CBSE Class 11 Mathematics Straight Lines Assignment Set 04

EXTRA/ HOT QUESTIONS

Question 1. Find the equation of the line through (4,-5) and parallel to the line joining the points (3,7) & (-2,4).
Answer: Slope of the line joining the points \((3, 7)\) and \((-2, 4)\) is given by:
\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 7}{-2 - 3} = \frac{-3}{-5} = \frac{3}{5}\)
Since the required line is parallel to this line, its slope will be the same, so \(m = \frac{3}{5}\).
Now, the equation of the line passing through the point \((4, -5)\) with slope \(m = \frac{3}{5}\) is:
\(y - y_1 = m(x - x_1)\)
\(y - (-5) = \frac{3}{5}(x - 4)\)
\(y + 5 = \frac{3}{5}(x - 4)\)
\(5(y + 5) = 3(x - 4)\)
\(5y + 25 = 3x - 12\)
\(3x - 5y - 37 = 0\)
In simple words: Parallel lines have equal slopes, so we first find the slope of the given points and use it with the given point-slope formula.

Exam Tip: Remember that parallel lines have identical slopes (\(m_1 = m_2\)).

Question 2. If A(1,4) , B(2,-3) and C(-1,-2) are the vertices of a triangle ABC . find a) The equation of the median through A b) The equation of the altitude through A c) The right bisector of side BC
Answer: Let the vertices be \(A(1, 4)\), \(B(2, -3)\), and \(C(-1, -2)\).
a) Equation of the median through A:
The median through \(A\) meets \(BC\) at its midpoint \(D\).
Coordinates of \(D = \left( \frac{2 + (-1)}{2}, \frac{-3 + (-2)}{2} \right) = \left( \frac{1}{2}, -\frac{5}{2} \right)\)
The equation of median \(AD\) passing through \(A(1, 4)\) and \(D\left(\frac{1}{2}, -\frac{5}{2}\right)\) is:
\(y - 4 = \frac{-\frac{5}{2} - 4}{\frac{1}{2} - 1}(x - 1)\)
\(y - 4 = \frac{-\frac{13}{2}}{-\frac{1}{2}}(x - 1)\)
\(y - 4 = 13(x - 1)\)
\(13x - y - 9 = 0\)
b) Equation of the altitude through A:
Slope of \(BC = \frac{-2 - (-3)}{-1 - 2} = \frac{1}{-3} = -\frac{1}{3}\)
Since the altitude is perpendicular to \(BC\), its slope is \(m = -\frac{1}{-1/3} = 3\).
Equation of the altitude passing through \(A(1, 4)\) with slope \(3\) is:
\(y - 4 = 3(x - 1)\)
\(3x - y + 1 = 0\)
c) The right bisector of side BC:
Midpoint of \(BC\) is \(D\left(\frac{1}{2}, -\frac{5}{2}\right)\).
Slope of the right bisector is perpendicular to \(BC\), so \(m = 3\).
Equation passing through \(D\left(\frac{1}{2}, -\frac{5}{2}\right)\) with slope \(3\) is:
\(y - \left(-\frac{5}{2}\right) = 3\left(x - \frac{1}{2}\right)\)
\(\frac{2y + 5}{2} = 3\left(\frac{2x - 1}{2}\right)\)
\(2y + 5 = 6x - 3\)
\(6x - 2y - 8 = 0 \implies 3x - y - 4 = 0\)
In simple words: A median connects a vertex to the midpoint of the opposite side. An altitude is perpendicular to the opposite side. A right bisector passes through the midpoint and is perpendicular to the side.

Exam Tip: Always clearly identify the midpoint and perpendicular slope conditions before applying the two-point form.

Question 3. Find the equation of the straight line which passes through (3,-2) and cuts off positive intercepts on the x axis and y axis which are in the ratio 4:3
Answer: Let the x-intercept be \(4a\) and the y-intercept be \(3a\) (since they are in the ratio 4:3 and are positive).
The intercept form of the line is:
\(\frac{x}{4a} + \frac{y}{3a} = 1\)
Multiplying by \(12a\):
\(3x + 4y = 12a\) --- (1)
Since the line passes through the point \((3, -2)\), substitute \(x = 3\) and \(y = -2\) into equation (1):
\(3(3) + 4(-2) = 12a\)
\(9 - 8 = 12a\)
\(12a = 1 \implies a = \frac{1}{12}\)
Substitute \(12a = 1\) into equation (1):
\(3x + 4y = 1\)
In simple words: Assume intercepts as multiples of a variable (like \(4a\) and \(3a\)), substitute the given point to find the variable, and substitute it back.

Exam Tip: When ratio is given as 4:3, always write intercepts as \(4a\) and \(3a\).

Question 4. Reduce the equation 3x-2y+4 =0 to intercept form. Hence find the length of the segment intercepted between the axes.
Answer: Given equation: \(3x - 2y + 4 = 0\)
Rearranging to the constant form: \(3x - 2y = -4\)
Dividing both sides by \(-4\) to make the right side 1:
\(\frac{3x}{-4} + \frac{-2y}{-4} = 1\)
\(\frac{x}{-\frac{4}{3}} + \frac{y}{2} = 1\)
This is the intercept form \(\frac{x}{a} + \frac{y}{b} = 1\), where x-intercept \(a = -\frac{4}{3}\) and y-intercept \(b = 2\).
The coordinates of the points where the line cuts the axes are \(A\left(-\frac{4}{3}, 0\right)\) and \(B(0, 2)\).
The length of the segment intercepted between the axes is the distance between \(A\) and \(B\):
\(AB = \sqrt{\left(0 - \left(-\frac{4}{3}\right)\right)^2 + (2 - 0)^2}\)
\(AB = \sqrt{\left(\frac{4}{3}\right)^2 + 2^2}\)
\(AB = \sqrt{\frac{16}{9} + 4} = \sqrt{\frac{16 + 36}{9}} = \sqrt{\frac{52}{9}} = \frac{2\sqrt{13}}{3} \units\)
In simple words: Convert the line equation so that the constant term is on the right side and equals 1, then use the distance formula between the two axis intercept points.

Exam Tip: The intercepts on the axes are directly \(a\) and \(b\) from the intercept form.

Question 5. Find the image of the point (1,2) in the line x-3y+4 =0
Answer: Let the given point be \(P(1, 2)\) and its image be \(Q(x_1, y_1)\) with respect to the line \(x - 3y + 4 = 0\).
The line segment \(PQ\) is bisected perpendicularly by the given line.
1. The midpoint of \(PQ\) is \(M\left(\frac{x_1 + 1}{2}, \frac{y_1 + 2}{2}\right)\). Since \(M\) lies on the line \(x - 3y + 4 = 0\):
\(\left(\frac{x_1 + 1}{2}\right) - 3\left(\frac{y_1 + 2}{2}\right) + 4 = 0\)
\(x_1 + 1 - 3y_1 - 6 + 8 = 0\)
\(x_1 - 3y_1 + 3 = 0\) --- (1)
2. The slope of line \(PQ\) is \(m_{PQ} = \frac{y_1 - 2}{x_1 - 1}\).
The slope of the given line \(x - 3y + 4 = 0\) is \(m = \frac{1}{3}\).
Since \(PQ\) is perpendicular to the line, the product of their slopes is \(-1\):
\(\left(\frac{y_1 - 2}{x_1 - 1}\right) \times \frac{1}{3} = -1\)
\(\frac{y_1 - 2}{3(x_1 - 1)} = -1\)
\(y_1 - 2 = -3x_1 + 3\)
\(3x_1 + y_1 - 5 = 0\) --- (2)
Solving equations (1) and (2) simultaneously:
From (2), \(y_1 = 5 - 3x_1\). Substitute in (1):
\(x_1 - 3(5 - 3x_1) + 3 = 0\)
\(x_1 - 15 + 9x_1 + 3 = 0\)
\(10x_1 - 12 = 0 \implies x_1 = \frac{12}{10} = \frac{6}{5}\)
Now, \(y_1 = 5 - 3\left(\frac{6}{5}\right) = 5 - \frac{18}{5} = \frac{7}{5}\)
Thus, the image point is \(\left(\frac{6}{5}, \frac{7}{5}\right)\).
In simple words: The image point is found using two conditions: the midpoint of the point and its image lies on the line, and the line joining them is perpendicular to the given line.

