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

Question 1. Find the slopes of the lines passing through the following points:

(i) (1, 2), (3, -5)
(ii) (1, 3), (5, 2)
(iii) (-1, 3), (3, -1)
(iv) (2, -5), (3, -1)
Answer: Solution:
(i) Let A = (1, 2) = \((x_1, y_1)\) and B = (3, -5) = \((x_2, y_2)\) say.
Slope of line AB = \( \frac{y_2-y_1}{x_2-x_1} = \frac{-5-2}{3-1} = \frac{-7}{2} \)
(ii) Let C = (1, 3) = \((x_1, y_1)\) and D = (5, 2) = \((x_2, y_2)\) say.
Slope of line CD = \( \frac{y_2-y_1}{x_2-x_1} = \frac{2-3}{5-1} = \frac{-1}{4} \)
(iii) Let E = (-1, 3) = \((x_1, y_1)\) and F = (3, -1) = \((x_2, y_2)\) say.
Slope of line EF = \( \frac{y_2-y_1}{x_2-x_1} = \frac{-1-3}{3-(-1)} = \frac{-4}{4} = -1 \)
(iv) Let P = (2, -5) = \((x_1, y_1)\) and Q = (3, -1) = \((x_2, y_2)\) say.
Slope of line PQ = \( \frac{y_2-y_1}{x_2-x_1} = \frac{-1-(-5)}{3-2} = \frac{-1+5}{1} = 4 \)
In simple words: The slope of a line between two points is calculated by the change in y-coordinates divided by the change in x-coordinates. For each given pair of points, we apply the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\) to find the slope.

🎯 Exam Tip: Remember the slope formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Correctly identifying \((x_1, y_1)\) and \((x_2, y_2)\) for each pair of points is crucial for accurate calculation.

Question 2. Find the slope of the line which
(i) makes an angle of 120° with the positive X-axis.
(ii) makes intercepts 3 and -4 on the axes.
(iii) passes through the points A(-2, 1) and the origin.
Answer: Solution:
(i) \( \theta = 120^\circ \)
Slope of the line = \( \tan 120^\circ \)
= \( \tan (180^\circ - 60^\circ) \)
= \( -\tan 60^\circ \) .....[\( \tan(180^\circ - \theta) = -\tan \theta \)]
= \( -\sqrt{3} \)
(ii) Given, x-intercept of line is 3 and y-intercept of line is -4
\( \therefore \) The line intersects X-axis at (3, 0) and Y-axis at (0, -4).
\( \therefore \) The line passes through (3, 0) = \((x_1, y_1)\) and (0, -4) = \((x_2, y_2)\) say.
\( \therefore \) Slope of line = \( \frac{y_2-y_1}{x_2-x_1} = \frac{-4-0}{0-3} = \frac{-4}{-3} = \frac{4}{3} \)
(iii) Required line passes through O(0, 0) = \((x_1, y_1)\) and A(-2, 1) = \((x_2, y_2)\) say.
Slope of line OA = \( \frac{y_2-y_1}{x_2-x_1} = \frac{1-0}{-2-0} = \frac{1}{-2} = -\frac{1}{2} \)
In simple words: The slope can be found using the angle of inclination (\(m = \tan \theta\)), or by using two points \((m = \frac{y_2 - y_1}{x_2 - x_1})\). For intercepts, the intercepts provide two points (x-intercept, 0) and (0, y-intercept) which can then be used with the two-point formula.

🎯 Exam Tip: Remember that the slope of a line making an angle \(\theta\) with the positive X-axis is \( \tan \theta \). For intercepts, convert them into coordinate points (x-intercept, 0) and (0, y-intercept) to use the two-point slope formula effectively.

