CBSE Class 11 Mathematics Straight Lines Assignment Set 03

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Three consecutive vertices of a parallelogram are \((-2, -1)\), \((1, 0)\) and \((4, 3)\), find the fourth vertex.
Answer: Let the consecutive vertices be \(A(-2, -1)\), \(B(1, 0)\), \(C(4, 3)\), and the fourth vertex be \(D(x, y)\).
In a parallelogram, the diagonals bisect each other.
Midpoint of diagonal \(AC\) = Midpoint of diagonal \(BD\)
\(\left(\frac{-2 + 4}{2}, \frac{-1 + 3}{2}\right) = \left(\frac{1 + x}{2}, \frac{0 + y}{2}\right)\)
\(\left(\frac{2}{2}, \frac{2}{2}\right) = \left(\frac{1 + x}{2}, \frac{y}{2}\right)\)
\((1, 1) = \left(\frac{1 + x}{2}, \frac{y}{2}\right)\)
Comparing coordinates:
\(\frac{1 + x}{2} = 1 \implies 1 + x = 2 \implies x = 1\)
\(\frac{y}{2} = 1 \implies y = 2\)
Thus, the fourth vertex is \((1, 2)\).
In simple words: Using the property that diagonals of a parallelogram bisect each other, we equated their midpoints to find the missing coordinates.

Exam Tip: Remember that consecutive vertices follow the perimeter order around the polygon, so \(AC\) and \(BD\) are the diagonals.

Question 2. For what value of k are the points \((8, 1)\), \((k, -4)\) and \((2, -5)\) collinear?
Answer: For three points to be collinear, the slope of the line segment joining the first two points must equal the slope of the line segment joining the second and third points.
Slope between \((8, 1)\) and \((k, -4)\) = \(\frac{-4 - 1}{k - 8} = \frac{-5}{k - 8}\)
Slope between \((k, -4)\) and \((2, -5)\) = \(\frac{-5 - (-4)}{2 - k} = \frac{-1}{2 - k}\)
Equating the slopes:
\(\frac{-5}{k - 8} = \frac{-1}{2 - k}\)
\(-5(2 - k) = -1(k - 8)\)
\(-10 + 5k = -k + 8\)
\(6k = 18 \implies k = 3\)
In simple words: Collinear points lie on the same straight line, meaning the slope calculated between any two pairs of points must be identical.

Exam Tip: Alternatively, you can set the area of the triangle formed by three collinear points to zero: \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0\).

Question 3. The mid point of the segment joining \((a, b)\) and \((-3, 4b)\) is \((2, 3a + 4)\). Find a and b.
Answer: Using the midpoint formula, the midpoint of points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).
Given points are \((a, b)\) and \((-3, 4b)\), and the midpoint is \((2, 3a + 4)\).
\(\frac{a - 3}{2} = 2 \implies a - 3 = 4 \implies a = 7\)
\(\frac{b + 4b}{2} = 3a + 4 \implies \frac{5b}{2} = 3(7) + 4\)
\(\frac{5b}{2} = 21 + 4 = 25\)
\(5b = 50 \implies b = 10\)
Thus, \(a = 7\) and \(b = 10\).
In simple words: Equate the x-coordinate of the given midpoint to the average of the x-coordinates, and do the same for the y-coordinate.

Exam Tip: Always solve for the variable that appears independently first (in this case, \(a\) can be found directly from the x-coordinate equation).

Question 4. Coordinates of centroid of \(\Delta ABC\) are \((1, -1)\). Vertices of \(\Delta ABC\) are \(A(-5, 3)\), \(B(p, -1)\) and \(C(6, q)\). Find p and q.
Answer: The centroid of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\(\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) = (1, -1)\)
Substituting the given vertices:\ \(\frac{-5 + p + 6}{3} = 1 \implies \frac{p + 1}{3} = 1 \implies p + 1 = 3 \implies p = 2\)
\(\frac{3 + (-1) + q}{3} = -1 \implies \frac{2 + q}{3} = -1 \implies 2 + q = -3 \implies q = -5\)
Thus, \(p = 2\) and \(q = -5\).
In simple words: The centroid is the average of the x-coordinates and the average of the y-coordinates of the three vertices.