Exam Tip: Form two equations: one from the midpoint substitution and another from the perpendicular slope condition \(m_1 m_2 = -1\).

Question 6. If the image of the point (2,1) in a line is (4,3) .Find the equation of the line.
Answer: Let the given point be \(P(2, 1)\) and its image be \(Q(4, 3)\).
The line whose equation we need to find is the perpendicular bisector of the line segment \(PQ\).
1. The midpoint of \(PQ\) is:
\(M = \left( \frac{2 + 4}{2}, \frac{1 + 3}{2} \right) = (3, 2)\)
2. The slope of the line segment \(PQ\) is:
\(m_{PQ} = \frac{3 - 1}{4 - 2} = \frac{2}{2} = 1\)
3. The slope of the perpendicular bisector line is the negative reciprocal of \(m_{PQ}\):
\(m = -\frac{1}{1} = -1\)
4. Using the point-slope form with point \((3, 2)\) and slope \(-1\):
\(y - 2 = -1(x - 3)\)
\(y - 2 = -x + 3\)
\(x + y - 5 = 0\)
In simple words: The line is the perpendicular bisector of the segment joining the point and its image. Find the midpoint and negative reciprocal slope.

Exam Tip: The required line always passes through the midpoint of the point and its image.

Question 7. Find the equation of a line passing through the point (-3,7) and the point of intersection of the lines 2x-3y+5 = 0 and 4x+9y = 7.
Answer: First, find the point of intersection of the two given lines:
\(2x - 3y + 5 = 0\) --- (1)
\(4x + 9y - 7 = 0\) --- (2)
Multiply equation (1) by 2:
\(4x - 6y + 10 = 0\) --- (3)
Subtract equation (3) from equation (2):
\((4x + 9y - 7) - (4x - 6y + 10) = 0\)
\(15y - 17 = 0 \implies y = \frac{17}{15}\) (Wait, let's re-verify with standard family of lines or direct substitution).
Let's solve normally:
From (1): \(2x = 3y - 5 \implies x = \frac{3y - 5}{2}\)
Substitute into (2):
\(4\left(\frac{3y - 5}{2}\right) + 9y = 7\)
\(2(3y - 5) + 9y = 7\)
\(6y - 10 + 9y = 7\)
\(15y = 17 \implies y = \frac{17}{15}\) (Wait, let us check the constant values in standard textbook problem: usually equations are chosen such that they yield integer intersection points. Let's re-verify the intersection or use family of lines).
Using family of lines passing through the intersection of \(2x - 3y + 5 = 0\) and \(4x + 9y - 7 = 0\):
\((2x - 3y + 5) + \lambda(4x + 9y - 7) = 0\) --- (A)
Since this line passes through \((-3, 7)\), substitute \(x = -3\) and \(y = 7\):
\((2(-3) - 3(7) + 5) + \lambda(4(-3) + 9(7) - 7) = 0\)
\((-6 - 21 + 5) + \lambda(-12 + 63 - 7) = 0\)
\(-22 + \lambda(44) = 0\)
\(44\lambda = 22 \implies \lambda = \frac{22}{44} = \frac{1}{2}\)
Substitute \(\lambda = \frac{1}{2}\) back into equation (A):
\((2x - 3y + 5) + \frac{1}{2}(4x + 9y - 7) = 0\)
Multiply entire equation by 2:
\(2(2x - 3y + 5) + (4x + 9y - 7) = 0\)
\(4x - 6y + 10 + 4x + 9y - 7 = 0\)
\(8x + 3y + 3 = 0\)
In simple words: Use the family of lines method: \(L_1 + \lambda L_2 = 0\), substitute the given point to find \(\lambda\), and simplify.

Exam Tip: The family of lines method \(L_1 + \lambda L_2 = 0\) saves time by avoiding the explicit calculation of the intersection point.

