Maharashtra Board Class 11 Maths Part 1 Chapter 5 Locus and Straight Line 5.4 Solutions

Question 1. Find the slope, x-intercept, y-intercept of each of the following lines.
(a) 2x + 3y - 6 = 0
(b) x + 2y = 0
Answer: Solution:
(a) Given equation of the line is 2x + 3y - 6 = 0
Comparing this equation with ax + by + c = 0, we get
a = 2, b = 3, c = -6
\[ \therefore \text{ Slope of the line } = \frac{-a}{b} = \frac{-2}{3} \]
\[ \text{x-intercept } = \frac{-c}{a} = \frac{-(-6)}{2} = 3 \]
\[ \text{y-intercept } = \frac{-c}{b} = \frac{-(-6)}{3} = 2 \]
(b) Given equation of the line is x + 2y = 0
Comparing this equation with ax + by + c = 0, we get
a = 1, b = 2, c = 0
\[ \therefore \text{ Slope of the line } = \frac{-a}{b} = \frac{-1}{2} \]
\[ \text{x-intercept } = \frac{-c}{a} = \frac{-0}{1} = 0 \]
\[ \text{y-intercept } = \frac{-c}{b} = \frac{-0}{2} = 0 \] In simple words: To find slope, x-intercept, and y-intercept, convert the line equation to the standard form ax + by + c = 0 and then apply the respective formulas: slope = -a/b, x-intercept = -c/a, and y-intercept = -c/b.

🎯 Exam Tip: Remember these fundamental formulas as they are crucial for understanding and analyzing linear equations in coordinate geometry.

Question 2. Write each of the following equations in ax + by + c = 0 form.
(a) y = 2x - 4
(b) y = 4
(c) \(\frac{x}{2} + \frac{y}{4} = 1\)
(d) \(\frac{x}{3} = \frac{y}{2}\)
Answer: Solution:
(a) y = 2x - 4
\[ \therefore \text{ 2x } - \text{ y } - \text{ 4 } = \text{ 0 is the equation in ax } + \text{ by } + \text{ c } = \text{ 0 form.} \]
(b) y = 4
\[ \therefore \text{ 0x } + \text{ 1y } - \text{ 4 } = \text{ 0 is the equation in ax } + \text{ by } + \text{ c } = \text{ 0 form.} \]
(c) \(\frac{x}{2} + \frac{y}{4} = 1\)
\[ \therefore \frac{2x + y}{4} = 1 \]
\[ \therefore \text{ 2x } + \text{ y } = \text{ 4 } \]
\[ \therefore \text{ 2x } + \text{ y } - \text{ 4 } = \text{ 0 is the equation in ax } + \text{ by } + \text{ c } = \text{ 0 form.} \]
(d) \(\frac{x}{3} = \frac{y}{2}\)
\[ \therefore \text{ 2x } = \text{ 3y } \]
\[ \therefore \text{ 2x } - \text{ 3y } + \text{ 0 } = \text{ 0 is the equation in ax } + \text{ by } + \text{ c } = \text{ 0 form.} \] In simple words: To convert any linear equation into the standard form ax + by + c = 0, rearrange all terms to one side of the equation, setting the other side to zero, ensuring coefficients a, b, and c are constants.

🎯 Exam Tip: Mastering algebraic manipulation to convert equations to standard form is essential for solving more complex problems involving lines.

Question 3. Show that the lines x - 2y - 7 = 0 and 2x - 4y + 5 = 0 are parallel to each other.
Answer: Solution:
Let m₁ be the slope of the line x - 2y - 7 = 0.
\[ \therefore \text{ m1 } = \frac{-1}{-2} = \frac{1}{2} \]
Let m₂ be the slope of the line 2x - 4y + 5 = 0.
\[ \therefore \text{ m2 } = \frac{-2}{-4} = \frac{1}{2} \]
Since, m₁ = m₂
\[ \therefore \text{ The given lines are parallel to each other.} \] In simple words: Two lines are parallel if and only if their slopes are equal. We calculate the slope of each given line using the formula m = -a/b and find they are both 1/2, confirming they are parallel.

🎯 Exam Tip: Always remember the condition for parallel lines (m₁ = m₂) and perpendicular lines (m₁m₂ = -1). This is a frequent concept in exams.