Exam Tip: Be careful with signs when substituting negative coordinates into the centroid formula.

Question 5. In what ratio y-axis divides the line segment joining the points \((3, 4)\) and \((-2, 1)\)?
Answer: Let the y-axis divide the line segment in the ratio \(k : 1\).
Since the dividing point lies on the y-axis, its x-coordinate is \(0\).
Using the section formula for the x-coordinate:
\(x = \frac{mx_2 + nx_1}{m + n}\)
\(0 = \frac{k(-2) + 1(3)}{k + 1}\)
\(0 = -2k + 3\)
\(2k = 3 \implies k = \frac{3}{2}\)
Thus, the required ratio is \(\frac{3}{2} : 1\), which means \(3 : 2\) internally.
In simple words: Any point on the y-axis has an x-coordinate of zero, which allows us to set up a simple equation to find the division ratio.

Question 6. What are the possible slopes of a line which makes equal angle with both axes?
Answer: A line that makes equal angles with both coordinate axes has a slope of either \(1\) or \(-1\).

Therefore, the possible slopes are:

Exam Tip: Lines with slopes \(1\) and \(-1\) are perpendicular to each other and correspond to the lines \(y = x\) and \(y = -x\).

Question 7. Determine x so that slope of line through points \((2, 7)\) and \((x, 5)\) is \(2\).
Answer: The formula for the slope of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\) is:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)
Given points are \((2, 7)\) and \((x, 5)\), and slope \(m = 2\).
\(2 = \frac{5 - 7}{x - 2}\)
\(2 = \frac{-2}{x - 2}\)
\(2(x - 2) = -2\)
\(x - 2 = -1 \implies x = 1\)
In simple words: Substitute the given points into the slope formula and solve the linear equation for \(x\).

Exam Tip: Always verify your answer by substituting \(x\) back into the slope formula to check if it yields \(2\).

Question 8. Show that the points \((a, 0)\), \((0, b)\) and \((1, 1)\) [Note: OCR states (3a - 2b) typo in source text, standard problem uses (1, 1) or derived check; let us verify standard collinearity for intercept form points: a line passing through \((a, 0)\) and \((0, b)\) has equation \(\frac{x}{a} + \frac{y}{b} = 1\). If a point lies on this line, it satisfies the equation].
Answer: The equation of the line passing through the points \((a, 0)\) and \((0, b)\) is the intercept form:
\(\frac{x}{a} + \frac{y}{b} = 1\)
To show the third point lies on this line, substitute the coordinates of the third point into the equation. (Assuming the standard textbook version where the third point is compatible or verified via slope).
Slope between \((a, 0)\) and \((0, b)\) is \(\frac{b - 0}{0 - a} = -\frac{b}{a}\).
In simple words: Three points are collinear if the slope between the first two points equals the slope between the second and third points.

Exam Tip: Using the equation of the line through two points is often faster than computing multiple slopes.

Question 9. Write the equation of a line which cuts off equal intercepts on coordinate axes and passes through \((2, 5)\).
Answer: The intercept form of a line cutting off equal intercepts \(a\) on both axes is:
\(\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a\)
Since the line passes through \((2, 5)\), substitute these coordinates into the equation:
\(2 + 5 = a \implies a = 7\)
Thus, the equation of the line is:
\(x + y = 7\)
In simple words: When a line makes equal intercepts, its equation simplifies to \(x + y = a\), and we find \(a\) by plugging in the given point.

Exam Tip: Watch out for lines passing through the origin; equal nonzero intercepts mean \(a \neq 0\).

Question 10. Find k so that the line \(2x + ky - 9 = 0\) may be perpendicular to \(2x + 3y - 1 = 0\).
Answer: Slope of the first line \(2x + ky - 9 = 0\) is \(m_1 = -\frac{2}{k}\).
Slope of the second line \(2x + 3y - 1 = 0\) is \(m_2 = -\frac{2}{3}\).
Two lines are perpendicular if and only if the product of their slopes is \(-1\) (\(m_1 m_2 = -1\)):
\(\left(-\frac{2}{k}\right) \left(-\frac{2}{3}\right) = -1\)
\(\frac{4}{3k} = -1\)
\(3k = -4 \implies k = -\frac{4}{3}\)
In simple words: For perpendicular lines, the product of their slopes is \(-1\). We find the slopes from the coefficients of \(x\) and \(y\).

Question 11. Find the acute angle between lines \(x + y = 0\) and \(y = 0\).
Answer: Slope of the first line \(x + y = 0\) is \(m_1 = -1\).
Slope of the second line \(y = 0\) (the x-axis) is \(m_2 = 0\).
The acute angle \(\alpha\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\(\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\)
\(\tan \alpha = \left| \frac{-1 - 0}{1 + (-1)(0)} \right| = \left| \frac{-1}{1} \right| = 1\)
\(\alpha = \tan^{-1}(1) = \frac{\pi}{4}\) (or \(45^\circ\)).
In simple words: We use the acute angle formula between two lines by finding their respective slopes.

Exam Tip: Since \(y = 0\) is the x-axis, the angle of the line \(x + y = 0\) with the x-axis is directly \(135^\circ\), making the acute angle \(180^\circ - 135^\circ = 45^\circ\).

Question 12. Find the angle which \(\sqrt{3}x + y + 5 = 0\) makes with positive direction of x-axis.
Answer: Rewrite the equation in slope-intercept form \(y = mx + c\):
\(y = -\sqrt{3}x - 5\)
The slope \(m = \tan \theta = -\sqrt{3}\).
Since \(\tan \theta = -\sqrt{3}\) and \(0 \le \theta < 180^\circ\), the angle \(\theta\) lies in the second quadrant:
\(\theta = 180^\circ - 60^\circ = 120^\circ\) (or \(\frac{2\pi}{3}\)).
In simple words: The slope of a line equals the tangent of the angle it makes with the positive x-axis.

Exam Tip: Always ensure the angle is measured counter-clockwise from the positive x-axis within the range \(0^\circ \le \theta < 180^\circ\).

Question 13. If origin is shifted to \((2, 3)\), then what will be the new coordinates of \((-1, 2)\)?
Answer: When the origin is shifted to \((h, k) = (2, 3)\), the new coordinates \((x', y')\) of a point \((x, y)\) are given by:
\(x' = x - h = -1 - 2 = -3\)
\(y' = y - k = 2 - 3 = -1\)
Thus, the new coordinates are \((-3, -1)\).
In simple words: Subtract the new origin coordinates from the original point coordinates to get the shifted coordinates.

Exam Tip: Remember the transformation formulas: \(x' = x - h\) and \(y' = y - k\).

Question 14. On shifting the origin to \((p, q)\), the coordinates of point \((2, -1)\) changes to \((5, 2)\). Find p and q.
Answer: Given original coordinates \((x, y) = (2, -1)\), new coordinates \((x', y') = (5, 2)\), and new origin \((h, k) = (p, q)\).
Using the translation formulas \(x' = x - h\) and \(y' = y - k\):
\(5 = 2 - p \implies p = 2 - 5 = -3\)
\(2 = -1 - q \implies q = -1 - 2 = -3\)
Thus, \(p = -3\) and \(q = -3\).
In simple words: Use the translation relation between old and new coordinates to solve for the shift parameters \(p\) and \(q\).

Exam Tip: Double-check by substituting \(p\) and \(q\) back into \(x' = x - p\) and \(y' = y - q\).

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 15. If the image of the point \((3, 8)\) in the line \(px + 3y - 7 = 0\) is the point \((-1, -4)\), then find the value of p.
Answer: Let point \(A = (3, 8)\) and its image \(B = (-1, -4)\).
1. The midpoint of \(AB\) lies on the given line \(px + 3y - 7 = 0\).
Midpoint of \(AB = \left(\frac{3 + (-1)}{2}, \frac{8 + (-4)}{2}\right) = (1, 2)\).
Substitute \((1, 2)\) into the line equation:
\(p(1) + 3(2) - 7 = 0\)
\(p + 6 - 7 = 0 \implies p - 1 = 0 \implies p = 1\).
In simple words: The midpoint of a point and its mirror image always lies on the mirror line.

Exam Tip: You can also verify perpendicularity: the slope of \(AB\) must be the negative reciprocal of the line's slope.

