6)$ and Angle bisector of B is y=x$y=x$. Because y=x$y=x$ bisects angle B, the geometric reflection of vertex A across the bisector line y=x$y=x$ must lie exactly on the line containing the side BC$BC$. Let the reflection of A(4,6)$A(4,6)$ be A'$A'$. Across y=x$y=x$, the coordinates swap: A' = (6,4)$A' = (6,4)$ Next, find the intersection point D$D$ of the bisector y=x$y=x$ and side AC$AC$ (2x-y=2$2x-y=2$). Substitute y=x$y=x$ into 2x-y=2$2x-y=2$: 2x - x = 2 Rightarrow x = 2 Rightarrow y = 2$$2x - x = 2 \Rightarrow x = 2 \Rightarrow y = 2$$ So, point D$D$ is (2,2)$(2,2)$.

### Step 1: Utilize Section Formula By the internal angle bisector theorem: fracADDC = fracABBC$$\frac{AD}{DC} = \frac{AB}{BC}$$ Given 2AB = BC$2AB = BC$, so fracABBC = frac12$\frac{AB}{BC} = \frac{1}{2}$. This means point D(2,2)$D(2,2)$ divides the segment AC$AC$ in the ratio 1:2$1:2$. Let C$C$ have coordinates (x_c, y_c)$(x_c, y_c)$. Applying the section formula for D(2,2)$D(2,2)$ dividing A(4,6)$A(4,6)$ and C(x_c, y_c)$C(x_c, y_c)$ in ratio 1:2$1:2$: 2 = frac1 cdot x_c + 2 cdot 41 + 2 Rightarrow 6 = x_c + 8 Rightarrow x_c = -2$$2 = \frac{1 \cdot x_c + 2 \cdot 4}{1 + 2} \Rightarrow 6 = x_c + 8 \Rightarrow x_c = -2$$ 2 = frac1 cdot y_c + 2 cdot 61 + 2 Rightarrow 6 = y_c + 12 Rightarrow y_c = -6$$2 = \frac{1 \cdot y_c + 2 \cdot 6}{1 + 2} \Rightarrow 6 = y_c + 12 \Rightarrow y_c = -6$$ So, C$C$ is (-2,-6)$(-2,-6)$. ### Step 2: Find Equation of BC The line BC$BC$ passes through point C(-2,-6)$C(-2,-6)$ and the reflection point A'(6,4)$A'(6,4)$. Find the slope of BC$BC$: m_BC = frac4 - (-6)6 - (-2) = frac108 = frac54$$m_{BC} = \frac{4 - (-6)}{6 - (-2)} = \frac{10}{8} = \frac{5}{4}$$ Equation of BC$BC$: y - 4 = frac54(x - 6)$$y - 4 = \frac{5}{4}(x - 6)$$ 4y - 16 = 5x - 30$$4y - 16 = 5x - 30$$ 5x - 4y - 14 = 0$$5x - 4y - 14 = 0$$ ### Step 3: Solve for Vertex B Vertex B(alpha, beta)$B(\alpha, \beta)$ is the intersection of line BC$BC$ and the angle bisector y=x$y=x$. Substitute y=x$y=x$ into 5x - 4y - 14 = 0$5x - 4y - 14 = 0$: 5x - 4x - 14 = 0 Rightarrow x = 14$$5x - 4x - 14 = 0 \Rightarrow x = 14$$ Thus, y = 14$y = 14$. Therefore, B$B$ is (14, 14)$(14, 14)$, implying alpha = 14$\alpha = 14$ and beta = 14$\beta = 14$. Calculate alpha + 2beta$\alpha + 2\beta$: alpha + 2beta = 14 + 2(14) = 42$$\alpha + 2\beta = 14 + 2(14) = 42$$ ### Pattern Recognition Reflection properties drastically simplify angle bisector questions. If you know the bisector equation, reflecting one vertex over it gives a coordinate on the opposing extended ray. This paired with the angle bisector proportion theorem locks the entire geometric frame. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