Question 4. If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Answer: Solution:
Let the line 3x + 4y = p cuts the X and Y-axes at points A and B respectively.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज OAB को दर्शाता है जो X-अक्ष, Y-अक्ष और एक रेखा 3x + 4y = p से बना है। बिंदु A (p/3, 0) X-अक्ष पर और बिंदु B (0, p/4) Y-अक्ष पर स्थित है, जहाँ रेखा निर्देशांक अक्षों को काटती है। यह त्रिभुज का शीर्ष मूलबिंदु O (0,0) है।
3x + 4y = p
\[ \therefore \frac{3x}{p} + \frac{4y}{p} = 1 \]
\[ \therefore \frac{x}{p/3} + \frac{y}{p/4} = 1 \]
The equation is of the form \(\frac{x}{a} + \frac{y}{b} = 1\), with a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
\[ \therefore \text{ A } = \text{ (a, 0) } = \left(\frac{p}{3}, 0\right) \text{ and B } = \text{ (0, b) } = \left(0, \frac{p}{4}\right) \]
\[ \therefore \text{ OA } = \frac{p}{3} \text{ and OB } = \frac{p}{4} \]
Given, A(ΔOAB) = 24 sq. units
\[ \therefore \frac{1}{2} \times \text{ OA } \times \text{ OB } = 24 \]
\[ \therefore \frac{1}{2} \times \frac{p}{3} \times \frac{p}{4} = 24 \]
\[ \therefore \text{ p}^2 = 576 \]
\[ \therefore \text{ p } = \pm24 \] In simple words: The line 3x + 4y = p forms a triangle with the coordinate axes. We find the x and y intercepts to determine the base and height of this right-angled triangle. Using the area formula, we solve for p.

🎯 Exam Tip: When a line forms a triangle with the coordinate axes, the intercepts on the axes serve as the base and height of the triangle. The area formula is \( \frac{1}{2} \times |\text{x-intercept}| \times |\text{y-intercept}| \).

Question 5. Find the co-ordinates of the circumcentre of the triangle whose vertices are A(-2, 3), B(6, -1), C(4, 3).
Answer: Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ΔABC.
Let F be the circumcentre of ΔABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज ABC को दर्शाता है जिसके शीर्ष A(-2, 3), B(6, -1) और C(4, 3) हैं। बिंदु F त्रिभुज का परिकेन्द्र है। FD भुजा BC का लंब समद्विभाजक है और FE भुजा AC का लंब समद्विभाजक है। D और E क्रमशः BC और AC के मध्यबिंदु हैं।
\[ \therefore \text{ D and E are the midpoints of side BC and AC respectively.} \]
\[ \therefore \text{ D } = \left(\frac{6+4}{2}, \frac{-1+3}{2}\right) = (5, 1) \]
\[ \text{and E } = \left(\frac{-2+4}{2}, \frac{3+3}{2}\right) = (1, 3) \]
Now, slope of BC = \(\frac{-1-3}{6-4} = \frac{-4}{2} = -2\)
\[ \therefore \text{ slope of FD } = \frac{1}{2} \text{ .....[FD } \perp \text{ BC]} \]
Since, FD passes through (5, 1) and has slope \(\frac{1}{2}\)
\[ \therefore \text{ Equation of FD is y } - \text{ 1 } = \frac{1}{2}\text{ (x } - \text{ 5)} \]
\[ \therefore \text{ 2(y } - \text{ 1) } = \text{ x } - \text{ 5} \]
\[ \therefore \text{ x } - \text{ 2y } - \text{ 3 } = \text{ 0 ......(i)} \]
Since, both the points A and C have same y co-ordinates i.e. 3
\[ \therefore \text{ the points A and C lie on the line y } = \text{ 3.} \]
Since, FE passes through E(1, 3).
\[ \therefore \text{ the equation of FE is x } = \text{ 1. .......(ii)} \]
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 - 2y - 3 = 0
\[ \therefore \text{ y } = -1 \]
\[ \therefore \text{ Co-ordinates of circumcentre F } = \text{ (1, -1).} \] In simple words: The circumcentre is the intersection of the perpendicular bisectors of the sides of a triangle. We find the midpoints and slopes of two sides (BC and AC), then determine the equations of their perpendicular bisectors (FD and FE). Solving these two equations gives the coordinates of the circumcentre.

🎯 Exam Tip: The circumcentre is equidistant from all three vertices. Using the distance formula (FA = FB = FC) is an alternative but often more complex method for finding the circumcentre.

Question 6. Find the equation of the line whose x-intercept is 3 and which is perpendicular to the line 3x - y + 23 = 0.
Answer: Solution:
Slope of the line 3x - y + 23 = 0 is 3.
\[ \therefore \text{ slope of the required line which is perpendicular to 3x } - \text{ y } + \text{ 23 } = \text{ 0 is } -\frac{1}{3}. \]
Since, the x-intercept of the required line is 3.
\[ \therefore \text{ it passes through (3, 0).} \]
\[ \therefore \text{ the equation of the required line is} \]
\[ \text{ y } - \text{ 0 } = -\frac{1}{3}\text{ (x } - \text{ 3)} \]
\[ \therefore \text{ 3y } = -\text{ x } - \text{ 3} \]
\[ \therefore \text{ x } - \text{ 3y } = \text{ 3} \] In simple words: First, find the slope of the given line. Since the required line is perpendicular, its slope will be the negative reciprocal. An x-intercept of 3 means the line passes through (3, 0). With a point and a slope, we use the point-slope form to find the equation of the line.

🎯 Exam Tip: Remember that if two lines are perpendicular, the product of their slopes is -1. An x-intercept 'a' means the line passes through (a, 0), and a y-intercept 'b' means it passes through (0, b).

Question 7. Find the distance of the point A(-2, 3) from the line 12x - 5y - 13 = 0.
Answer: Solution:
Let p be the perpendicular distance of the point A(-2, 3) from the line 12x - 5y - 13 = 0
Here, a = 12, b = -5, c = -13, x₁ = -2, y₁ = 3
\[ \text{ p } = \left|\frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}\right| \]
\[ = \left|\frac{12(-2) - 5(3) - 13}{\sqrt{12^2 + (-5)^2}}\right| \]
\[ = \left|\frac{-24 - 15 - 13}{\sqrt{144 + 25}}\right| \]
\[ = \left|\frac{-52}{\sqrt{169}}\right| \]
\[ = \left|\frac{-52}{13}\right| \]
\[ = 4 \text{ units} \] In simple words: To find the perpendicular distance from a point (x₁, y₁) to a line ax + by + c = 0, use the formula: \( \text{distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \). Substitute the given point and line coefficients into this formula and calculate the result.

🎯 Exam Tip: The absolute value in the distance formula ensures that the distance is always positive. Make sure to correctly identify a, b, c from the line equation and x₁, y₁ from the point.

Question 8. Find the distance between parallel lines 9x + 6y - 1 = 0 and 9x + 6y - 32 = 0.
Answer: Solution:
Equations of the given parallel lines are 9x + 6y - 7 = 0 and 9x + 6y - 32 = 0.
Here, a = 9, b = 6, C₁ = -7 and C₂ = -32
\[ \therefore \text{ Distance between the parallel lines } = \frac{|C_1 - C_2|}{\sqrt{a^2 + b^2}} \]
\[ = \frac{|-7 - (-32)|}{\sqrt{9^2 + 6^2}} \]
\[ = \frac{|-7 + 32|}{\sqrt{81 + 36}} \]
\[ = \frac{25}{\sqrt{117}} \]
\[ = \frac{25}{\sqrt{117}} \text{ units} \] In simple words: The distance between two parallel lines, ax + by + c₁ = 0 and ax + by + c₂ = 0, is found using the formula \( \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \). We identify the coefficients a, b, c₁, and c₂ from the given parallel line equations and plug them into the formula.

🎯 Exam Tip: Ensure that the 'a' and 'b' coefficients are identical for both parallel lines before applying the formula. If they are not, you might need to multiply one equation by a constant to make them match.

