### 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

### Related Formula textCentroid divides the line joining Orthocentre and Circumcentre in 2:1$\text{Centroid divides the line joining Orthocentre and Circumcentre in } 2:1$ Distance in parametric form: x = x_1 + rcostheta, y = y_1 + rsintheta$x = x_1 + r\cos\theta, y = y_1 + r\sin\theta$ ### Core Logic

Let Orthocentre C(-6, -8)$C(-6, -8)$ and Circumcentre B(3, 4)$B(3, 4)$. Centroid A(a, b)$A(a, b)$ divides CB$CB$ in 2:1$2:1$. a = frac2(3) + 1(-6)2+1 = 0$a = \frac{2(3) + 1(-6)}{2+1} = 0$ b = frac2(4) + 1(-8)2+1 = 0$b = \frac{2(4) + 1(-8)}{2+1} = 0$ So, P(2a+3, 7b+5) = (3, 5)$P(2a+3, 7b+5) = (3, 5)$. The line along which distance is measured is parallel to x - 2y - 1 = 0$x - 2y - 1 = 0$, giving slope m = tantheta = frac12$m = \tan\theta = \frac{1}{2}$. Using parametric coordinates from P(3,5)$P(3,5)$: x = 3 + rcostheta, quad y = 5 + rsintheta$x = 3 + r\cos\theta, \quad y = 5 + r\sin\theta$ Substitute into the target line 2x + 3y - 4 = 0$2x + 3y - 4 = 0$: 2(3 + rcostheta) + 3(5 + rsintheta) - 4 = 0$2(3 + r\cos\theta) + 3(5 + r\sin\theta) - 4 = 0$ r(2costheta + 3sintheta) = -17$r(2\cos\theta + 3\sin\theta) = -17$ From tantheta = 1/2$\tan\theta = 1/2$, we get sintheta = frac1sqrt5$\sin\theta = \frac{1}{\sqrt{5}}$ and costheta = frac2sqrt5$\cos\theta = \frac{2}{\sqrt{5}}$. rleft(2left(frac2sqrt5right) + 3left(frac1sqrt5right)right) = -17$r\left(2\left(\frac{2}{\sqrt{5}}\right) + 3\left(\frac{1}{\sqrt{5}}\right)\right) = -17$ rleft(frac7sqrt5right) = -17 implies |r| = frac17sqrt57$r\left(\frac{7}{\sqrt{5}}\right) = -17 \implies |r| = \frac{17\sqrt{5}}{7}$ ### Pattern Recognition Standard Euler line property: O, G, C$O, G, C$ are collinear and G$G$ divides OC$OC$ in 2:1$2:1$. Use parametric equation to find intersection distance directly without finding the intersection point. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Straight Lines