### Related Formula textDistance of (x, y) text from ax+by+c=0 text is d = frac|ax + by + c|sqrta^2 + b^2$$\text{Distance of } (x, y) \text{ from } ax+by+c=0 \text{ is } d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}}$$ ### Core Logic The locus of a point P(x, y)$P(x, y)$ equidistant from lines x + 2y + 7 = 0$x + 2y + 7 = 0$ and 2x - y + 8 = 0$2x - y + 8 = 0$ is the pair of angle bisectors: frac|x + 2y + 7|sqrt1^2 + 2^2 = frac|2x - y + 8|sqrt2^2 + (-1)^2$$\frac{|x + 2y + 7|}{\sqrt{1^2 + 2^2}} = \frac{|2x - y + 8|}{\sqrt{2^2 + (-1)^2}}$$ fracx + 2y + 7sqrt5 = pm frac2x - y + 8sqrt5$$\frac{x + 2y + 7}{\sqrt{5}} = \pm \frac{2x - y + 8}{\sqrt{5}}$$ ### Step 1: Generating the Combined Equation Squaring both sides eliminates the pm$\pm$ and generates the combined equation of the bisectors: (x + 2y + 7)^2 - (2x - y + 8)^2 = 0$$(x + 2y + 7)^2 - (2x - y + 8)^2 = 0$$ Using a^2 - b^2 = (a - b)(a + b)$a^2 - b^2 = (a - b)(a + b)$: [ (x + 2y + 7) - (2x - y + 8) ] [ (x + 2y + 7) + (2x - y + 8) ] = 0$$[ (x + 2y + 7) - (2x - y + 8) ] [ (x + 2y + 7) + (2x - y + 8) ] = 0$$ (-x + 3y - 1)(3x + y + 15) = 0$$(-x + 3y - 1)(3x + y + 15) = 0$$ (x - 3y + 1)(3x + y + 15) = 0$$(x - 3y + 1)(3x + y + 15) = 0$$ ### Step 2: Expanding the Equation Multiply out the terms: 3x^2 + xy + 15x - 9xy - 3y^2 - 45y + 3x + y + 15 = 0$$3x^2 + xy + 15x - 9xy - 3y^2 - 45y + 3x + y + 15 = 0$$ 3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0$$3x^2 - 3y^2 - 8xy + 18x - 44y + 15 = 0$$ ### Step 3: Comparing Coefficients The standard form given is x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$. Divide our derived equation by 3 to match the leading coefficients: x^2 - y^2 - frac83xy + 6x - frac443y + 5 = 0$$x^2 - y^2 - \frac{8}{3}xy + 6x - \frac{44}{3}y + 5 = 0$$ Now, compare coefficients: 2h = -frac83 Rightarrow h = -frac43$$2h = -\frac{8}{3} \Rightarrow h = -\frac{4}{3}$$ 2g = 6 Rightarrow g = 3$$2g = 6 \Rightarrow g = 3$$ 2f = -frac443 Rightarrow f = -frac223$$2f = -\frac{44}{3} \Rightarrow f = -\frac{22}{3}$$ c = 5$c = 5$ ### Step 4: Final Calculation Substitute into the expression g + c + h - f$g + c + h - f$: 3 + 5 - frac43 - left(-frac223right) = 8 + frac183 = 8 + 6 = 14$$3 + 5 - \frac{4}{3} - \left(-\frac{22}{3}\right) = 8 + \frac{18}{3} = 8 + 6 = 14$$ ### Pattern Recognition Locus of equidistant points from two lines is their pair of angle bisectors. Equating squares d_1^2 = d_2^2$d_1^2 = d_2^2$ directly yields the joint equation of bisectors without needing explicit line separation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines

### Related Formula y - y_1 = tan(theta)(x - x_1)$$y - y_1 = \tan(\theta)(x - x_1)$$ ### Core Logic

The initial line makes an angle of 30^circ$30^{\circ}$ with the positive x-axis. It is rotated clockwise by 15^circ$15^{\circ}$ about the point A(9, 0)$A(9, 0)$. The new angle made by the line with the positive direction of the x-axis is 30^circ - 15^circ = 15^circ$30^{\circ} - 15^{\circ} = 15^{\circ}$. ### Step 1: Equation of the new line The equation of the line passing through A(9, 0)$A(9, 0)$ with a slope of tan 15^circ$\tan 15^{\circ}$ is: textEq^n: y - 0 = tan 15^circ (x - 9)$$\text{Eq}^n: y - 0 = \tan 15^{\circ} (x - 9)$$ We know that tan 15^circ = 2 - sqrt3$\tan 15^{\circ} = 2 - \sqrt{3}$. y = (2 - sqrt3) (x - 9)$$y = (2 - \sqrt{3}) (x - 9)$$ ### Step 2: Rearranging to match options Dividing by (2 - sqrt3)$(2 - \sqrt{3})$: fracy2 - sqrt3 = x - 9$$\frac{y}{2 - \sqrt{3}} = x - 9$$ Notice that 2 - sqrt3 = -(sqrt3 - 2)$2 - \sqrt{3} = -(\sqrt{3} - 2)$. Thus: frac-ysqrt3 - 2 = x - 9$$\frac{-y}{\sqrt{3} - 2} = x - 9$$ fracysqrt3 - 2 + x = 9$$\frac{y}{\sqrt{3} - 2} + x = 9$$ ### Pattern Recognition A clockwise rotation decreases the angle of inclination. Calculate the new angle, find its tangent, and carefully algebraicize the denominator to match the given option forms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines