CBSE Class 11 Mathematics Straight Lines Worksheet Set 03

Worksheet with questions and answers for Straight Lines for CBSE Class 11 Mathematics. These worksheets have been prepared by teachers from the best schools in India. These worksheets have been designed with detailed explanation which will help the students to clear their doubts and improve understanding of Straight lines chapter.

Question 1. If the \(\Delta ABC\) with vertices \(A(2,3)\), \(B(4, -1)\) and \(C(1,2)\). Find the equation and length of altitude from the vertex \(A\).
Answer: The gradient of the line \(BC\) is calculated as follows:
\(m_{BC} = \frac{2 - (-1)}{1 - 4} = \frac{3}{-3} = -1\)
Because the altitude \(AD\) is perpendicular to \(BC\):
\(\therefore\) slope of \(AD = 1\) - (negative reciprocal of \(-1\))
Applying the point-slope formula with the point \(A(2,3)\) and slope \(m = 1\), we find the equation of the altitude \(AD\):
\(y - 3 = 1(x - 2)\)
\(\Rightarrow y - 3 = x - 2\)
\(\Rightarrow x - y + 1 = 0\) ans.
Next, we find the equation of the side \(BC\) using the two-point form with points \(B(4,-1)\) and \(C(1,2)\):
\(y + 1 = \frac{2 - (-1)}{1 - 4}(x - 4)\)
\(\Rightarrow y + 1 = -1(x - 4)\)
\(\Rightarrow y + 1 = -x + 4\)
\(\Rightarrow x + y - 3 = 0\)
The length of \(AD\) represents the perpendicular distance from vertex \(A(2,3)\) to the line \(BC\). Using the perpendicular distance formula, we get:
\(AD = \frac{|2 + 3 - 3|}{\sqrt{1^2 + 1^2}}\)
\(AD = \frac{2}{\sqrt{2}}\)
\(AD = \sqrt{2}\) ans.

In simple words: To find the altitude's equation, we find the slope of BC and take its negative reciprocal to get the slope of AD. Then, we write the equation of AD, find the equation of BC, and calculate the shortest distance from point A to BC.

Exam Tip: Always remember that perpendicular lines have slopes that are negative reciprocals of each other (\(m_1 \cdot m_2 = -1\)). This is a crucial shortcut for finding altitudes quickly.

Question 2. Find the perpendicular distance from the origin to the line joining the points \((\cos \theta , \sin \theta)\) and \((\cos \phi , \sin \phi)\).
Answer: The equation of the straight line \(AB\) passing through these points is written in two-point form as:
\(y - \sin \theta = \left(\frac{\sin \phi - \sin \theta}{\cos \phi - \cos \theta}\right) (x - \cos \theta)\)
Using the sum-to-product trigonometric identities:
\(\Rightarrow y - \sin \theta = \left(\frac{2\cos\left(\frac{\phi+\theta}{2}\right)\cdot\sin\left(\frac{\phi-\theta}{2}\right)}{-2\sin\left(\frac{\phi+\theta}{2}\right)\cdot\sin\left(\frac{\phi-\theta}{2}\right)}\right) \cdot (x - \cos \theta)\)
\(\Rightarrow -y \sin \left(\frac{\phi+\theta}{2}\right) + \sin \theta \cdot \sin \left(\frac{\phi+\theta}{2}\right) = x \cos \left(\frac{\phi+\theta}{2}\right) - \cos \theta \cdot \cos \left(\frac{\phi+\theta}{2}\right)\)
Rearranging the terms:
\(\Rightarrow x \cos \left(\frac{\phi+\theta}{2}\right) + y \sin \left(\frac{\phi+\theta}{2}\right) - \cos \theta \cdot \cos \left(\frac{\phi+\theta}{2}\right) - \sin \theta \cdot \sin \left(\frac{\phi+\theta}{2}\right) = 0\)
\(\Rightarrow x \cos \left(\frac{\phi+\theta}{2}\right) + y \sin \left(\frac{\phi+\theta}{2}\right) - \left\{\cos \theta \cdot \cos \left(\frac{\phi+\theta}{2}\right) + \sin \theta \cdot \sin \left(\frac{\phi+\theta}{2}\right)\right\} = 0\)
Using the identity \(\cos A \cos B + \sin A \sin B = \cos(A - B)\):
\(\Rightarrow x \cos \left(\frac{\phi+\theta}{2}\right) + y \sin \left(\frac{\phi+\theta}{2}\right) - \cos \left(\theta - \frac{\phi+\theta}{2}\right) = 0\)
This gives the simplified equation of the line \(AB\):
\(\Rightarrow x \cos \left(\frac{\phi+\theta}{2}\right) + y \sin \left(\frac{\phi+\theta}{2}\right) - \cos \left(\frac{\theta-\phi}{2}\right) = 0\)
Next, we calculate the shortest distance from the origin \((0,0)\) to the line \(AB\) using the point-to-line distance formula:
\(d = \frac{\left|0 + 0 - \cos\left(\frac{\theta-\phi}{2}\right)\right|}{\sqrt{\cos^2\left(\frac{\phi+\theta}{2}\right) + \sin^2\left(\frac{\phi+\theta}{2}\right)}}\)
Using the Pythagorean identity \(\cos^2 \alpha + \sin^2 \alpha = 1\):
\(d = \frac{\cos\left(\frac{\theta-\phi}{2}\right)}{1}\)
\(\therefore\) required distance \(= \cos \left(\frac{\theta-\phi}{2}\right)\) ans.