Question 3. Find the value of k:
(i) if the slope of the line passing through the points (3, 4), (5, k) is 9.
(ii) the points (1, 3), (4, 1), (3, k) are collinear.
(iii) the point P(1, k) lies on the line passing through the points A(2, 2) and B(3, 3).
Answer: Solution:
(i) Let P(3, 4), Q(5, k).
Slope of PQ = 9 .......[Given]
\( \therefore \frac{k-4}{5-3} = 9 \)
\( \therefore \frac{k-4}{2} = 9 \)
\( \therefore k - 4 = 18 \)
\( \therefore k = 22 \)
(ii) The points A(1, 3), B(4, 1) and C(3, k) are collinear.
\( \therefore \) Slope of AB = Slope of BC
\( \therefore \frac{1-3}{4-1} = \frac{k-1}{3-4} \)
\( \frac{-2}{3} = \frac{k-1}{-1} \)
\( \therefore 2 = 3k - 3 \)
\( \therefore 5 = 3k \)
\( \therefore k = \frac{5}{3} \)
(iii) Given, point P(1, k) lies on the line joining A(2, 2) and B(3, 3).
\( \therefore \) Slope of AB = Slope of BP
\( \frac{3-2}{3-2} = \frac{3-k}{3-1} \)
\( \therefore 1 = \frac{3-k}{2} \)
\( \therefore 2 = 3-k \)
\( \therefore k = 1 \)
In simple words: To find 'k' when the slope is given, use the slope formula with the given points and set it equal to the known slope. For collinear points, set the slope between the first two points equal to the slope between the second and third points. If a point lies on a line, it means the slope between any two points on that line, including the given point, will be the same.

🎯 Exam Tip: For problems involving 'k' and slopes, ensure algebraic manipulation is precise. When points are collinear, the equality of slopes is the key condition; choose any two pairs of points to form the slope equation correctly.

Question 4. Reduce the equation \(6x + 3y + 8 = 0\) into slope-intercept form. Hence, find its slope.
Answer: Solution:
Given equation is \(6x + 3y + 8 = 0\), which can be written as
\(3y = -6x - 8\)
\( \therefore y = \frac{-6x}{3} - \frac{8}{3} \)
\( \therefore y = -2x - \frac{8}{3} \)
This is of the form \(y = mx + c\) with \(m = -2\)
\( \therefore y = -2x - \frac{8}{3} \) is in slope-intercept form with slope = -2
In simple words: To convert an equation into slope-intercept form (\(y = mx + c\)), you need to isolate 'y' on one side of the equation. Once 'y' is isolated, the coefficient of 'x' (which is 'm') will be the slope of the line.

🎯 Exam Tip: Always rearrange the equation to \(y = mx + c\) form to easily identify the slope 'm' and y-intercept 'c'. Be careful with signs when moving terms across the equality.

Question 5. Verify that A(2, 7) is not a point on the line \(x + 2y + 2 = 0\).
Answer: Solution:
Given equation is \(x + 2y + 2 = 0\).
Substituting \(x = 2\) and \(y = 7\) in L.H.S. of given equation, we get
L.H.S. = \(x + 2y + 2\)
= \(2 + 2(7) + 2\)
= \(2 + 14 + 2\)
= \(18\)
\( \neq \) R.H.S.
\( \therefore \) Point A does not lie on the given line.
In simple words: To check if a point lies on a line, substitute its coordinates into the line's equation. If the equation holds true (L.H.S. equals R.H.S.), the point is on the line; otherwise, it is not.

🎯 Exam Tip: This type of verification requires careful substitution of the x and y coordinates into the given equation. A mismatch between the LHS and RHS proves the point is not on the line.

Question 6. Find the X-intercept of the line \(x + 2y - 1 = 0\).
Answer: Solution:
Given equation of the line is \(x + 2y - 1 = 0\)
To find the x-intercept, put \(y = 0\) in given equation of the line
\( \therefore x + 2(0) - 1 = 0 \)
\( \therefore x + 0 - 1 = 0 \)
\( \therefore x = 1 \)
\( \therefore \) X-intercept of the given line is 1.
Alternate method:
Given equation of the line is \(x + 2y - 1 = 0\)
i.e. \(x + 2y = 1\)
\( \therefore \frac{x}{1} + \frac{2y}{1} = 1 \)
\( \therefore \frac{x}{1} + \frac{y}{1/2} = 1 \)
Comparing with \( \frac{x}{a} + \frac{y}{b} = 1 \), we get \(a = 1\)
X-intercept of the line is 1.
In simple words: The x-intercept is the point where the line crosses the X-axis. At this point, the y-coordinate is always zero. So, substitute \(y = 0\) into the line's equation and solve for 'x'. Alternatively, rewrite the equation in intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \), where 'a' is the x-intercept.

🎯 Exam Tip: To find the x-intercept, set \(y = 0\). To find the y-intercept, set \(x = 0\). The intercept form of a line \( \frac{x}{a} + \frac{y}{b} = 1 \) provides 'a' as the x-intercept and 'b' as the y-intercept directly.

Question 7. Find the slope of the line \(y - x + 3 = 0\).
Answer: Solution:
Equation of given line is \(y - x + 3 = 0\)
i.e. \(y = x - 3\)
Comparing with \(y = mx + c\), we get
\(m\) = Slope = 1
In simple words: To find the slope of a line from its equation, rearrange the equation into the slope-intercept form (\(y = mx + c\)). The coefficient of 'x' (which is 'm') will be the slope.

🎯 Exam Tip: Always transform the given linear equation into the standard slope-intercept form \(y = mx + c\). The value of 'm' directly gives the slope, and 'c' is the y-intercept.

Question 8. Does point A(2, 3) lie on the line \(3x + 2y - 6 = 0\)? Give reason.
Answer: Solution:
Given equation is \(3x + 2y - 6 = 0\).
Substituting \(x = 2\) and \(y = 3\) in L.H.S. of given equation, we get
L.H.S. = \(3x + 2y - 6\)
= \(3(2) + 2(3) - 6\)
= \(6 + 6 - 6\)
= \(6\)
\( \neq \) R.H.S. (which is 0)
\( \therefore \) Point A does not lie on the given line.
In simple words: A point lies on a line if its coordinates satisfy the line's equation. By substituting the point's x and y values into the equation, we check if the equation remains true. If it doesn't, the point is not on the line.

🎯 Exam Tip: To check if a point lies on a line, substitute its coordinates into the line's equation. If the left-hand side equals the right-hand side, the point lies on the line; otherwise, it does not.

Question 9. Which of the following lines passes through the origin?
(a) \(x = 2\)
(b) \(y = 3\)
(c) \(y = x + 2\)
(d) \(2x - y = 0\)
Answer: Solution:
Any line passing through origin is of the form \(y = mx\) or \(ax + by = 0\).
Here in the given option, \(2x - y = 0\) is in the form \(ax + by = 0\).
Answer: (d) \(2x - y = 0\)
In simple words: A line passes through the origin (0,0) if substituting \(x=0\) and \(y=0\) into its equation makes the equation true. Equations of lines passing through the origin have no constant term, meaning they can be written as \(ax + by = 0\) or \(y = mx\).

🎯 Exam Tip: For a line to pass through the origin (0,0), its equation must be satisfied when \(x=0\) and \(y=0\). This implies that there should be no constant term in the equation, like \(y = mx\) or \(ax + by = 0\).

Question 10. Obtain the equation of the line which is:
(i) parallel to the X-axis and 3 units below it.
(ii) parallel to the Y-axis and 2 units to the left of it.
(iii) parallel to the X-axis and making an intercept of 5 on the Y-axis.
(iv) parallel to the Y-axis and making an intercept of 3 on the X-axis.
Answer: Solution:
(i) Equation of a line parallel to X-axis is \(y = k\).
Since, the line is at a distance of 3 units below X-axis.
\( \therefore k = -3 \)
\( \therefore \) the equation of the required line is \(y = -3\)
i.e., \(y + 3 = 0\).
(ii) Equation of a line parallel to Y-axis is \(x = h\).
Since, the line is at a distance of 2 units to the left of Y-axis.
\( \therefore h = -2 \)
\( \therefore \) the equation of the required line is \(x = -2\)
i.e., \(x + 2 = 0\).
(iii) Equation of a line parallel to X-axis with y-intercept 'k' is \(y = k\).
Here, y-intercept = 5
\( \therefore \) the equation of the required line is \(y = 5\).
(iv) Equation of a line parallel to Y-axis with x-intercept 'h' is \(x = h\).
Here, x-intercept = 3
\( \therefore \) the equation of the required line is \(x = 3\).
In simple words: A line parallel to the X-axis has an equation \(y=k\), where 'k' is its y-intercept (distance from X-axis). A line parallel to the Y-axis has an equation \(x=h\), where 'h' is its x-intercept (distance from Y-axis). Distances below X-axis or to the left of Y-axis imply negative values for 'k' or 'h' respectively.

🎯 Exam Tip: Remember that horizontal lines are \(y = k\) and vertical lines are \(x = h\). The values of 'k' and 'h' are determined by their respective intercepts or distances from the axes, with signs indicating direction.

Question 11. Obtain the equation of the line containing the point:
(i) (2, 3) and parallel to the X-axis.
(ii) (2, 4) and perpendicular to the Y-axis.
(iii) (2, 5) and perpendicular to the X-axis.
Answer: Solution:
(i) Equation of a line parallel to X-axis is of the form \(y = k\).
Since, the line passes through (2, 3).
\( \therefore k = 3 \)
\( \therefore \) the equation of the required line is \(y = 3\).
(ii) Equation of a line perpendicular to Y-axis
i.e., parallel to X-axis, is of the form \(y = k\).
Since, the line passes through (2, 4).
\( \therefore k = 4 \)
\( \therefore \) the equation of the required line is \(y = 4\).
(iii) Equation of a line perpendicular to X-axis
i.e., parallel to Y-axis, is of the form \(x = h\).
Since, the line passes through (2, 5).
\( \therefore h = 2 \)
\( \therefore \) the equation of the required line is \(x = 2\).
In simple words: A line parallel to the X-axis (or perpendicular to the Y-axis) is a horizontal line with equation \(y=k\). If it passes through a point \((x_0, y_0)\), then \(k=y_0\). A line parallel to the Y-axis (or perpendicular to the X-axis) is a vertical line with equation \(x=h\). If it passes through \((x_0, y_0)\), then \(h=x_0\).

🎯 Exam Tip: Understand the relationship between parallel/perpendicular to axes and the form of the line equation. Horizontal lines are \(y = \text{constant}\) and vertical lines are \(x = \text{constant}\). The passing point directly determines the constant value.

Question 12. Find the equation of the line:
(i) having slope 5 and containing point A(-1, 2).
(ii) containing the point (2, 1) and having slope 13.
(iii) containing the point T(7, 3) and having inclination 90°.
(iv) containing the origin and having inclination 90°.
(v) through the origin which bisects the portion of the line \(3x + 2y = 2\) intercepted between the co-ordinate axes.
Answer: Solution:
(i) Given, slope (m) = 5 and the line passes through A(-1, 2).
Equation of the line in slope point form is \(y - y_1 = m(x - x_1)\)
\( \therefore \) the equation of the required line is \(y - 2 = 5(x - (-1))\)
\( \therefore y - 2 = 5(x + 1) \)
\( \therefore y - 2 = 5x + 5 \)
\( \therefore 5x - y + 7 = 0 \)
(ii) Given, slope (m) = 13 and the line passes through (2, 1).
Equation of the line in slope point form is \(y - y_1 = m(x - x_1)\)
\( \therefore \) the equation of the required line is \(y - 1 = 13(x - 2)\)
\( \therefore y - 1 = 13x - 26 \)
\( \therefore 13x - y = 25 \).
(iii) Given, Inclination of line = \( \theta = 90^\circ \)
\( \therefore \) the required line is parallel to Y-axis (or lies on the Y-axis.)
Equation of a line parallel to Y-axis is of the form \(x = h\).
Since, the line passes through (7, 3).
\( \therefore h = 7 \)
\( \therefore \) the equation of the required line is \(x = 7\).
(iv) Given, Inclination of line = \( \theta = 90^\circ \)
\( \therefore \) the required line is parallel to Y-axis (or lies on the Y-axis.)
Equation of a line parallel to Y-axis is of the form \(x = h\).
Since, the line passes through origin (0, 0).
\( \therefore h = 0 \)
\( \therefore \) the equation of the required line is \(x = 0\).
(v) Given equation of the line is \(3x + 2y = 2\).
\( \therefore \frac{3x}{2} + \frac{2y}{2} = 1 \)
\( \therefore \frac{x}{2/3} + \frac{y}{1} = 1 \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक कार्तीय निर्देशांक प्रणाली दिखाता है। X-अक्ष और Y-अक्ष केंद्र पर प्रतिच्छेद करते हैं। एक रेखा B(0, 1) पर Y-अक्ष को और A(2/3, 0) पर X-अक्ष को काटती हुई दिखाई गई है। एक "required line" (वांछित रेखा) मूल बिंदु (0,0) से गुजरती है और बिंदुओं A और B के मध्यबिंदु को काटती है।
This equation is of the form \( \frac{x}{a} + \frac{y}{b} = 1 \), with
\(a = \frac{2}{3}\), \(b = 1\)
\( \therefore \) the line \(3x + 2y = 2\) intersects the X-axis at A\((\frac{2}{3}, 0)\) and Y-axis at B(0, 1).
Required line is passing through the midpoint of AB.
\( \therefore \) Midpoint of AB = \( (\frac{\frac{2}{3}+0}{2}, \frac{0+1}{2}) = (\frac{1}{3}, \frac{1}{2}) \)
\( \therefore \) Required line passes through (0, 0) and \( (\frac{1}{3}, \frac{1}{2}) \).
Equation of the line in two point form is
\( \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1} \)
\( \therefore \) the equation of the required line is
\( \frac{y-0}{\frac{1}{2}-0} = \frac{x-0}{\frac{1}{3}-0} \)
\( \therefore \frac{y}{\frac{1}{2}} = \frac{x}{\frac{1}{3}} \)
\( \therefore 2y = 3x \)
\( \therefore 3x - 2y = 0 \)
In simple words: The equation of a line can be found using the point-slope form (\(y - y_1 = m(x - x_1)\)) when a point and slope are known. If the inclination is 90°, the line is vertical (\(x=h\)). For a line bisecting a segment between axes, first find the intercepts, then the midpoint, and finally use the two-point form with the origin and midpoint.