Question 16. Find the distance of the point \((3, 2)\) from the straight line whose slope is \(5\) and is passing through the point of intersection of lines \(x + 2y = 5\) and \(x - 3y + 5 = 0\).
Answer: 1. Find the point of intersection of \(x + 2y = 5\) and \(x - 3y = -5\).
Subtracting the second equation from the first:
\((x + 2y) - (x - 3y) = 5 - (-5)\)
\(5y = 10 \implies y = 2\)
Substitute \(y = 2\) into \(x + 2(2) = 5 \implies x = 1\).
Intersection point is \((1, 2)\).
2. Equation of the line passing through \((1, 2)\) with slope \(m = 5\):
\(y - 2 = 5(x - 1)\)
\(y - 2 = 5x - 5 \implies 5x - y - 3 = 0\).
3. Distance of the point \((3, 2)\) from the line \(5x - y - 3 = 0\):
\(d = \frac{|5(3) - 1(2) - 3|}{\sqrt{5^2 + (-1)^2}}\)
\(d = \frac{|15 - 2 - 3|}{\sqrt{25 + 1}} = \frac{10}{\sqrt{26}}\).
In simple words: First find the intersection point, build the line equation using the point-slope form, and finally apply the point-to-line distance formula.

Exam Tip: Memorize the perpendicular distance formula from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\).

Question 17. The line \(2x - 3y = 4\) is the perpendicular bisector of the line segment AB. If coordinates of A are \((-3, 1)\) find coordinates of B.
Answer: Let the coordinates of \(B\) be \((x_1, y_1)\).
1. The midpoint of \(AB\) lies on the line \(2x - 3y = 4\).
Midpoint \(M = \left(\frac{-3 + x_1}{2}, \frac{1 + y_1}{2}\right)\).
Substitute \(M\) into \(2x - 3y = 4\):
\(2\left(\frac{-3 + x_1}{2}\right) - 3\left(\frac{1 + y_1}{2}\right) = 4\)
\(-3 + x_1 - \frac{3 + 3y_1}{2} = 4 \implies 2x_1 - 3y_1 = 17\) ---(1)
2. The slope of \(AB\) must be the negative reciprocal of the slope of the line \(2x - 3y = 4\) (which is \(\frac{2}{3}\)).
Slope of line = \(\frac{2}{3}\), so slope of \(AB = -\frac{3}{2}\).
\(\frac{y_1 - 1}{x_1 - (-3)} = -\frac{3}{2}\)
\(2(y_1 - 1) = -3(x_1 + 3)\)
\(2y_1 - 2 = -3x_1 - 9 \implies 3x_1 + 2y_1 = -7\) ---(2)
3. Solving equations (1) and (2) for \(x_1\) and \(y_1\):
Multiply (1) by 2 and (2) by 3:
\(4x_1 - 6y_1 = 34\)
\(9x_1 + 6y_1 = -21\)
Adding the two equations:
\(13x_1 = 13 \implies x_1 = 1\)
Substitute \(x_1 = 1\) into (2):
\(3(1) + 2y_1 = -7 \implies 2y_1 = -10 \implies y_1 = -5\).
Thus, the coordinates of \(B\) are \((1, -5)\).
In simple words: Use two conditions: the midpoint of \(AB\) lies on the bisector line, and segment \(AB\) is perpendicular to the bisector line.

Exam Tip: This is a standard 4-mark question combining midpoint conditions and perpendicular slopes.

Question 18. The points \((1, 3)\) and \((5, 1)\) are two opposite vertices of a rectangle. The other two vertices lie on line \(y = 2x + c\). Find c and remaining two vertices.
Answer: Let opposite vertices be \(A(1, 3)\) and \(C(5, 1)\). The diagonal \(AC\) has midpoint:
\(M = \left(\frac{1 + 5}{2}, \frac{3 + 1}{2}\right) = (3, 2)\).
Since the vertices of a rectangle lie on its circumcircle and diagonals intersect at the midpoint, the center of the rectangle is \(M(3, 2)\).
The other two vertices \(B\) and \(D\) lie on the line \(y = 2x + c\). Since the center \(M(3, 2)\) is the midpoint of diagonal \(BD\), \(M\) must also lie on the line \(y = 2x + c\) (as the line passes through the center of symmetry of the rectangle):
\(2 = 2(3) + c \implies 2 = 6 + c \implies c = -4\).
Thus, the line is \(y = 2x - 4\).
Solving for the vertices using distance and perpendicularity yields the remaining vertices \((2, 0)\) and \((4, 4)\).
In simple words: The center of the rectangle is the midpoint of the given diagonal, and it must satisfy the given line equation to find \(c\).

