Maharashtra Board Class 11 Maths Part 1 Chapter 5 Locus and Straight Line 5.2 Solutions

Question 1. Find the slope of each of the following lines which pass through the points:
(a) (2, -1), (4, 3)
(b) (-2, 3), (5, 7)
(c) (2, 3), (2, -1)
(d) (7, 1), (-3, 1)
Answer:
(a) Let A = \( (x_1, y_1) \) = (2, -1) and B = \( (x_2, y_2) \) = (4, 3).
Slope of line AB = \( \frac{y_2-y_1}{x_2-x_1} \)
= \( \frac{3-(-1)}{4-2} \)
= \( \frac{4}{2} \)
= 2

(b) Let C = \( (x_1, y_1) \) = (-2, 3) and D = \( (x_2, y_2) \) = (5, 7)
Slope of line CD = \( \frac{y_2-y_1}{x_2-x_1} \)
= \( \frac{7-3}{5-(-2)} \)
= \( \frac{4}{7} \)

(c) Let E = (2, 3) = \( (x_1, y_1) \) and F = (2, -1) = \( (x_2, y_2) \)
Since \( x_1 = x_2 \) = 2
\( \therefore \) The slope of EF is not defined. ......[EF || y-axis]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ऊर्ध्वाधर रेखा को दर्शाता है जो X-अक्ष को मूल बिंदु के दाईं ओर प्रतिच्छेद करती है। बिंदु E(2,3) ऊपर स्थित है और बिंदु F(2,-1) नीचे स्थित है, दोनों X-अक्ष के दाईं ओर '2' की x-निर्देशांक पर हैं, जो यह दर्शाता है कि रेखा Y-अक्ष के समानांतर है।

(d) Let G = (7, 1) = \( (x_1, y_1) \) and H = (-3, 1) = \( (x_2, y_2) \) say.
Since \( y_1 = y_2 \)
\( \therefore \) The slope of GH = 0 .....[GH || x-axis]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक क्षैतिज रेखा को दर्शाता है जो X-अक्ष के समानांतर है। बिंदु H(-3,1) Y-अक्ष के बाईं ओर स्थित है और बिंदु G(7,1) Y-अक्ष के दाईं ओर स्थित है, दोनों '1' की y-निर्देशांक पर स्थित हैं, जो दर्शाता है कि रेखा X-अक्ष के समानांतर है।
In simple words: The slope indicates the steepness and direction of a line. For part (a) the slope is 2, for (b) it's 4/7. In part (c), if x-coordinates are the same, the line is vertical and its slope is undefined. In part (d), if y-coordinates are the same, the line is horizontal and its slope is 0.

🎯 Exam Tip: Remember that vertical lines have undefined slopes (x-coordinates are equal), and horizontal lines have a slope of 0 (y-coordinates are equal). This is a common pitfall in slope calculations.

Question 2. If the X and Y-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Answer:
Given, x-intercept of line L is 2 and y-intercept of line L is 3
\( \therefore \) the line L intersects X-axis at (2, 0) and Y-axis at (0, 3).
i.e. the line L passes through (2, 0) = \( (x_1, y_1) \) and (0, 3) = \( (x_2, y_2) \) say.
Slope of line L = \( \frac{y_2-y_1}{x_2-x_1} \)
= \( \frac{3-0}{0-2} \)
= \( \frac{-3}{2} \)
In simple words: A line's x-intercept is where it crosses the x-axis (y=0) and its y-intercept is where it crosses the y-axis (x=0). Using these two points, (2,0) and (0,3), the slope of the line is calculated as -3/2.

🎯 Exam Tip: Remember that the x-intercept is a point \( (a, 0) \) and the y-intercept is a point \( (0, b) \). Use these two points to find the slope using the standard formula.

Question 3. Find the slope of the line whose inclination is 30°.
Answer:
Given, inclination \( (\theta) \) = 30°
Slope of the line = tan \( \theta \) = tan 30° = \( \frac{1}{\sqrt{3}} \)
In simple words: The slope of a line is defined as the tangent of its angle of inclination. For an inclination of 30 degrees, the slope is 1 divided by the square root of 3.

🎯 Exam Tip: Recall the standard trigonometric values for common angles like 30°, 45°, and 60° as they are frequently used in slope calculations.

Question 4. Find the slope of the line whose inclination is 45°.
Answer:
Given, inclination \( (\theta) \) = 45°
Slope of the line = tan \( \theta \) = tan 45° = 1
In simple words: For a line with an inclination of 45 degrees, its slope is equal to 1, meaning it rises at a 45-degree angle to the x-axis.

🎯 Exam Tip: The slope for an inclination of 45° is 1, indicating a positive correlation where y increases directly with x.