Question 9. Find the equation of the line passing through the point of intersection of lines x + y - 2 = 0 and 2x - 3y + 4 = 0 and making intercept 3 on the X-axis.
Answer: Solution:
Given equations of lines are
x + y - 2 = 0 ......(i)
and 2x - 3y - 4 = 0 ......(ii)
Multiplying equation (i) by 3, we get
3x - 3y - 6 = 0 .....(iii)
Adding equation (ii) and (iii), we get
5x - 2 = 0
\[ \therefore \text{ x } = \frac{2}{5} \]
Substituting x = \(\frac{2}{5}\) in equation (i), we get
\[ \frac{2}{5} + \text{ y } - \text{ 2 } = \text{ 0} \]
\[ \therefore \text{ y } = \text{ 2 } - \frac{2}{5} = \frac{8}{5} \]
\[ \therefore \text{ The required line passes through point } \left(\frac{2}{5}, \frac{8}{5}\right). \]
Also, the line makes intercept of 3 on X-axis
\[ \therefore \text{ it also passes through point (3, 0).} \]
\[ \therefore \text{ required equation of line passing through points } \left(\frac{2}{5}, \frac{8}{5}\right) \text{ and (3, 0) is} \]
\[ \frac{y - \frac{8}{5}}{0 - \frac{8}{5}} = \frac{x - \frac{2}{5}}{3 - \frac{2}{5}} \]
\[ \frac{\frac{5y - 8}{5}}{\frac{-8}{5}} = \frac{\frac{5x - 2}{5}}{\frac{13}{5}} \]
\[ \frac{5y - 8}{-8} = \frac{5x - 2}{13} \]
\[ \therefore \text{ 13(5y } - \text{ 8) } = \text{ -8(5x } - \text{ 2)} \]
\[ \therefore \text{ 65y } - \text{ 104 } = \text{ -40x } + \text{ 16} \]
\[ \therefore \text{ 40x } + \text{ 65y } - \text{ 120 } = \text{ 0} \]
\[ \therefore \text{ 8x } + \text{ 13y } - \text{ 24 } = \text{ 0 which is the equation of the required line.} \] In simple words: First, solve the system of two linear equations to find their point of intersection. This point, along with the given x-intercept (which implies another point (3,0)), allows us to find the equation of the required line using the two-point form formula.

🎯 Exam Tip: For problems involving the intersection of lines, accurately solving simultaneous equations is critical. Pay attention to fractions and algebraic simplification steps to avoid errors.