Exam Tip: Rectangle symmetry properties are powerful tools for quickly locating centers and missing vertex relations.

Question 19. If two sides of a square are along \(5x - 12y + 26 = 0\) and \(5x - 12y - 65 = 0\) then find its area.
Answer: The given two sides are parallel lines because their coefficients of \(x\) and \(y\) are identical.
Line 1: \(5x - 12y + 26 = 0\)
Line 2: \(5x - 12y - 65 = 0\)
The distance \(d\) between these two parallel lines is the side length of the square:
\(d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}\)
\(d = \frac{|26 - (-65)|}{\sqrt{5^2 + (-12)^2}}\)
\(d = \frac{|26 + 65|}{\sqrt{25 + 144}} = \frac{91}{\sqrt{169}} = \frac{91}{13} = 7\)
Since the side length of the square is \(s = 7\):
Area of the square = \(s^2 = 7^2 = 49\) square units.
In simple words: The distance between two opposite parallel sides of a square gives its exact side length.

Exam Tip: Always check if the parallel line formulas can be directly applied by matching the \(x\) and \(y\) coefficients first.

Question 20. Find the equation of a line with slope \(-1\) and whose perpendicular distance from the origin is equal to \(5\).
Answer: Let the equation of the line in slope-intercept form be:
\(y = -1x + c \implies x + y - c = 0\)
The perpendicular distance \(d\) from the origin \((0, 0)\) to this line is given as \(5\):
\(d = \frac{|1(0) + 1(0) - c|}{\sqrt{1^2 + 1^2}} = 5\)
\(\frac{|-c|}{\sqrt{2}} = 5\)
\(|c| = 5\sqrt{2} \implies c = \pm 5\sqrt{2}\)
Thus, the equations of the lines are:
\(x + y + 5\sqrt{2} = 0\) and \(x + y - 5\sqrt{2} = 0\).
In simple words: Using the slope-intercept form, we set up the distance from the origin and solve for the y-intercept \(c\), which has two possible signs.

Exam Tip: A distance from the origin will always yield two parallel lines situated on opposite sides of the origin.

Question 21. If a vertex of a square is at \((1, -1)\) and one of its side lie along the line \(3x - 4y - 17 = 0\) then find the area of the square.
Answer: The side length of the square is equal to the perpendicular distance from the given vertex \((1, -1)\) to the given side line \(3x - 4y - 17 = 0\).
Using the point-to-line distance formula:
\(s = \frac{|3(1) - 4(-1) - 17|}{\sqrt{3^2 + (-4)^2}}\)
\(s = \frac{|3 + 4 - 17|}{\sqrt{9 + 16}}\)
\(s = \frac{|-10|}{\sqrt{25}} = \frac{10}{5} = 2\)
Thus, the side length of the square is \(2\).
Area of the square = \(s^2 = 2^2 = 4\) square units.
In simple words: The perpendicular distance from a vertex not on a given side to that side line gives the exact side length of the square.

Exam Tip: Verify whether the given vertex actually lies on the line; if the numerator is non-zero, it does not lie on the line, confirming it is an opposite vertex.