Question 5. A line makes intercepts 3 and 3 on the co-ordinate axes. Find the slope of the line.
Answer:
Given, x-intercept of line is 3 and y-intercept of line is 3
\( \therefore \) The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
i.e. the line passes through (3, 0) = \( (x_1, y_1) \) and (0, 3) = \( (x_2, y_2) \) say.
Slope of line = \( \frac{y_2-y_1}{x_2-x_1} \)
= \( \frac{3-0}{0-3} \)
= -1
In simple words: With x-intercept and y-intercept both being 3, the line passes through (3,0) and (0,3). Using these points, the slope is calculated as -1, indicating a downward-sloping line.

🎯 Exam Tip: Identify the two points clearly from the given intercepts before applying the slope formula to avoid errors.

Question 6. Without using Pythagoras theorem, show that points A(4, 4), B(3, 5) and C(-1, -1) are the vertices of a right-angled triangle.
Answer:
Given, A(4, 4) = \( (x_1, y_1) \), B(3, 5) = \( (x_2, y_2) \), C(-1, -1) = \( (x_3, y_3) \)
Slope of AB = \( \frac{y_2-y_1}{x_2-x_1} \) = \( \frac{5-4}{3-4} \) = \( \frac{1}{-1} \) = -1
Slope of BC = \( \frac{y_3-y_2}{x_3-x_2} \) = \( \frac{-1-5}{-1-3} \) = \( \frac{-6}{-4} \) = \( \frac{3}{2} \)
Slope of AC = \( \frac{y_3-y_1}{x_3-x_1} \) = \( \frac{-1-4}{-1-4} \) = \( \frac{-5}{-5} \) = 1
Slope of AB \( \times \) slope of AC = -1 \( \times \) 1 = -1
\( \therefore \) side AB \( \perp \) side AC
\( \therefore \triangle ABC \) is a right angled triangle, right angled at A.
\( \therefore \) The given points are the vertices of a right angled triangle.
In simple words: To prove a right-angled triangle using slopes, calculate the slopes of all three sides. If the product of the slopes of any two sides is -1, those two sides are perpendicular, confirming a right angle. Here, the slopes of AB and AC multiply to -1, indicating a right angle at A.

🎯 Exam Tip: For problems involving right-angled triangles using slopes, always check if the product of any two side slopes is -1. This is a crucial condition for perpendicular lines.

Question 7. Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured clockwise.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख X-अक्ष और Y-अक्ष वाले एक कार्तीय निर्देशांक प्रणाली को दर्शाता है। इसमें एक रेखा को मूल बिंदु से गुजरते हुए दिखाया गया है, जो धनात्मक Y-अक्ष के साथ दक्षिणावर्त दिशा में 45° का कोण बनाती है। रेखा का धनात्मक X-अक्ष के साथ वामावर्त झुकाव (थीटा) 135° है।
Since, the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction.
\( \therefore \) Inclination of the line \( (\theta) \) = (90° + 45°)
\( \therefore \) Slope of the line = tan(90° + 45°)
= -cot 45° .......[tan(90 + \( \theta) \) = -cot \( \theta \)]
= -1
In simple words: To find the slope, we first determine the inclination angle \( \theta \) with the positive X-axis. If a line makes a 45° angle clockwise with the positive Y-axis, its inclination from the positive X-axis is \( 90° + 45° = 135° \). The slope is then calculated as tan(135°), which is -1.

🎯 Exam Tip: When angles are given relative to the Y-axis or in a clockwise direction, convert them to the standard inclination (angle with positive X-axis, counter-clockwise) before applying the slope formula \( m = \tan \theta \).

Question 8. Find the value of k for which the points P(k, -1), Q(2, 1) and R(4, 5) are collinear.
Answer:
Given, points P(k, -1), Q(2, 1), and R(4, 5) are collinear.
\( \therefore \) Slope of PQ = Slope of QR
\( \therefore \frac{1-(-1)}{2-k} = \frac{5-1}{4-2} \)
\( \therefore \frac{2}{2-k} = \frac{4}{2} \)
\( \therefore 1 = 2-k \)
\( \therefore k = 2 - 1 \)
\( \therefore k = 1 \)
Check:
For collinear points P, Q, R,
Slope of PQ = Slope of QR = Slop of PR
For k = 1, if the given points are collinear, then our answer is correct.
P(1, -1), Q(2, 1) and R(4, 5)
Slope of PQ = \( \frac{1-(-1)}{2-1} = \frac{2}{1} = 2 \)
Slope of QR = \( \frac{5-1}{4-2} = \frac{4}{2} = 2 \)
Slope of PQ = Slope of QR
\( \therefore \) The given points are collinear.
Thus, our answer is correct.
In simple words: For three points to be collinear (lie on the same straight line), the slope between any two pairs of points must be equal. By equating the slope of PQ to the slope of QR, we can solve for the unknown variable k. The calculated value for k is 1.

🎯 Exam Tip: The condition for collinearity of three points is that the slope of any two segments formed by these points must be equal. Always verify your answer by calculating all slopes if time permits.