Question 10. D(-1, 8), E(4, -2), F(-5, -3) are midpoints of sides BC, CA and AB of ΔABC. Find
(i) equations of sides of ΔABC.
(ii) co-ordinates of the circumcentre of ΔABC.
Answer: Solution:
(i) Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज ABC को दर्शाता है जिसके शीर्ष A(x₁, y₁), B(x₂, y₂) और C(x₃, y₃) हैं। D(-1, 8) भुजा BC का मध्यबिंदु है, E(4, -2) भुजा CA का मध्यबिंदु है, और F(-5, -3) भुजा AB का मध्यबिंदु है।
\[ \text{ D } = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right) \]
\[ (-1, 8) = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right) \]
\[ \therefore \text{ x}_2 + \text{ x}_3 = -2 \text{ ......(i)} \]
\[ \text{and y}_2 + \text{ y}_3 = 16 \text{ ......(ii)} \]
\[ \text{Also, E } = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) \]
\[ (4, -2) = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) \]
\[ \therefore \text{ x}_1 + \text{ x}_3 = 8 \text{ ......(iii)} \]
\[ \text{and y}_1 + \text{ y}_3 = -4 \text{ ......(iv)} \]
\[ \text{Similarly, F } = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
\[ (-5, -3) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
\[ \therefore \text{ x}_1 + \text{ x}_2 = -10 \text{ ......(v)} \]
\[ \text{and y}_1 + \text{ y}_2 = -6 \text{ ......(vi)} \]
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = -4
\[ \therefore \text{ x}_1 + \text{ x}_2 + \text{ x}_3 = -2 \text{ .....(vii)} \]
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = -10
Solving (v) and (vii), we get x₃ = 8
For y-coordinates:
Adding (ii), (iv) and (vi), we get
2y₁ + 2y₂ + 2y₃ = 6
\[ \therefore \text{ y}_1 + \text{ y}_2 + \text{ y}_3 = 3 \text{ .....(viii)} \]
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
\[ \therefore \text{ Vertices of ΔABC are A(0, -13), B(-10, 7), C(8, 9)} \]
a. Equation of side AB is
\[ \frac{y + 13}{7 + 13} = \frac{x - 0}{-10 - 0} \]
\[ \therefore \frac{y + 13}{20} = \frac{x}{-10} \]
\[ \therefore \frac{y + 13}{2} = -x \]
\[ \therefore \text{ 2x } + \text{ y } + \text{ 13 } = \text{ 0} \]
b. Equation of side BC is
\[ \frac{y - 7}{9 - 7} = \frac{x + 10}{8 + 10} \]
\[ \therefore \frac{y - 7}{2} = \frac{x + 10}{18} \]
\[ \therefore \text{ y } - \text{ 7 } = \frac{x + 10}{9} \]
\[ \therefore \text{ 9(y } - \text{ 7) } = \text{ x } + \text{ 10} \]
\[ \therefore \text{ 9y } - \text{ 63 } = \text{ x } + \text{ 10} \]
\[ \therefore \text{ x } - \text{ 9y } + \text{ 73 } = \text{ 0} \]
c. Equation of side AC is
\[ \frac{y + 13}{9 + 13} = \frac{x - 0}{8 - 0} \]
\[ \therefore \frac{y + 13}{22} = \frac{x}{8} \]
\[ \therefore \text{ 8(y } + \text{ 13) } = \text{ 22x} \]
\[ \therefore \text{ 4(y } + \text{ 13) } = \text{ 11x} \]
\[ \therefore \text{ 4y } + \text{ 52 } = \text{ 11x} \]
\[ \therefore \text{ 11x } - \text{ 4y } - \text{ 52 } = \text{ 0} \]
(ii) Here, A(0, -13), B(-10, 7), C(8, 9) are the vertices of ΔABC.
Let F be the circumcentre of ΔABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज ABC को दर्शाता है जिसके शीर्ष A(0, -13), B(-10, 7) और C(8, 9) हैं। बिंदु F त्रिभुज का परिकेन्द्र है। FD भुजा BC का लंब समद्विभाजक है और FE भुजा AC का लंब समद्विभाजक है। D और E क्रमशः BC और AC के मध्यबिंदु हैं।
\[ \therefore \text{ D and E are the midpoints of side BC and AC.} \]
\[ \therefore \text{ D } = \left(\frac{-10+8}{2}, \frac{7+9}{2}\right) = (-1, 8) \]
\[ \text{and E } = \left(\frac{0+8}{2}, \frac{-13+9}{2}\right) = (4, -2) \]
Now, slope of BC = \(\frac{7-9}{-10-8} = \frac{-2}{-18} = \frac{1}{9}\)
\[ \therefore \text{ slope of FD } = -9 \text{ ......[FD } \perp \text{ BC]} \]
Since, FD passes through (-1, 8) and has slope -9
\[ \therefore \text{ Equation of FD is y } - \text{ 8 } = -9\text{(x } + \text{ 1)} \]
\[ \therefore \text{ y } - \text{ 8 } = -9\text{x } - \text{ 9} \]
\[ \therefore \text{ y } = -9\text{x } - \text{ 1 .....(i)} \]
Also, slope of AC = \(\frac{-13-9}{0-8} = \frac{-22}{-8} = \frac{11}{4}\)
\[ \therefore \text{ Slope of FE } = -\frac{4}{11} \text{ ....[FE } \perp \text{ AC]} \]
Since, FE passes through (4, -2) and has slope \( -\frac{4}{11} \)
\[ \therefore \text{ Equation of FE is y } + \text{ 2 } = -\frac{4}{11}\text{ (x } - \text{ 4)} \]
\[ \therefore \text{ 11(y } + \text{ 2) } = -4\text{(x } - \text{ 4)} \]
\[ \therefore \text{ 11y } + \text{ 22 } = -4\text{x } + \text{ 16} \]
\[ \therefore \text{ 4x } + \text{ 11y } = -6 \text{ ....(ii)} \]
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of y in (ii), we get
4x + 11(-9x - 1) = -6
4x - 99x - 11 = -6
-95x = 5
\[ \therefore \text{ x } = -\frac{5}{95} = -\frac{1}{19} \]
Substituting the value of x in (i), we get
\[ \text{ y } = -9\left(-\frac{1}{19}\right) - 1 = \frac{9}{19} - 1 = \frac{9-19}{19} = -\frac{10}{19} \]
\[ \therefore \text{ Co-ordinates of circumcentre F } = \left(-\frac{1}{19}, -\frac{10}{19}\right) \] In simple words: First, use the midpoint formula for each side (D, E, F) to set up a system of equations. Solve these equations to find the coordinates of the vertices A, B, and C. Once the vertices are known, find the equations of two perpendicular bisectors of the sides. The intersection point of these bisectors is the circumcentre of the triangle.

🎯 Exam Tip: This is a multi-step problem. Break it down into finding vertices, then midpoints, then slopes, then perpendicular bisector equations, and finally solving for their intersection. Accuracy in each step is vital.