CBSE Class 11 Mathematics Straight Lines Assignment Set 02

Chapter: Straight Lines and Conic Sections

Q1. What point on x-axis is equidistant from the points A (1, 3) and B (2,-5)
Answer: Let the required point on the x-axis be \(P(x, 0)\).
Since \(P\) is equidistant from \(A(1, 3)\) and \(B(2, -5)\), we have \(PA = PB\)
\(PA^2 = PB^2\)
\((x - 1)^2 + (0 - 3)^2 = (x - 2)^2 + (0 - (-5))^2\)
\(x^2 - 2x + 1 + 9 = x^2 - 4x + 4 + 25\)
\(-2x + 10 = -4x + 29\)
\(4x - 2x = 29 - 10\)
\(2x = 19\)
\(x = \frac{19}{2}\)
Thus, the required point is \(\left(\frac{19}{2}, 0\right)\).
In simple words: A point on the x-axis has a y-coordinate of zero. We equate the square of the distances from the given two points to solve for x.

Exam Tip: Always equate the squares of distances instead of square roots to simplify calculations easily.

Q2. If the points (a, 0), (0, b) and (1, 1) are collinear, show that \(\frac{1}{a} + \frac{1}{b} = 1\)
Answer: Since the three points \((a, 0)\), \((0, b)\), and \((1, 1)\) are collinear, the area of the triangle formed by them must be zero.
Area \(= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0\)
\(\frac{1}{2} |a(b - 1) + 0(1 - 0) + 1(0 - b)| = 0\)
\(ab - a - b = 0\)
\(ab = a + b\)
Dividing both sides by \(ab\), we get:
\(\frac{ab}{ab} = \frac{a}{ab} + \frac{b}{ab}\)
\(1 = \frac{1}{b} + \frac{1}{a}\)
\(\frac{1}{a} + \frac{1}{b} = 1\). Hence proved.
In simple words: When three points lie on the same straight line, the area of the triangle formed by them is always zero.

Exam Tip: You can also solve this by showing that the slope of the line joining the first two points is equal to the slope of the line joining the second and third points.

Q3. For what values of x, the area of the triangle formed the points (5, -1), (x, 4) and (6, 3) is 5.5 sq. units?
Answer: The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
Area \(= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
Given vertices: \((5, -1)\), \((x, 4)\), \((6, 3)\) and Area \(= 5.5\)
\(\frac{1}{2} |5(4 - 3) + x(3 - (-1)) + 6(-1 - 4)| = \frac{11}{2}\)
\(|5(1) + x(4) + 6(-5)| = 11\)
\(|5 + 4x - 30| = 11\)
\(|4x - 25| = 11\)
Removing the modulus, we get two cases:
Case 1: \(4x - 25 = 11 \implies 4x = 36 \implies x = 9\)
Case 2: \(4x - 25 = -11 \implies 4x = 14 \implies x = \frac{14}{4} = \frac{7}{2}\)
Thus, \(x = \frac{7}{2}\) or \(9\).
In simple words: Because of the absolute value (modulus), area calculations always yield two possible values for an unknown coordinate.

Exam Tip: Do not forget to consider both positive and negative values when removing the modulus sign.

Q4. A line cut off intercepts -3 and 4 on x and y-axis respectively. Find the slope and equation of the line.
Answer: Given x-intercept \(a = -3\) and y-intercept \(b = 4\).
The intercept form of a line is:
\(\frac{x}{a} + \frac{y}{b} = 1\)
\(\frac{x}{-3} + \frac{y}{4} = 1\)
Multiplying by 12 throughout:
\(-4x + 3y = 12\)
\(4x - 3y + 12 = 0\)
To find the slope, convert the equation into slope-intercept form (\(y = mx + c\)):
\(3y = 4x + 12\)
\(y = \frac{4}{3}x + 4\)
Thus, slope \(m = \frac{4}{3}\) and equation is \(4x - 3y + 12 = 0\).
In simple words: The intercept form directly gives the equation when the points where the line cuts the axes are known.

Exam Tip: Slope of a line given in intercept form \(\frac{x}{a} + \frac{y}{b} = 1\) is always \(-\frac{b}{a}\).

Q5. Find the equation of a line parallel to x-axis at a distance of 3 units above x-axis.
Answer: Any line parallel to the x-axis has a constant y-coordinate for all points on it.
Since it is at a distance of 3 units above the x-axis, every point on this line has a y-coordinate equal to 3.
Therefore, the equation of the line is \(y = 3\).
In simple words: Lines parallel to the x-axis have equations of the form y = k.

Exam Tip: If the line were below the x-axis, the equation would be \(y = -3\).

Q6. Find the value of c and m so that the line y = m x +c may pass through the points (-2, 3) and (4,-3)
Answer: Since the line \(y = mx + c\) passes through \((-2, 3)\), substitute \(x = -2\) and \(y = 3\):
\(3 = m(-2) + c \implies -2m + c = 3\) ---(Equation 1)
Since the line passes through \((4, -3)\), substitute \(x = 4\) and \(y = -3\):
\(-3 = m(4) + c \implies 4m + c = -3\) ---(Equation 2)
Subtracting Equation 1 from Equation 2:
\((4m + c) - (-2m + c) = -3 - 3\)
\(6m = -6 \implies m = -1\)
Substitute \(m = -1\) into Equation 1:
\(-2(-1) + c = 3\)
\(2 + c = 3 \implies c = 1\)
Thus, \(c = 1\) and \(m = -1\).
In simple words: Substitute the given coordinate points into the equation to form simultaneous linear equations in terms of m and c.

Exam Tip: Always verify your values of m and c by substituting them back into both original point equations.

Q7. Mid-points of the sides of a triangle are (2, 2), (2, 3) and (4, 6). Find the equations of the sides of a triangle.
Answer: Let the vertices of the triangle be \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\).
Let the given mid-points be \(D(2, 2)\), \(E(2, 3)\), and \(F(4, 6)\).
Using the mid-point relations:
\(\frac{x_1 + x_2}{2} = 2 \implies x_1 + x_2 = 4\) ---(1)
\(\frac{x_2 + x_3}{2} = 2 \implies x_2 + x_3 = 4\) ---(2)
\(\frac{x_3 + x_1}{2} = 4 \implies x_3 + x_1 = 8\) ---(3)
Adding (1), (2), and (3): \(2(x_1 + x_2 + x_3) = 16 \implies x_1 + x_2 + x_3 = 8\).
From this, \(x_3 = 4\), \(x_1 = 4\), \(x_2 = 0\).
Similarly for y-coordinates:
\(\frac{y_1 + y_2}{2} = 2 \implies y_1 + y_2 = 4\) ---(4)
\(\frac{y_2 + y_3}{2} = 3 \implies y_2 + y_3 = 6\) ---(5)
\(\frac{y_3 + y_1}{2} = 6 \implies y_3 + y_1 = 12\) ---(6)
Adding (4), (5), and (6): \(2(y_1 + y_2 + y_3) = 22 \implies y_1 + y_2 + y_3 = 11\).
From this, \(y_3 = 7\), \(y_1 = 5\), \(y_2 = -1\).
Vertices are \(A(4, 5)\), \(B(0, -1)\), and \(C(4, 7)\).
Now, find the equation of side \(AB\) passing through \((4, 5)\) and \((0, -1)\):
\(y - 5 = \frac{-1 - 5}{0 - 4}(x - 4) \implies y - 5 = \frac{-6}{-4}(x - 4) \implies y - 5 = \frac{3}{2}(x - 4)\)
\(2y - 10 = 3x - 12 \implies 3x - 2y = 2\)
Equation of side \(BC\) passing through \((0, -1)\) and \((4, 7)\):
\(y - (-1) = \frac{7 - (-1)}{4 - 0}(x - 0) \implies y + 1 = 2x \implies 2x - y = 1\)
Equation of side \(CA\) passing through \((4, 7)\) and \((4, 5)\):
Since x-coordinates are equal, the line is vertical: \(x = 4\).
Thus, the equations of the sides are \(3x - 2y = 2\), \(2x - y = 1\), and \(x = 4\).
In simple words: First find the triangle's vertices using the mid-point formulas, then find the equations of the lines joining each pair of vertices.

Exam Tip: Sides of a triangle are straight lines connecting its vertices. Use the two-point form to find their equations.

Q8. Find the equations of the medians of the triangle whose vertices are (2, 0), (0, 2) and (4, 6).
Answer: Let vertices be \(A(2, 0)\), \(B(0, 2)\), and \(C(4, 6)\).
Let \(D\) be the mid-point of \(BC\):
\(D = \left(\frac{0 + 4}{2}, \frac{2 + 6}{2}\right) = (2, 4)\)
Median \(AD\) passes through \(A(2, 0)\) and \(D(2, 4)\). Since x-coordinates are both 2, the equation is \(x = 2\).
Let \(E\) be the mid-point of \(AC\):
\(E = \left(\frac{2 + 4}{2}, \frac{0 + 6}{2}\right) = (3, 3)\)
Median \(BE\) passes through \(B(0, 2)\) and \(E(3, 3)\):
\(y - 2 = \frac{3 - 2}{3 - 0}(x - 0) \implies y - 2 = \frac{1}{3}x \implies 3y - 6 = x \implies x - 3y + 6 = 0\)
Let \(F\) be the mid-point of \(AB\):
\(F = \left(\frac{2 + 0}{2}, \frac{0 + 2}{2}\right) = (1, 1)\)
Median \(CF\) passes through \(C(4, 6)\) and \(F(1, 1)\):
\(y - 6 = \frac{1 - 6}{1 - 4}(x - 4) \implies y - 6 = \frac{-5}{-3}(x - 4) \implies y - 6 = \frac{5}{3}(x - 4)\)
\(3y - 18 = 5x - 20 \implies 5x - 3y = 2\)
Thus, equations of medians are \(x = 2\), \(5x - 3y = 2\), and \(x - 3y + 6 = 0\).
In simple words: A median connects a vertex to the mid-point of the opposite side. Find the mid-point first, then apply the two-point form.

Exam Tip: Double check that your mid-points are calculated correctly before finding the line equations.

Q9. Show that the points (a, 0) (0, b) and (3a, -2b) are collinear Also find the equation of the line containing them.
Answer: Let points be \(P(a, 0)\), \(Q(0, b)\), and \(R(3a, -2b)\).
Slope of line \(PQ\) \((m_1) = \frac{b - 0}{0 - a} = -\frac{b}{a}\).
Slope of line \(QR\) \((m_2) = \frac{-2b - b}{3a - 0} = \frac{-3b}{3a} = -\frac{b}{a}\).
Since \(m_1 = m_2\), the points are collinear.
Now, the equation of the line passing through \((a, 0)\) and \((0, b)\) is given by the intercept form:
\(\frac{x}{a} + \frac{y}{b} = 1\)
\(bx + ay = ab\).
In simple words: Collinearity is proven by showing that the slope between the first two points equals the slope between the second and third points.

Exam Tip: Intercept form \(\frac{x}{a} + \frac{y}{b} = 1\) is the quickest way to find a line equation when x and y intercepts are known.

Q10. Find the new coordinates of the point (3,-5) if origin is shifted to the point (2,3) by a transformation of axes.
Answer: Let old coordinates be \((x, y) = (3, -5)\) and the new origin be \((h, k) = (2, 3)\).
Transformation formulas are:
\(X = x - h\)
\(Y = y - k\)
\(X = 3 - 2 = 1\)
\(Y = -5 - 3 = -8\)
Thus, the new coordinates are \((1, -8)\).
In simple words: To find new coordinates after shifting the origin, subtract the new origin coordinates from the old coordinates: \(X = x - h\), \(Y = y - k\).

Exam Tip: Be very careful with negative signs when subtracting coordinates.

Q11. Find the equation of the line such that segment intercepted by the axes is divided by the points (-5, 4) in the ratio 1:2.
Answer: Let the x-intercept be \(a\) and y-intercept be \(b\).
The points where the line meets the coordinate axes are \(A(a, 0)\) and \(B(0, b)\).
The point \((-5, 4)\) divides the line segment \(AB\) internally in the ratio \(1:2$.
Using the section formula:\br /> \(x = \frac{m_2x_1 + m_1x_2}{m_1 + m_2} \implies -5 = \frac{2(a) + 1(0)}{1 + 2} \implies -5 = \frac{2a}{3} \implies 2a = -15 \implies a = -\frac{15}{2}\)
\(y = \frac{m_2y_1 + m_1y_2}{m_1 + m_2} \implies 4 = \frac{2(0) + 1(b)}{1 + 2} \implies 4 = \frac{b}{3} \implies b = 12\)
Substitute \(a\) and \(b\) into the intercept form equation:
\(\frac{x}{-\frac{15}{2}} + \frac{y}{12} = 1\)
\(-\frac{2x}{15} + \frac{y}{12} = 1\)
Multiplying by 60 throughout:
\(-8x + 5y = 60\)
\(8x - 5y + 60 = 0\).
In simple words: Use the axis intercept coordinates \((a,0)\) and \((0,b)\) in the section formula to determine \(a\) and \(b\).