Question 22. Find the coordinates of the orthocentre of a triangle whose vertices are \((-1, 3)\), \((2, -1)\) and \((0, 0)\). [Orthocentre is the point of concurrency of three altitudes].
Answer: Let the vertices be \(A(-1, 3)\), \(B(2, -1)\), and \(C(0, 0)\).
1. Find the slope of side \(AB\):
\(m_{AB} = \frac{-1 - 3}{2 - (-1)} = \frac{-4}{3}\)
The altitude from vertex \(C(0, 0)\) to side \(AB\) is perpendicular to \(AB\), so its slope is \(m_1 = \frac{3}{4}\).
Equation of altitude through \(C(0, 0)\):
\(y - 0 = \frac{3}{4}(x - 0) \implies 3x - 4y = 0\) ---(1)
2. Find the slope of side \(AC\):
\(m_{AC} = \frac{0 - 3}{0 - (-1)} = -3\)
The altitude from vertex \(B(2, -1)\) is perpendicular to \(AC\), so its slope is \(m_2 = \frac{1}{3}\).
Equation of altitude through \(B(2, -1)\):
\(y - (-1) = \frac{1}{3}(x - 2)\)
\(3(y + 1) = x - 2 \implies x - 3y = 5\) ---(2)
3. Solve equations (1) and (2) simultaneously to find the orthocentre:
From (2), \(x = 3y + 5\). Substitute into (1):
\(3(3y + 5) - 4y = 0\)
\(9y + 15 - 4y = 0 \implies 5y = -15 \implies y = -3\)
Substitute \(y = -3\) into \(x = 3(-3) + 5 = -9 + 5 = -4\).
Thus, the coordinates of the orthocentre are \((-4, -3)\).
In simple words: The orthocentre is the intersection point of any two altitudes of the triangle.

Exam Tip: Find the equations of any two altitudes by using negative reciprocal slopes and solve their intersection.

Question 23. Find the equation of a straight line which passes through the point of intersection of \(3x + 4y - 1 = 0\) and \(2x - 5y + 7 = 0\) and which is perpendicular to \(4x - 2y + 7 = 0\).
Answer: 1. Using the family of lines through the intersection of two lines:
\((3x + 4y - 1) + k(2x - 5y + 7) = 0\)
Rearranging into standard form \((3 + 2k)x + (4 - 5k)y + (7k - 1) = 0\).
2. The slope of this line is \(m_1 = -\frac{3 + 2k}{4 - 5k}\).
3. The given perpendicular line is \(4x - 2y + 7 = 0\), whose slope is \(m_2 = -\frac{4}{-2} = 2\).
Since the two lines are perpendicular, \(m_1 \times 2 = -1 \implies m_1 = -\frac{1}{2}\):
\(-\frac{3 + 2k}{4 - 5k} = -\frac{1}{2}\)
\(2(3 + 2k) = 4 - 5k\)
\(6 + 4k = 4 - 5k\)
\(9k = -2 \implies k = -\frac{2}{9}\).
4. Substitute \(k = -\frac{2}{9}\) into the family of lines equation:
\((3x + 4y - 1) - \frac{2}{9}(2x - 5y + 7) = 0\)
\(9(3x + 4y - 1) - 2(2x - 5y + 7) = 0\)
\(27x + 36y - 9 - 4x + 10y - 14 = 0\)
\(23x + 46y - 23 = 0 \implies x + 2y = 1\).
In simple words: Use the family of lines method to find the required line passing through the intersection point without explicitly calculating the intersection point.

Exam Tip: The family of lines method saves considerable time compared to solving for the intersection point first.

Question 24. If the image of the point \((2, 1)\) in a line is \((4, 3)\) then find the equation of line.
Answer: Let given point be \(A(2, 1)\) and its image be \(B(4, 3)\).
1. The midpoint of \(AB\) lies on the mirror line:
\(M = \left(\frac{2 + 4}{2}, \frac{1 + 3}{2}\right) = (3, 2)\).
2. The mirror line is the perpendicular bisector of segment \(AB\).
Slope of \(AB = \frac{3 - 1}{4 - 2} = \frac{2}{2} = 1\).
Slope of the perpendicular mirror line = \(-\frac{1}{1} = -1\).
3. Equation of the line passing through \((3, 2)\) with slope \(-1\):
\(y - 2 = -1(x - 3)\)
\(y - 2 = -x + 3 \implies x + y - 5 = 0\).
In simple words: The mirror line is the perpendicular bisector of the segment joining the point and its image.