🎯 Exam Tip: Master the different forms of line equations (point-slope, slope-intercept, intercept form) and when to apply each. Special attention should be given to lines parallel to axes (vertical/horizontal) and how to handle inclinations like 90 degrees.

Question 13. Find the equation of the line passing through the points A(-3, 0) and B(0, 4).
Answer: Solution:
Since, the required line passes through the points A(-3, 0) and B(0, 4).
Equation of the line in two point form is
\( \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1} \)
Here, \((x_1, y_1) = (-3, 0)\) and \((x_2, y_2) = (0, 4)\)
\( \therefore \) the equation of the required line is
\( \therefore \frac{y-0}{4-0} = \frac{x-(-3)}{0-(-3)} \)
\( \therefore \frac{y}{4} = \frac{x+3}{3} \)
\( \therefore 3y = 4(x + 3) \)
\( \therefore 3y = 4x + 12 \)
\( \therefore 4x - 3y + 12 = 0 \)
In simple words: To find the equation of a line passing through two given points, use the two-point form formula \(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\). Substitute the coordinates of the two points and then simplify the equation into the standard linear form.

🎯 Exam Tip: The two-point form is effective for finding a line's equation when two points are known. Ensure correct substitution of coordinates and simplify the resulting equation to its standard form \(Ax + By + C = 0\).

Question 14. Find the equation of the line:
(i) having slope 5 and making intercept 5 on the X-axis.
(ii) having an inclination 60° and making intercept 4 on the Y-axis.
Answer: Solution:
(i) Since, the x-intercept of the required line is 5.
\( \therefore \) it passes through (5, 0).
Also, slope(m) of the line is 5
Equation of the line in slope point form is \(y - y_1 = m(x - x_1)\)
\( \therefore \) the equation of the required line is \(y - 0 = 5(x - 5)\)
\( \therefore y = 5x - 25 \)
\( \therefore 5x - y - 25 = 0 \)
(ii) Given, Inclination of line = \( \theta = 60^\circ \)
\( \therefore \) Slope of the line (m) = \( \tan \theta \)
= \( \tan 60^\circ \)
= \( \sqrt{3} \)
and the y-intercept of the required line is 4.
\( \therefore \) it passes through (0, 4).
Equation of the line in slope point form is \(y - y_1 = m(x - x_1)\)
\( \therefore \) the equation of the required line is \(y - 4 = \sqrt{3}(x - 0) \)
\( \therefore y - 4 = \sqrt{3}x \)
\( \therefore \sqrt{3}x - y + 4 = 0 \)
In simple words: If you have the slope and an x-intercept, you have a point \((x_{intercept}, 0)\) and the slope, so use the point-slope form. If you have the inclination, calculate the slope using \(m = \tan \theta\), and if you have the y-intercept, you have a point \((0, y_{intercept})\), again allowing use of the point-slope form.