Exam Tip: Intercepts on axes are always of the form \((a, 0)\) and \((0, b)\).

Q12. Find the equations of lines which cut off intercepts on the axes whose sum and product are 1 and -6 respectively.
Answer: Let intercepts be \(a\) and \(b\).
Given: \(a + b = 1\) and \(ab = -6\).
We know that \((a - b)^2 = (a + b)^2 - 4ab = (1)^2 - 4(-6) = 1 + 24 = 25\).
Therefore, \(a - b = \pm 5\).
Case 1: If \(a + b = 1\) and \(a - b = 5\):
Adding both equations: \(2a = 6 \implies a = 3\).
Subtracting equations: \(2b = -4 \implies b = -2\).
Equation of line 1: \(\frac{x}{3} + \frac{y}{-2} = 1 \implies 2x - 3y = 6\).
Case 2: If \(a + b = 1\) and \(a - b = -5\):
Adding both equations: \(2a = -4 \implies a = -2\).
Subtracting equations: \(2b = 6 \implies b = 3\).
Equation of line 2: \(\frac{x}{-2} + \frac{y}{3} = 1 \implies -3x + 2y = 6 \implies 3x - 2y + 6 = 0\).
Thus, the equations are \(2x - 3y = 6\) and \(3x - 2y + 6 = 0\).
In simple words: Form a quadratic equation or use sum-product relations to find individual intercept values, then write the intercept form.

Exam Tip: When sum and product are given, use the identity \((a-b)^2 = (a+b)^2 - 4ab\) to find the difference of intercepts.

Q13. Find the angle between the lines joining the points (3,-1) and (2,3) and the points (5,2) and (9,3)
Answer: Let line 1 pass through \((3, -1)\) and \((2, 3)\).
Slope of line 1 (\(m_1\)) \(= \frac{3 - (-1)}{2 - 3} = \frac{4}{-1} = -4\).
Let line 2 pass through \((5, 2)\) and \((9, 3)\).
Slope of line 2 (\(m_2\)) \(= \frac{3 - 2}{9 - 5} = \frac{1}{4}\).
Since \(m_1 \cdot m_2 = (-4) \cdot \left(\frac{1}{4}\right) = -1\), the two lines are perpendicular.
Therefore, the angle between the lines is \(90^\circ\).
In simple words: If the product of the slopes of two lines is -1, the lines are perpendicular, and the angle between them is 90 degrees.

Exam Tip: Always check for perpendicularity (\(m_1 m_2 = -1\)) or parallelism (\(m_1 = m_2\)) first to save time.

Q14. Find the angle between the lines whose equations are ax +by +c =0 and (a +b) x = (a-b) y
Answer: Equation 1: \(ax + by + c = 0 \implies y = -\frac{a}{b}x - \frac{c}{b}\).
Slope \(m_1 = -\frac{a}{b}\).
Equation 2: \((a + b)x = (a - b)y \implies y = \frac{a + b}{a - b}x\).
Slope \(m_2 = \frac{a + b}{a - b}\).
Let \(\theta\) be the angle between the lines. Then:
\(\tan\theta = \left| \frac{m_2 - m_1}{1 + m_1m_2} \right|\)
\(m_2 - m_1 = \frac{a + b}{a - b} - \left(-\frac{a}{b}\right) = \frac{a + b}{a - b} + \frac{a}{b} = \frac{b(a + b) + a(a - b)}{b(a - b)} = \frac{ab + b^2 + a^2 - ab}{b(a - b)} = \frac{a^2 + b^2}{b(a - b)}\)
\(1 + m_1m_2 = 1 + \left(-\frac{a}{b}\right)\left(\frac{a + b}{a - b}\right) = 1 - \frac{a(a + b)}{b(a - b)} = \frac{b(a - b) - a(a + b)}{b(a - b)} = \frac{ab - b^2 - a^2 - ab}{b(a - b)} = \frac{-(a^2 + b^2)}{b(a - b)}\)
\(\tan\theta = \left| \frac{\frac{a^2 + b^2}{b(a - b)}}{\frac{-(a^2 + b^2)}{b(a - b)}} \right| = |-1| = 1\)
\(\theta = 45^\circ\).
In simple words: Find the slopes of both lines and substitute them into the angle between two lines formula.

Exam Tip: Take care with algebraic signs when simplifying complex fractional slope expressions.

Q15. Find the equation of a line which passes through origin and making an angle \(60^\circ\) with the line \(x + \sqrt{3}y + 5 = 0\)
Answer: The given line is \(x + \sqrt{3}y + 5 = 0 \implies \sqrt{3}y = -x - 5 \implies y = -\frac{1}{\sqrt{3}}x - \frac{5}{\sqrt{3}}\).
Its slope \(m_1 = -\frac{1}{\sqrt{3}}\).
Let the slope of the required line passing through the origin be \(m_2 = m\).
Since the angle between them is \(60^\circ\):
\(\tan 60^\circ = \left| \frac{m - \left(-\frac{1}{\sqrt{3}}\right)}{1 + m\left(-\frac{1}{\sqrt{3}}\right)} \right|\)
\(\sqrt{3} = \pm \frac{m + \frac{1}{\sqrt{3}}}{1 - \frac{m}{\sqrt{3}}} = \pm \frac{\sqrt{3}m + 1}{\sqrt{3} - m}\)
Case 1: \(\sqrt{3} = \frac{\sqrt{3}m + 1}{\sqrt{3} - m}\)
\(\sqrt{3}(\sqrt{3} - m) = \sqrt{3}m + 1\)
\(3 - \sqrt{3}m = \sqrt{3}m + 1\)
\(2\sqrt{3}m = 2 \implies m = \frac{1}{\sqrt{3}}\)
Equation of line: \(y = \frac{1}{\sqrt{3}}x \implies x - \sqrt{3}y = 0\) (or \(x = \sqrt{3}y\)).
Case 2: \(\sqrt{3} = -\frac{\sqrt{3}m + 1}{\sqrt{3} - m}\)
\(\sqrt{3}(\sqrt{3} - m) = -\sqrt{3}m - 1\)
\(3 - \sqrt{3}m = -\sqrt{3}m - 1\)
\(3 = -1\) (Not possible)
Wait, let's re-evaluate Case 2 properly:
\(\sqrt{3}(\sqrt{3} - m) = -(\sqrt{3}m + 1) \implies 3 - \sqrt{3}m = -\sqrt{3}m - 1 \implies 3 = -1\) (invalid).
Let us check the other slope where line passes through origin and is vertical (\(x = 0\)), or recompute:
Let slope be \(m\). Angle formula: \(\tan 60^\circ = \left| \frac{m - m_1}{1 + mm_1} \right| \implies \sqrt{3} = \frac{m + \frac{1}{\sqrt{3}}}{1 - \frac{m}{\sqrt{3}}}\) or \(\sqrt{3} = -\frac{m + \frac{1}{\sqrt{3}}}{1 - \frac{m}{\sqrt{3}}}\).
If \(m = \infty\) (vertical line \(x = 0\)), slope of given line is \(-\frac{1}{\sqrt{3}}\) (angle \(30^\circ\) with negative x-axis). The angle between vertical line and this line is \(90^\circ - 30^\circ = 60^\circ\).
Thus, the two lines are \(x = 0\) and \(x = \sqrt{3}y\).
In simple words: Use the angle between two lines formula with \(\tan 60^\circ\) to find possible slopes, remembering that a vertical line has an undefined slope.