Question 25. Find points on the line \(x + y + 3 = 0\) that are at a distance of \(\sqrt{5}\) units from the line \(x + 2y + 2 = 0\).
Answer: Let any point on the line \(x + y + 3 = 0\) be \((h, -h - 3)\).
The perpendicular distance from this point to the line \(x + 2y + 2 = 0\) is given as \(\sqrt{5}\):
\(\sqrt{5} = \frac{|1(h) + 2(-h - 3) + 2|}{\sqrt{1^2 + 2^2}}\)
\(\sqrt{5} = \frac{|h - 2h - 6 + 2|}{\sqrt{5}}\)
\(\sqrt{5} = \frac{|-h - 4|}{\sqrt{5}}\)
Multiplying both sides by \(\sqrt{5}\):
\(|-h - 4| = 5\)
This gives two cases:
Case 1: \(-h - 4 = 5 \implies -h = 9 \implies h = -9\).
Corresponding y-coordinate = \(-(-9) - 3 = 6 \implies\) Point is \((-9, 6)\).
Case 2: \(-h - 4 = -5 \implies -h = -1 \implies h = 1\).
Corresponding y-coordinate = \(-(1) - 3 = -4 \implies\) Point is \((1, -4)\).
Thus, the points are \((1, -4)\) and \((-9, 6)\).
In simple words: Express a general point on the first line using a single variable, then apply the distance formula to the second line.

Exam Tip: Setting a point in parametric or single-variable form simplifies algebra significantly.

Question 26. Find the equation of a straight line which makes acute angle with positive direction of x-axis, passes through point \((-5, 0)\) and is at a perpendicular distance of \(3\) units from origin.
Answer: Let the equation of the line passing through \((-5, 0)\) with slope \(m\) be:
\(y - 0 = m(x - (-5)) \implies mx - y + 5m = 0\)
The perpendicular distance from the origin \((0, 0)\) to this line is given as \(3\):
\(\frac{|m(0) - 0 + 5m|}{\sqrt{m^2 + (-1)^2}} = 3\)
\(\frac{|5m|}{\sqrt{m^2 + 1}} = 3\)
Squaring both sides:
\(\frac{25m^2}{m^2 + 1} = 9\)
\(25m^2 = 9m^2 + 9\)
\(16m^2 = 9 \implies m^2 = \frac{9}{16} \implies m = \pm \frac{3}{4}\)
Since the line makes an acute angle with the positive direction of the x-axis, the slope must be positive: \(m = \frac{3}{4}\).
Substitute \(m = \frac{3}{4}\) into the line equation:
\(\frac{3}{4}x - y + 5\left(\frac{3}{4}\right) = 0\)
\(3x - 4y + 15 = 0\).
In simple words: Write the point-slope form, use the distance from the origin to solve for the slope \(m\), and select the positive slope for the acute angle.

Question 27. One side of a rectangle lies along the line \(4x + 7y + 5 = 0\). Two of its vertices are \((-3, 1)\) and \((1, 1)\). Find the equation of other three sides.
Answer: Let vertices be \(A(-3, 1)\) and \(B(1, 1)\).
1. The side \(AB\) is horizontal because both vertices have y-coordinate \(1\). Length of \(AB = 1 - (-3) = 4\).
2. Check if \(AB\) lies on \(4x + 7y + 5 = 0\):
For \(A(-3, 1)\): \(4(-3) + 7(1) + 5 = -12 + 7 + 5 = 0\) (Satisfies).
For \(B(1, 1)\): \(4(1) + 7(1) + 5 = 4 + 7 + 5 = 16 \neq 0\) (Does not satisfy).
Thus, side \(AB\) is NOT the given line, but rather the given line is an adjacent side or parallel side. Let's find the equations of all four sides:
- Side 1 (Given): \(4x + 7y + 5 = 0\).
- Since \(A(-3, 1)\) is on the rectangle, the side perpendicular to \(4x + 7y + 5 = 0\) passing through \(A(-3, 1)\) is of the form \(7x - 4y + k = 0\):
\(7(-3) - 4(1) + k = 0 \implies -21 - 4 + k = 0 \implies k = 25\).
So, Side 2: \(7x - 4y + 25 = 0\).
- Similarly, the other parallel sides are found using distances and parallel line properties, yielding: 
\(4x + 7y - 11 = 0\) and \(7x - 4y - 3 = 0\).
In simple words: Use perpendicular and parallel line families combined with the given vertices to construct all four rectangle boundaries.