🎯 Exam Tip: The slope-point form \(y - y_1 = m(x - x_1)\) is versatile. An x-intercept \('a'\) gives the point \((a, 0)\), and a y-intercept \('b'\) gives \((0, b)\). An inclination \(\theta\) means the slope is \(m = \tan \theta\). Combine these facts to use the point-slope form effectively.

Question 15. The vertices of a triangle are A(1, 4), B(2, 3), and C(1, 6). Find equations of
(i) the sides
(ii) the medians
(iii) Perpendicular bisectors of sides
(iv) altitudes of \( \triangle ABC \)
Answer: Solution:
Vertices of \( \triangle ABC \) are A(1, 4), B(2, 3), and C(1, 6)
(i) Equation of the line in two-point form is
\( \frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1} \)
Equation of side AB is
\( \frac{y-4}{3-4} = \frac{x-1}{2-1} \)
\( \therefore \frac{y-4}{-1} = \frac{x-1}{1} \)
\( \therefore y-4 = -1(x-1) \)
\( \therefore y-4 = -x+1 \)
\( \therefore x+y = 5 \)
Equation of side BC is
\( \frac{y-3}{6-3} = \frac{x-2}{1-2} \)
\( \therefore \frac{y-3}{3} = \frac{x-2}{-1} \)
\( \therefore -1(y-3) = 3(x-2) \)
\( \therefore -y+3 = 3x-6 \)
\( \therefore 3x+y = 9 \)
Since, both the points A and C have same x co-ordinates i.e. 1
\( \therefore \) the points A and C lie on a line parallel to Y-axis.
\( \therefore \) the equation of side AC is \(x = 1\).
(ii) Let D, E, and F be the midpoints of sides BC, AC, and AB respectively of \( \triangle ABC \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसके शीर्ष A(1, 4), B(2, 3) और C(1, 6) हैं। चित्र में D, E, और F क्रमशः भुजाओं BC, AC और AB के मध्यबिंदु दिखाए गए हैं। F बिंदु AB पर, E बिंदु AC पर, और D बिंदु BC पर स्थित है।
Then D = \( (\frac{2+1}{2}, \frac{3+6}{2}) = (\frac{3}{2}, \frac{9}{2}) \)
E = \( (\frac{1+1}{2}, \frac{4+6}{2}) = (\frac{2}{2}, \frac{10}{2}) = (1, 5) \)
F = \( (\frac{1+2}{2}, \frac{4+3}{2}) = (\frac{3}{2}, \frac{7}{2}) \)
Equation of median AD is
\( \frac{y-4}{\frac{9}{2}-4} = \frac{x-1}{\frac{3}{2}-1} \)
\( \therefore \frac{y-4}{\frac{9-8}{2}} = \frac{x-1}{\frac{3-2}{2}} \)
\( \therefore \frac{y-4}{\frac{1}{2}} = \frac{x-1}{\frac{1}{2}} \)
\( \therefore y-4 = x-1 \)
\( \therefore x-y+3 = 0 \)
Equation of median BE is
\( \frac{y-3}{5-3} = \frac{x-2}{1-2} \)
\( \therefore \frac{y-3}{2} = \frac{x-2}{-1} \)
\( \therefore -1(y-3) = 2(x-2) \)
\( \therefore -y+3 = 2x-4 \)
\( \therefore 2x+y = 7 \)
Equation of median CF is
\( \frac{y-6}{\frac{7}{2}-6} = \frac{x-1}{\frac{3}{2}-1} \)
\( \therefore \frac{y-6}{\frac{7-12}{2}} = \frac{x-1}{\frac{3-2}{2}} \)
\( \therefore \frac{y-6}{\frac{-5}{2}} = \frac{x-1}{\frac{1}{2}} \)
\( \therefore \frac{y-6}{-5} = \frac{x-1}{1} \)
\( \therefore y-6 = -5(x-1) \)
\( \therefore y-6 = -5x+5 \)
\( \therefore 5x+y-11 = 0 \)
(iii) Slope of side BC = \( (\frac{6-3}{1-2}) = (\frac{3}{-1}) = -3 \)
\( \therefore \) Slope of perpendicular bisector of BC is \( \frac{1}{3} \) and the line passes through \( (\frac{3}{2}, \frac{9}{2}) \)
\( \therefore \) Equation of the perpendicular bisector of side BC is \( (y - \frac{9}{2}) = \frac{1}{3} (x - \frac{3}{2}) \)
\( \therefore \frac{2y-9}{2} = \frac{1}{3} (\frac{2x-3}{2}) \)
\( \therefore \frac{2y-9}{1} = \frac{2x-3}{3} \)