Exam Tip: When one slope is undefined, always test for a vertical line (\(x = c\)).

Q16. Find the equation of a line that passes through the intersection of \(4x + 3y = 6\), and \(3x + 4y = 8\) and whose slope is 1.
Answer: First, find the point of intersection of the two lines by solving them simultaneously:
\(4x + 3y = 6\) ---(1)
\(3x + 4y = 8\) ---(2)
Multiply (1) by 4 and (2) by 3:
\(16x + 12y = 24$
\(9x + 12y = 24$
Subtracting both: \(7x = 0 \implies x = 0\).
Substitute \(x = 0\) in (1): \(4(0) + 3y = 6 \implies 3y = 6 \implies y = 2\).
Intersection point is \((0, 2)\).
Now, find the equation of the line passing through \((0, 2)\) with slope \(m = 1\):
\(y - y_1 = m(x - x_1)\)
\(y - 2 = 1(x - 0)\)
\(x - y + 2 = 0\).
In simple words: Find the intersection point by solving the two equations, then use the point-slope form with slope 1.

Exam Tip: Alternatively, use the family of lines formula: \((4x + 3y - 6) + k(3x + 4y - 8) = 0\).

Q17. Find the equation of the line passing through the midpoint of the line segment joining the point (1, 3) and (2,-1) and parallel to the line \(3x - y = 7\)
Answer: Midpoint of the line segment joining \((1, 3)\) and \((2, -1)\) is:
\(M = \left(\frac{1 + 2}{2}, \frac{3 + (-1)}{2}\right) = \left(\frac{3}{2}, 1\right)\)
The given parallel line is \(3x - y = 7\), so its slope is \(m = 3\).
Since the required line is parallel, its slope is also \(m = 3\).
Equation of the line passing through \(\left(\frac{3}{2}, 1\right)\) with slope 3:
\(y - 1 = 3\left(x - \frac{3}{2}\right)\)
\(y - 1 = 3x - \frac{9}{2}\)
Multiplying by 2:
\(2y - 2 = 6x - 9\)
\(6x - 2y = 7 \implies 6x - 2y = 7\) (Note: OCR text had a comma typo, correct standard form is \(6x - 2y = 7\)).
In simple words: Parallel lines share the exact same slope. Find the midpoint first, then use the point-slope form.

Exam Tip: Parallel lines have identical slopes (\(m_1 = m_2\)).

Q18. Find the equation of the line passing through the midpoint of the line segment joining the point (3, 4) and (5,-2) and perpendicular to the line \(x + 3y = 8\)
Answer: Midpoint of the line segment joining \((3, 4)\) and \((5, -2)\) is:
\(M = \left(\frac{3 + 5}{2}, \frac{4 + (-2)}{2}\right) = (4, 1)\)
The given line is \(x + 3y = 8 \implies 3y = -x + 8 \implies y = -\frac{1}{3}x + \frac{8}{3}\).
Its slope is \(m_1 = -\frac{1}{3}\).
Since the required line is perpendicular to this line, its slope \(m_2\) is the negative reciprocal:
\(m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{1}{3}} = 3\)
Equation of the line passing through \((4, 1)\) with slope 3:
\(y - 1 = 3(x - 4)\)
\(y - 1 = 3x - 12\)
\(3x - y = 11\).
In simple words: Perpendicular lines have negative reciprocal slopes (\(m_1 \cdot m_2 = -1\)).

Exam Tip: Negative reciprocal of \(-\frac{1}{3}\) is \(+\frac{3}{1} = 3\).

Q19. Find the equation of the line perpendicular to the line \(3x - y = 5\) and at a distance of 4 units from the origin.
Answer: Any line perpendicular to \(3x - y = 5\) is of the form \(x + 3y + k = 0\) (interchanging coefficients of x and y and changing one sign).
We are given that the perpendicular distance of this line from the origin \((0, 0)\) is 4 units.
Using the perpendicular distance formula from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\):
\(d = \left| \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} \right|\)
\(4 = \left| \frac{1(0) + 3(0) + k}{\sqrt{1^2 + 3^2}} \right|\)
\(4 = \left| \frac{k}{\sqrt{10}} \right|\)
\(k = \pm 4\sqrt{10}\)
Thus, the required equation of the line is \(x + 3y \pm 4\sqrt{10} = 0\).
In simple words: A line perpendicular to \(Ax + By + C = 0\) has the form \(Bx - Ay + k = 0\). Use the distance from origin formula to find k.

Exam Tip: Distance from origin to \(Ax + By + C = 0\) is simply \(\frac{|C|}{\sqrt{A^2 + B^2}}\).

Q20. Find what the equation \(x^2 + xy - 3y^2 - y + 2 = 0\) becomes when the origin is shifted to the point \((1, 1)\)
Answer: Let old coordinates be \((x, y)\) and new coordinates be \((X, Y)\).
Since the origin is shifted to \((1, 1)\), the transformation relations are:
\(x = X + 1\)
\(y = Y + 1\)
Substitute these into the given equation \(x^2 + xy - 3y^2 - y + 2 = 0\):
\((X + 1)^2 + (X + 1)(Y + 1) - 3(Y + 1)^2 - (Y + 1) + 2 = 0\)
\((X^2 + 2X + 1) + (XY + X + Y + 1) - 3(Y^2 + 2Y + 1) - Y - 1 + 2 = 0\)
\(X^2 + 2X + 1 + XY + X + Y + 1 - 3Y^2 - 6Y - 3 - Y - 1 + 2 = 0\)
Grouping terms by degree:
\(X^2 + XY - 3Y^2 + (2X + X) + (Y - 6Y - Y) + (1 + 1 - 3 - 1 + 2) = 0\)
\(X^2 + XY - 3Y^2 + 3X - 6Y = 0\)
Wait, let us check the original equation transcription and constant terms carefully:
Original equation: \(x^2 + xy - 3y^2 - y + 2 = 0\).
Let's re-verify constant terms: \(1 + 1 - 3 - 1 + 2 = 0\).
So the transformed equation is \(X^2 + XY - 3Y^2 + 3X - 6Y = 0\).
In simple words: Replace every \(x\) with \(X + h\) and every \(y\) with \(Y + k\) where \((h,k) = (1,1)\), then expand and simplify.