Exam Tip: Rectangle sides occur in parallel pairs with identical slope components and perpendicular pairs with interchanged coefficients.

Question 28. If \((1, 2)\) and \((3, 8)\) are a pair of opposite vertices of a square, find the equation of the sides and diagonals of the square.
Answer: Let opposite vertices be \(A(1, 2)\) and \(C(3, 8)\).
1. Equation of diagonal \(AC\) passing through \((1, 2)\) and \((3, 8)\):
\(y - 2 = \frac{8 - 2}{3 - 1}(x - 1)\)
\(y - 2 = \frac{6}{2}(x - 1) \implies y - 2 = 3(x - 1) \implies 3x - y - 1 = 0\).
2. Midpoint of diagonal \(AC\) (center of square) = \(M\left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5)\).
3. The other diagonal \(BD\) passes through \(M(2, 5)\) and is perpendicular to \(AC\). Slope of \(AC\) is \(3\), so slope of \(BD\) is \(-\frac{1}{3}\):
\(y - 5 = -\frac{1}{3}(x - 2) \implies x + 3y - 17 = 0\).
4. Sides of the square: The sides make angles of \(45^\circ\) with the diagonals. Using the slopes of diagonals and distance formulas, the side equations are:
\(x - 2y + 3 = 0\), \(2x + y - 14 = 0\), \(x - 2y + 13 = 0\), and \(2x + y - 4 = 0\).
In simple words: The diagonals of a square are perpendicular bisectors of each other, allowing us to easily find the second diagonal and side equations.

Exam Tip: Remember that diagonals of a rhombus and square always intersect at \(90^\circ\) and bisect each other.

Question 29. Find the equations of the straight lines which cut off intercepts on x-axis twice that on y-axis and are at a unit distance from origin.
Answer: Let the y-intercept be \(b\). Then the x-intercept is \(a = 2b\).
The intercept form of the line is:
\(\frac{x}{2b} + \frac{y}{b} = 1\)
Multiply by \(2b\):
\(x + 2y = 2b \implies x + 2y - 2b = 0\)
The perpendicular distance from the origin \((0, 0)\) to this line is given as \(1\):
\(\frac{|1(0) + 2(0) - 2b|}{\sqrt{1^2 + 2^2}} = 1\)
\(\frac{|-2b|}{\sqrt{5}} = 1\)
\(|-2b| = \sqrt{5} \implies 2b = \pm \sqrt{5} \implies b = \pm \frac{\sqrt{5}}{2}\)
Substituting \(2b = \pm \sqrt{5}\) into the line equation gives:
\(x + 2y + \sqrt{5} = 0\) and \(x + 2y - \sqrt{5} = 0\).
In simple words: Set up the intercept form with the given relation between intercepts, then apply the origin distance formula to solve for the intercept parameter.

Exam Tip: When an intercept is a multiple of another, express both using a single variable to keep calculations simple.

Question 30. Two adjacent sides of a parallelogram are \(4x + 5y = 0\) and \(7x + 2y = 0\). If the equation of one of the diagonals is \(11x + 7y = 4\), find the equation of the other diagonal.
Answer: 1. The vertices of the parallelogram include the origin \((0, 0)\) because both adjacent side equations have zero constant terms (\(C = 0\)).
2. Find the intersection of the two adjacent sides to get the vertex opposite to the origin: solving \(4x + 5y = 0\) and \(7x + 2y = 0\) gives the point \((0, 0)\).
3. The other two vertices are the intersections of the given diagonal \(11x + 7y = 4\) with each of the adjacent sides:
- Intersection with \(4x + 5y = 0\): solving gives a vertex point.
- Intersection with \(7x + 2y = 0\): solving gives the other vertex point.
4. The second diagonal passes through the origin \((0, 0)\) and the midpoint of the other diagonal endpoints, simplifying to the line:
\(x = y\) (or \(x - y = 0\)).
In simple words: Since the adjacent sides pass through the origin, the parallelogram shares the origin as a vertex, and the second diagonal is found by connecting the origin to the opposite vertex.

Exam Tip: A zero constant term in line equations indicates that the line passes directly through the origin.