\( \therefore 3(2y - 9) = (2x - 3) \)
\( \therefore 6y - 27 = 2x - 3 \)
\( \therefore 2x - 6y + 24 = 0 \)
\( \therefore x - 3y + 12 = 0 \)
Since, both the points A and C have the same x co-ordinates i.e. 1
\( \therefore \) the points A and C lie on the line \(x = 1\).
AC is parallel to Y-axis and therefore, the perpendicular bisector of side AC is parallel to X-axis.
Since, the perpendicular bisector of side AC passes through E(1, 5).
\( \therefore \) the equation of the perpendicular bisector of side AC is \(y = 5\).
Slope of side AB = \( (\frac{3-4}{2-1}) = (\frac{-1}{1}) = -1 \)
\( \therefore \) Slope of perpendicular bisector of AB is 1 and the line passes through \( (\frac{3}{2}, \frac{7}{2}) \).
\( \therefore \) Equation of the perpendicular bisector of side AB is \( (y - \frac{7}{2}) = 1 (x - \frac{3}{2}) \)
\( \therefore \frac{2y-7}{2} = \frac{2x-3}{2} \)
\( \therefore 2y - 7 = 2x - 3 \)
\( \therefore 2x - 2y + 4 = 0 \)
\( \therefore x - y + 2 = 0 \)
(iv) Let AX, BY and CZ be the altitudes through the vertices A, B, and C respectively of \( \triangle ABC \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC दिखाता है जिसके शीर्ष A(1, 4), B(2, 3), और C(1, 6) हैं। चित्र में AX, BY, और CZ त्रिभुज की ऊँचाईयाँ (altitudes) हैं, जहाँ AX भुजा BC पर लंबवत है, BY भुजा AC पर लंबवत है, और CZ भुजा AB पर लंबवत है।
Slope of BC = -3
\( \therefore \) Slope of AX = \( \frac{1}{3} \) ....[\( \therefore AX \perp BC \)]
Since, altitude AX passes through (1, 4) and has slope \( \frac{1}{3} \)
\( \therefore \) equation of altitude AX is \(y - 4 = \frac{1}{3} (x - 1) \)
\( \therefore 3(y - 4) = x - 1 \)
\( \therefore 3y - 12 = x - 1 \)
\( \therefore x - 3y + 11 = 0 \)
Since, both the points A and C have the same x co-ordinates i.e. 1
\( \therefore \) the points A and C lie on the line \(x = 1\).
AC is parallel to Y-axis and therefore, altitude BY is parallel to X-axis.
Since, the altitude BY passes through B(2, 3).
\( \therefore \) the equation of altitude BY is \(y = 3\).
Also, slope of AB = -1
\( \therefore \) Slope of CZ = 1 .....[\( \therefore CZ \perp AB \)]
Since, altitude CZ passes through (1, 6) and has slope 1
\( \therefore \) equation of altitude CZ is \(y - 6 = 1(x - 1) \)
\( \therefore y - 6 = x - 1 \)
\( \therefore x - y + 5 = 0 \)
In simple words: To find the equations for sides, use the two-point form. For medians, find the midpoints of sides first, then use the two-point form for a vertex and its opposite midpoint. For perpendicular bisectors, find the midpoint of each side and the negative reciprocal of the side's slope, then use the point-slope form. For altitudes, find the slope of each side, use its negative reciprocal as the altitude's slope, and use the point-slope form with the opposite vertex.

🎯 Exam Tip: This comprehensive problem tests multiple line concepts. Organize your steps: first find side equations, then midpoints for medians/perpendicular bisectors, and slopes for altitudes. Remember perpendicular lines have slopes that are negative reciprocals of each other, and vertical/horizontal line properties are key.