Exam Tip: Substitute carefully using brackets to avoid sign errors during expansion.

Q21. Find the equations of the medians of the triangle ABC whose vertices are A (2, 5), B (-4, 9) and C (-2,-1)
Answer: Let vertices be \(A(2, 5)\), \(B(-4, 9)\), and \(C(-2, -1)\).
Let \(D\) be the midpoint of \(BC\):
\(D = \left(\frac{-4 + (-2)}{2}, \frac{9 + (-1)}{2}\right) = (-3, 4)\)
Equation of median \(AD\) passing through \(A(2, 5)\) and \(D(-3, 4)\):
\(y - 5 = \frac{4 - 5}{-3 - 2}(x - 2) \implies y - 5 = \frac{-1}{-5}(x - 2) \implies y - 5 = \frac{1}{5}(x - 2)\)
\(5y - 25 = x - 2 \implies x - 5y + 23 = 0\).
Let \(E\) be the midpoint of \(AC\):
\(E = \left(\frac{2 + (-2)}{2}, \frac{5 + (-1)}{2}\right) = (0, 2)\)
Equation of median \(BE\) passing through \(B(-4, 9)\) and \(E(0, 2)\):
\(y - 2 = \frac{9 - 2}{-4 - 0}(x - 0) \implies y - 2 = \frac{7}{-4}x \implies -4y + 8 = 7x \implies 7x + 4y - 8 = 0\).
Let \(F\) be the midpoint of \(AB\):
\(F = \left(\frac{2 + (-4)}{2}, \frac{5 + 9}{2}\right) = (-1, 7)\)
Equation of median \(CF\) passing through \(C(-2, -1)\) and \(F(-1, 7)\):
\(y - (-1) = \frac{7 - (-1)}{-1 - (-2)}(x - (-1)) \implies y + 1 = \frac{8}{1}(x + 1)\)
\(y + 1 = 8x + 8 \implies 8x - y + 7 = 0\).
Thus, the equations of medians are \(x - 5y + 23 = 0\), \(7x + 4y - 8 = 0\), and \(8x - y + 7 = 0\) (or \(8x - y + 15 = 0\) depending on vertex pairings).
In simple words: Find midpoints of all three sides, then write the line equation joining each vertex to the opposite midpoint.

Exam Tip: Label vertices and midpoints clearly to avoid mixing up coordinate points.

Q22. Find the image of the point (-8, 12) with respect to the line mirror \(4x + 7y + 13 = 0\).
Answer: Let the given point be \(P(x_1, y_1) = (-8, 12)\) and its image be \(Q(x_2, y_2)\).
The line mirror is \(4x + 7y + 13 = 0\).
Using the reflection formula for a point across a line:
\(\frac{x_2 - x_1}{A} = \frac{y_2 - y_1}{B} = \frac{-2(Ax_1 + By_1 + C)}{A^2 + B^2}\)
Here \(A = 4\), \(B = 7\), \(C = 13\).
\(Ax_1 + By_1 + C = 4(-8) + 7(12) + 13 = -32 + 84 + 13 = 65\)
\(A^2 + B^2 = 4^2 + 7^2 = 16 + 49 = 65\)
\(\frac{-2(Ax_1 + By_1 + C)}{A^2 + B^2} = \frac{-2(65)}{65} = -2\)
Therefore:
\(\frac{x_2 - (-8)}{4} = -2 \implies x_2 + 8 = -8 \implies x_2 = -16\)
\(\frac{y_2 - 12}{7} = -2 \implies y_2 - 12 = -14 \implies y_2 = -2\)
Thus, the image point is \((-16, -2)\).
In simple words: Use the standard reflection formula to directly find the coordinates of the mirror image of a point.

Exam Tip: This formula is a massive time-saver compared to finding the foot of the perpendicular manually.

Q23. Find the equation of the line passing through the intersection of the lines \(3x - 4y + 1 = 0\) and \(5x + y - 1 = 0\) and cutting off equal intercepts on the coordinate axes.
Answer: Any line passing through the intersection of the two given lines is given by the family of lines equation:
\((3x - 4y + 1) + k(5x + y - 1) = 0\)
\((3 + 5k)x + (-4 + k)y + (1 - k) = 0\) ---(1)
Since the line cuts off equal intercepts on the axes, its x-intercept and y-intercept are equal.
The x-intercept is \(-\frac{1 - k}{3 + 5k}\) and the y-intercept is \(-\frac{1 - k}{-4 + k}\).
Equating the intercepts:
\(-\frac{1 - k}{3 + 5k} = -\frac{1 - k}{-4 + k}\)
This gives either \(1 - k = 0 \implies k = 1\), or:
\(\frac{1}{3 + 5k} = \frac{1}{-4 + k} \implies 3 + 5k = -4 + k \implies 4k = -7 \implies k = -7/4\).
If \(k = 1\), substituting into (1):
\((3 + 5)x + (-4 + 1)y + (1 - 1) = 0 \implies 8x - 3y = 0\) (passes through origin, so intercepts are both 0).
If \(k = -7/4\), multiplying by 4:
\((12 - 35)x + (-16 - 7)y + (4 + 7) = 0\)
\(-23x - 23y + 11 = 0 \implies 23x + 23y = 11\).
In simple words: Use the family of lines method, then equate the expressions for x and y intercepts to find the parameter k.

Exam Tip: Lines cutting equal non-zero intercepts have a slope of -1.

Q24. Find the coordinates of the foot of the perpendicular from a point (-1, 3) to the line \(3x - 4y = 16\)
Answer: Let the given point be \((x_1, y_1) = (-1, 3)\) and the foot of the perpendicular be \((h, k)\).
Using the foot of perpendicular formula:
\(\frac{h - x_1}{A} = \frac{k - y_1}{B} = \frac{-(Ax_1 + By_1 + C)}{A^2 + B^2}\)
Here line is \(3x - 4y - 16 = 0\), so \(A = 3\), \(B = -4\), \(C = -16\).
\(Ax_1 + By_1 + C = 3(-1) - 4(3) - 16 = -3 - 12 - 16 = -31\)
\(A^2 + B^2 = 3^2 + (-4)^2 = 9 + 16 = 25\)
\(\frac{-(Ax_1 + By_1 + C)}{A^2 + B^2} = \frac{-(-31)}{25} = \frac{31}{25}\)
Therefore:
\(\frac{h - (-1)}{3} = \frac{31}{25} \implies h + 1 = \frac{93}{25} \implies h = \frac{93}{25} - 1 = \frac{68}{25}\)
\(\frac{k - 3}{-4} = \frac{31}{25} \implies k - 3 = \frac{-124}{25} \implies k = 3 - \frac{124}{25} = \frac{75 - 124}{25} = \frac{-49}{25}\)
Thus, the coordinates of the foot of the perpendicular are \(\left(\frac{68}{25}, -\frac{49}{25}\right)\).
In simple words: The foot of perpendicular formula instantly yields the coordinates without needing to solve simultaneous equations for the perpendicular line.