Question 8. Find the equation of straight lines which are perpendicular to the line 12x+5y = 17 and at a distance of 2 units from the point (-4,1)
Answer: The given line is \(12x + 5y = 17\).
Any line perpendicular to this line is of the form:
\(5x - 12y + k = 0\) --- (1)
We are given that the distance of this line from the point \((-4, 1)\) is 2 units.
Using the perpendicular distance formula from a point to a line:
\(\text{Distance} = \left| \frac{5(-4) - 12(1) + k}{\sqrt{5^2 + (-12)^2}} \right| = 2\)
\(\left| \frac{-20 - 12 + k}{\sqrt{25 + 144}} \right| = 2\)
\(\left| \frac{k - 32}{\sqrt{169}} \right| = 2\)
\(\frac{|k - 32|}{13} = 2\)
\(|k - 32| = 26\)
Removing the modulus, we get two cases:
Case 1: \(k - 32 = 26 \implies k = 32 + 26 = 58\)
Case 2: \(k - 32 = -26 \implies k = 32 - 26 = 6\)
Substituting the values of \(k\) into equation (1), the required equations are:
\(5x - 12y + 58 = 0\) and \(5x - 12y + 6 = 0\)
In simple words: Swap coefficients of \(x\) and \(y\) and change the sign of one to get a perpendicular line, then use the distance formula to find the unknown constant.

Exam Tip: Perpendicular line to \(Ax + By + C = 0\) is always of the form \(Bx - Ay + k = 0\).

Question 9. The points A(2,3) B(4,-1) & C(-1,2) are the vertices of a triangle. Find the length of perpendicular from A to BC and hence the area of ABC
Answer: Let the vertices be \(A(2, 3)\), \(B(4, -1)\), and \(C(-1, 2)\).
1. First, find the equation of the line \(BC\):
Slope of \(BC = \frac{2 - (-1)}{-1 - 4} = \frac{3}{-5} = -\frac{3}{5}\)
Equation of line \(BC\) passing through \(B(4, -1)\):
\(y - (-1) = -\frac{3}{5}(x - 4)\)
\(5(y + 1) = -3(x - 4)\)
\(5y + 5 = -3x + 12\)
\(3x + 5y - 7 = 0\)
2. The length of the perpendicular from \(A(2, 3)\) to the line \(3x + 5y - 7 = 0\) is the height \(h\):
\(h = \left| \frac{3(2) + 5(3) - 7}{\sqrt{3^2 + 5^2}} \right|\)
\(h = \left| \frac{6 + 15 - 7}{\sqrt{9 + 25}} \right| = \frac{14}{\sqrt{34}} \units\)
3. Length of base \(BC\) is:
\(BC = \sqrt{(-1 - 4)^2 + (2 - (-1))^2} = \sqrt{(-5)^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}\)
4. Area of triangle \(ABC\):
\(\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}\)
\(\text{Area} = \frac{1}{2} \times BC \times h = \frac{1}{2} \times \sqrt{34} \times \frac{14}{\sqrt{34}} = 7 \text{ sq. units}\)
In simple words: Find the line equation of side BC, calculate the perpendicular distance from A as height, and then use the standard area formula \(\frac{1}{2} \times \text{base} \times \text{height}\).

Exam Tip: The length of perpendicular from A to BC acts directly as the height of the triangle.

Question 10. Find the equation of straight line whose intercepts on the axes are thrice as long as those made by 2x + 11y = 6
Answer: Given line: \(2x + 11y = 6\)
Dividing by 6 to convert to intercept form:
\(\frac{2x}{6} + \frac{11y}{6} = 1\)
\(\frac{x}{3} + \frac{y}{\frac{6}{11}} = 1\)
The intercepts of this line are \(a_1 = 3\) and \(b_1 = \frac{6}{11}\).
The required line has intercepts which are thrice as long as these:
\(a = 3 \times 3 = 9\)
\(b = 3 \times \frac{6}{11} = \frac{18}{11}\)
Using the intercept form for the required line:
\(\frac{x}{9} + \frac{y}{\frac{18}{11}} = 1\)
\(\frac{x}{9} + \frac{11y}{18} = 1\)
Multiplying the entire equation by 18:
\(2x + 11y = 18\)
In simple words: Find the intercepts of the given line by converting it into intercept form, multiply them by 3, and write the new equation using the intercept form.

Exam Tip: Always convert equations to intercept form \(\frac{x}{a} + \frac{y}{b} = 1\) to directly read off the axis intercepts \(a\) and \(b\).