Exam Tip: Ensure the line equation is in general form \(Ax + By + C = 0\) before applying the formula.

Q25. Find the equation of the line parallel to y- axis and drawn through the point of intersection of \(x - 7y + 5 = 0\) and \(3x + 7y = 7\)
Answer: First, find the intersection point by adding the two equations:
\((x - 7y + 5) + (3x + 7y - 7) = 0\)
\(4x - 2 = 0 \implies x = \frac{1}{2}\)
Substitute \(x = \frac{1}{2}\) into \(3x + 7y = 7\):
\(3\left(\frac{1}{2}\right) + 7y = 7 \implies \frac{3}{2} + 7y = 7 \implies 7y = 7 - \frac{3}{2} = \frac{11}{2} \implies y = \frac{11}{14}\)
Intersection point is \(\left(\frac{1}{2}, \frac{11}{14}\right)\).
A line parallel to the y-axis has a constant x-coordinate for all its points.
Since it passes through \(x = \frac{1}{2}\), the equation of the line is \(x = \frac{1}{2}\).
In simple words: Lines parallel to the y-axis have the equation \(x = c\). Find the x-coordinate of the intersection point.

Exam Tip: You only need to solve for the x-coordinate of the intersection point since the y-axis parallel line depends entirely on x.

Q26. Find the equation of parabola with focus at \((5, 0)\) and directrix \(x + 5 = 0\), Also find the length of latus rectum.
Answer: Given focus \(S(5, 0)\) and directrix \(x = -5\).
The distance from focus to directrix is \(2a = 5 - (-5) = 10 \implies a = 5\).
Since the focus lies on the x-axis and directrix is vertical, the parabola opens to the right.
The standard equation is \(y^2 = 4ax\).
Substituting \(a = 5\):
\(y^2 = 4(5)x \implies y^2 = 20x\)
Length of latus rectum \(= 4a = 4(5) = 20\).
In simple words: The distance between focus and directrix equals 2a. Use standard parabola forms once the orientation is known.

Exam Tip: Length of latus rectum for any standard parabola is always equal to \(4a\).

Q27. For the parabola \(y^2 = -12x\), Find the coordinates of focus, the equation of directrix and length of latus rectum.
Answer: Comparing \(y^2 = -12x\) with standard left-opening parabola \(y^2 = -4ax\):
\(4a = 12 \implies a = 3\)
Since it opens to the left:
- Focus is at \((-a, 0) = (-3, 0)\)
- Equation of directrix is \(x = a \implies x = 3$
- Length of latus rectum \(= 4a = 12\).
In simple words: Match the given equation with standard conic forms to read off parameters like a.

Exam Tip: A negative sign in \(y^2 = -4ax\) indicates the parabola opens towards the negative x-axis.

Q28. Find the equation of parabola with vertex at origin and having directrix \(y = 2\)
Answer: Since the directrix is \(y = 2\) (a horizontal line above the origin), the parabola opens downwards with vertex at \((0, 0)\).
Here \(a = 2\).
The standard equation for a downward-opening parabola is \(x^2 = -4ay\).
Substituting \(a = 2\):
\(x^2 = -4(2)y \implies x^2 = -8y\).
In simple words: A horizontal directrix above the origin means the parabola opens downwards, making the equation \(x^2 = -4ay\).

Exam Tip: Focus is always opposite to the directrix across the origin.

Q29. Find the equation of the circle which passes through the points (5,-8), (2,-9) and (2, 1). Find also the coordinate of its centre and radius.
Answer: Let the general equation of a circle be \(x^2 + y^2 + 2gx + 2fy + c = 0\).
Since it passes through \((5, -8)\):
\(5^2 + (-8)^2 + 2g(5) + 2f(-8) + c = 0 \implies 25 + 64 + 10g - 16f + c = 0 \implies 10g - 16f + c = -89\) ---(1)
Since it passes through \((2, -9)\):
\(2^2 + (-9)^2 + 2g(2) + 2f(-9) + c = 0 \implies 4 + 81 + 4g - 18f + c = 0 \implies 4g - 18f + c = -85\) ---(2)
Since it passes through \((2, 1)\):
\(2^2 + 1^2 + 2g(2) + 2f(1) + c = 0 \implies 4 + 1 + 4g + 2f + c = 0 \implies 4g + 2f + c = -5\) ---(3)
Subtracting (2) from (1):
\(6g + 2f = -4 \implies 3g + f = -2\) ---(4)
Subtracting (3) from (2):
\(-20f = -80 \implies f = 4\)
Substitute \(f = 4\) into (4):
\(3g + 4 = -2 \implies 3g = -6 \implies g = -2\)
Substitute \(g = -2\) and \(f = 4\) into (3):
\(4(-2) + 2(4) + c = -5 \implies -8 + 8 + c = -5 \implies c = -5\)
Thus, the equation of the circle is \(x^2 + y^2 - 4x + 8y - 5 = 0\).
Centre \((-g, -f) = (2, -4)\).
Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + 4^2 - (-5)} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5\).
In simple words: Substitute the three points into the general circle equation to form three linear equations in terms of g, f, and c.

Exam Tip: Centre is given by \((-g, -f)\) and radius by \(\sqrt{g^2 + f^2 - c}\).

Q30. Find the equation of the circle which passes through the points (-2,-3) and has its centre on negative side of x-axis and is of radius 5 units.
Answer: Since the centre lies on the negative side of the x-axis, let the centre be \((-h, 0)\) where \(h > 0\).
Given radius \(r = 5\).
The equation of the circle is:
\((x - (-h))^2 + (y - 0)^2 = 5^2\)
\((x + h)^2 + y^2 = 25\) ---(1)
Since the circle passes through \((-2, -3)\), substitute these coordinates into (1):
\((-2 + h)^2 + (-3)^2 = 25\)
\(4 - 4h + h^2 + 9 = 25\)
\(h^2 - 4h + 13 = 25\)
\(h^2 - 4h - 12 = 0\)
\((h - 6)(h + 2) = 0\)
Since the centre is on the negative x-axis, \(h = 6\) (giving centre \((-6, 0)\)).
Substitute \(h = 6\) into the equation:
\((x + 6)^2 + y^2 = 25\)
\(x^2 + 12x + 36 + y^2 = 25\)
\(x^2 + y^2 + 12x + 11 = 0\).
In simple words: Assume centre coordinates based on given geometric conditions, then substitute the passing point to find unknowns.

Exam Tip: A point on the negative x-axis always has coordinates of the form \((-a, 0)\) where \(a > 0\).

Q31. Find the equation of the circle whose centre is (3,-2) and which passes through the intersection of the line \(5x + 7y = 3\) and \(2x - 3y = 7\),
Answer: First, find the intersection point of \(5x + 7y = 3\) and \(2x - 3y = 7\).
Multiply first equation by 3 and second by 7:
\(15x + 21y = 9$
\(14x - 21y = 49$
Adding both: \(29x = 58 \implies x = 2\).
Substitute \(x = 2\) into \(2x - 3y = 7\):
\(2(2) - 3y = 7 \implies 4 - 3y = 7 \implies -3y = 3 \implies y = -1\).
The intersection point is \((2, -1)\).
The circle has centre \((3, -2)\) and passes through \((2, -1)\).
Radius \(r\) is the distance between centre \((3, -2)\) and point \((2, -1)\):
\(r = \sqrt{(2 - 3)^2 + (-1 - (-2))^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}\)
Equation of the circle with centre \((h, k) = (3, -2)\) and radius \(r = \sqrt{2}\):
\((x - 3)^2 + (y - (-2))^2 = (\sqrt{2})^2\)
\(x^2 - 6x + 9 + y^2 + 4y + 4 = 2\)
\(x^2 + y^2 - 6x + 4y + 11 = 0\).
In simple words: Find the point of intersection to act as a point on the circle, then compute the radius using the distance formula from the centre.

Exam Tip: Radius is always the distance between the known centre and any point on the circumference.

Q32. Find the equation of the circle which passes through the points (3,-2), (-2, 0) and having the line segment \(2x - y = 3\) as its diameter.
Answer: Let the circle equation be \(x^2 + y^2 + 2gx + 2fy + c = 0\).
Since it passes through \((3, -2)\):
\(9 + 4 + 6g - 4f + c = 0 \implies 6g - 4f + c = -13\) ---(1)
Since it passes through \((-2, 0)\):
\(4 + 0 - 4g + 0 + c = 0 \implies -4g + c = -4 \implies c = 4g - 4\) ---(2)
Also, the centre of the circle \((-g, -f)\) must lie on its diameter line \(2x - y = 3\):
\(2(-g) - (-f) = 3 \implies -2g + f = 3 \implies f = 2g + 3\) ---(3)
Substitute \(c = 4g - 4\) into (1):
\(6g - 4f + (4g - 4) = -13 \implies 10g - 4f = -9\) ---(4)
Substitute \(f = 2g + 3\) into (4):
\(10g - 4(2g + 3) = -9\)
\(10g - 8g - 12 = -9\)
\(2g = 3 \implies g = \frac{3}{2}\)
Find \(f\) and \(c\):
\(f = 2\left(\frac{3}{2}\right) + 3 = 3 + 3 = 6\)
\(c = 4\left(\frac{3}{2}\right) - 4 = 6 - 4 = 2\)
Thus, the equation of the circle is \(x^2 + y^2 + 3x + 12y + 2 = 0\).
In simple words: The centre of any circle always lies on its diameter line. Use this property along with passing points to determine all circle coefficients.

Exam Tip: Centre coordinates \((-g, -f)\) satisfy the equation of the diameter line.

Q33. Find the equation of the circle which passes through the points (-3, 1), (6, 4) and (2, -6).
Answer: Let the circle equation be \(x^2 + y^2 + 2gx + 2fy + c = 0\).
For \((-3, 1)\):
\(9 + 1 - 6g + 2f + c = 0 \implies -6g + 2f + c = -10\) ---(1)
For \((6, 4)\):
\(36 + 16 + 12g + 8f + c = 0 \implies 12g + 8f + c = -52\) ---(2)
For \((2, -6)\):
\(4 + 36 + 4g - 12f + c = 0 \implies 4g - 12f + c = -40\) ---(3)
Subtracting (1) from (2):
\(18g + 6f = -42 \implies 3g + f = -7\) ---(4)
Subtracting (3) from (2):
\(8g + 20f = -12 \implies 2g + 5f = -3\) ---(5)
Solving simultaneous equations (4) and (5):
Multiply (4) by 5: \(15g + 5f = -35\)
Subtract (5): \((15g + 5f) - (2g + 5f) = -35 - (-3) \implies 13g = -32 \implies g = -\frac{32}{13}\).
Substitute \(g\) into (4): \(3\left(-\frac{32}{13}\right) + f = -7 \implies -\frac{96}{13} + f = -7 \implies f = \frac{96}{13} - 7 = \frac{96 - 91}{13} = \frac{5}{13}\).
Find \(c\) using (1):
\(-6\left(-\frac{32}{13}\right) + 2\left(\frac{5}{13}\right) + c = -10 \implies \frac{192}{13} + \frac{10}{13} + c = -10 \implies \frac{202}{13} + c = -10 \implies c = -10 - \frac{202}{13} = -\frac{332}{13}\)
Multiplying the entire circle equation by 13:
\(13x^2 + 13y^2 - 64x + 10y - 332 = 0\).
In simple words: Substitute each point into the general circle equation and solve the resulting system of linear equations for g, f, and c.

Exam Tip: Keep fractions exact throughout the simultaneous equation steps to avoid rounding errors.

Q34. Find the coordinate of the foci, the vertices, the eccentricity, the length of the latus rectum, the axis of the hyperbola \(y^2 - 16x^2 = 16\)
Answer: Rewriting the hyperbola equation in standard form by dividing by 16:
\(\frac{y^2}{16} - \frac{x^2}{1} = 1\)
Here \(a^2 = 16 \implies a = 4\) and \(b^2 = 1 \implies b = 1\). This is a vertical hyperbola.
- Vertices: \((0, \pm a) = (0, \pm 4)\).
- Eccentricity (\(e\)): \(e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{16}} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4}\).
- Foci: \((0, \pm ae) = \left(0, \pm 4 \cdot \frac{\sqrt{17}}{4}\right) = (0, \pm\sqrt{17})\).
- Length of latus rectum: \(\frac{2b^2}{a} = \frac{2(1)}{4} = \frac{1}{2}\).
- Transverse axis length \(= 2a = 8\), Conjugate axis length \(= 2b = 2\).
In simple words: Convert the equation to standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) to identify a and b.

Exam Tip: For vertical hyperbolas, foci and vertices lie along the y-axis.

Q35. Find the equation of the hyperbola whose vertices are \((\pm 5, 0)\) and foci \((\pm 7, 0)\)
Answer: Since vertices are on the x-axis at \((\pm 5, 0)\), \(a = 5\).
Since foci are at \((\pm 7, 0)\), \(ae = 7 \implies 5e = 7 \implies e = \frac{7}{5}\).
We know the relation \(b^2 = a^2(e^2 - 1)\):
\(b^2 = 5^2 \left(\left(\frac{7}{5}\right)^2 - 1\right) = 25 \left(\frac{49}{25} - 1\right) = 25 \left(\frac{24}{25}\right) = 24\)
The standard equation of a horizontal hyperbola is:
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
\(\frac{x^2}{25} - \frac{y^2}{24} = 1\).
In simple words: Use vertex data to find a and foci data to find e, then apply the hyperbola parameter relation \(b^2 = a^2(e^2 - 1)\).

Exam Tip: Remember that for hyperbolas, \(e > 1\) always.

Q36. If the major axis and eccentricity of an ellipse are respectively 8 and \(\frac{1}{2}\), then find equation of the ellipse
Answer: Given length of major axis \(2a = 8 \implies a = 4\).
Given eccentricity \(e = \frac{1}{2}\).
We know the relation for an ellipse: \(b^2 = a^2(1 - e^2)\).
\(b^2 = 4^2 \left(1 - \left(\frac{1}{2}\right)^2\right) = 16 \left(1 - \frac{1}{4}\right) = 16 \left(\frac{3}{4}\right) = 12\)
Assuming standard horizontal ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\):
\(\frac{x^2}{16} + \frac{y^2}{12} = 1\)
Multiplying by 48 throughout:
\(3x^2 + 4y^2 = 48\).
In simple words: Major axis length gives \(2a\). Use \(b^2 = a^2(1 - e^2)\) to find \(b^2\), then write the ellipse equation.

Exam Tip: Major axis length is always \(2a\), never \(a\).

Q37. Find the coordinate of the foci, the vertices, the eccentricity, the length of the latus rectum, the major axis and the minor axis of the ellipse \(\frac{x^2}{169} + \frac{y^2}{144} = 1\)
Answer: Given ellipse equation: \(\frac{x^2}{169} + \frac{y^2}{144} = 1\).
Here \(a^2 = 169 \implies a = 13\) and \(b^2 = 144 \implies b = 12\) (since \(a > b\), this is a horizontal ellipse).
- Vertices: \((\pm a, 0) = (\pm 13, 0)\).
- Eccentricity (\(e\)): \(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}\).
- Foci: \((\pm ae, 0) = \left(\pm 13 \cdot \frac{5}{13}, 0\right) = (\pm 5, 0)\).
- Length of latus rectum: \(\frac{2b^2}{a} = \frac{2(144)}{13} = \frac{288}{13}\).
- Major axis length \(= 2a = 2(13) = 26\).
- Minor axis length \(= 2b = 2(12) = 24\).
In simple words: Identify \(a^2\) and \(b^2\) where \(a\) is the denominator of the larger term, then apply standard ellipse formulas.

Exam Tip: For an ellipse, eccentricity is always between 0 and 1.

Q38. Find the point to which the origin should be shifted after shifting of origin so that the equation \(x^2 - 12x + 4 = 0\) will have no first degree term.
Answer: Let the origin be shifted to \((h, k)\).
Transformation equations are \(x = X + h\) and \(y = Y + k\).
Substitute \(x = X + h\) into \(x^2 - 12x + 4 = 0\):
\((X + h)^2 - 12(X + h) + 4 = 0\)
\(X^2 + 2hX + h^2 - 12X - 12h + 4 = 0\)
\(X^2 + (2h - 12)X + (h^2 - 12h + 4) = 0\)
For the transformed equation to have no first-degree term in \(X\), the coefficient of \(X\) must be set to zero:
\(2h - 12 = 0 \implies 2h = 12 \implies h = 6\)
Since there is no y-term in the original equation, \(k\) can be any real value.
Thus, the new origin is \((6, k)\) where \(k\) is any real number.
In simple words: Substitute \(x = X + h\), expand, and set the coefficient of the linear term to zero to solve for h.

Exam Tip: Eliminating first-degree terms is achieved by equating the linear term coefficient to zero.

Q39. Prove that the area of a triangle is invariant under the translation of the axes.
Answer: Let the vertices of a triangle with respect to the old coordinate system be \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\).
The area of the triangle in the old system is:
\(\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
When the origin is shifted to \((h, k)\), let the new coordinates be \((X_i, Y_i)\) such that \(x_i = X_i + h\) and \(y_i = Y_i + k\) for \(i = 1, 2, 3\).
Substitute these into the area formula:
\(\Delta' = \frac{1}{2} |(X_1 + h)((Y_2 + k) - (Y_3 + k)) + (X_2 + h)((Y_3 + k) - (Y_1 + k)) + (X_3 + h)((Y_1 + k) - (Y_2 + k))|\)
Notice that the constants \(+k\) and \(-k\) cancel out in every parenthesis:
\(\Delta' = \frac{1}{2} |(X_1 + h)(Y_2 - Y_3) + (X_2 + h)(Y_3 - Y_1) + (X_3 + h)(Y_1 - Y_2)|\)
Expanding this expression:
\(\Delta' = \frac{1}{2} |X_1(Y_2 - Y_3) + h(Y_2 - Y_3) + X_2(Y_3 - Y_1) + h(Y_3 - Y_1) + X_3(Y_1 - Y_2) + h(Y_1 - Y_2)|\)
Grouping the terms containing \(h\):
\(h[(Y_2 - Y_3) + (Y_3 - Y_1) + (Y_1 - Y_2)] = h[0] = 0\)
Thus, the expression simplifies to:
\(\Delta' = \frac{1}{2} |X_1(Y_2 - Y_3) + X_2(Y_3 - Y_1) + X_3(Y_1 - Y_2)| = \Delta\)
Since \(\Delta' = \Delta\), the area of the triangle remains unchanged (invariant) under the translation of axes. Hence proved.
In simple words: When shifting axes, the shift constants \(h\) and \(k\) cancel out during the area determinant expansion, proving the area is unchanged.

Exam Tip: Translation only changes coordinate locations, not the relative distances or geometric area between points.