In simple words: First, we find the equation of the line that connects the two given points. Then, we use the standard formula to find how far this line is from the origin \((0,0)\).

Exam Tip: Use the sum-to-product trigonometric identities carefully to simplify the slope. Expressing the line in its normal form \(x \cos\alpha + y \sin\alpha = p\) directly gives the perpendicular distance \(p\) from the origin.

Question 3. Prove that the product of the lengths of the perpendicular drawn from the points \((\sqrt{a^2 - b^2}, 0)\) and \((-\sqrt{a^2 - b^2}, 0)\) to the line \(\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1\) is \(b^2\).
Answer: We aim to prove that \(pq = b^2\).
The given equation of the line is:
\(\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1\)
Multiplying the entire equation by \(ab\), we get:
\(bx \cos\theta + ay \sin\theta - ab = 0\)
Let \(p\) and \(q\) be the perpendicular distances from the points \((\sqrt{a^2 - b^2}, 0)\) and \((-\sqrt{a^2 - b^2}, 0)\) respectively to the line. Using the perpendicular distance formula, we get:
\(p = \frac{|b\sqrt{a^2-b^2} \cos\theta - ab|}{\sqrt{b^2 \cos^2\theta + a^2 \sin^2\theta}}\)
\(q = \frac{|-b\sqrt{a^2-b^2} \cos\theta - ab|}{\sqrt{b^2 \cos^2\theta + a^2 \sin^2\theta}}\)
Multiplying these two distances together:
\(pq = \frac{|b\sqrt{a^2-b^2} \cos\theta - ab| \cdot |-b\sqrt{a^2-b^2} \cos\theta - ab|}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
Using the identity \(|-x| = |x|\) and \(|x||y| = |xy|\), we can simplify the numerator:
\(pq = \frac{\left| \left(b\sqrt{a^2-b^2} \cos\theta\right)^2 - (ab)^2 \right|}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
\(pq = \frac{\left| b^2(a^2 - b^2)\cos^2\theta - a^2b^2 \right|}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
Factoring out \(b^2\) from the absolute value in the numerator:
\(pq = \frac{b^2 \left| (a^2 - b^2)\cos^2\theta - a^2 \right|}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
Expanding the term inside the absolute value:
\(pq = \frac{b^2 \left| a^2 \cos^2\theta - b^2 \cos^2\theta - a^2 \right|}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
\(pq = \frac{b^2 \left| a^2(\cos^2\theta - 1) - b^2 \cos^2\theta \right|}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
Since \(\cos^2\theta - 1 = -\sin^2\theta\):
\(pq = \frac{b^2 \left| -a^2 \sin^2\theta - b^2 \cos^2\theta \right|}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
Using the absolute value property \(|-x| = |x|\):
\(pq = \frac{b^2 (a^2 \sin^2\theta + b^2 \cos^2\theta)}{b^2 \cos^2\theta + a^2 \sin^2\theta}\)
The terms in the numerator and denominator cancel out, leaving:
\(pq = b^2\)
Hence proved.

In simple words: First, we write the perpendicular distance from both given points to the line. When we multiply these two distances, the algebraic terms simplify and cancel out perfectly to leave just \(b^2\).

Exam Tip: Be careful with the algebraic identity \((x-y)(x+y) = x^2 - y^2\) inside the absolute values. Keeping \(b^2\) factored out early makes the cancellation of the trigonometric terms much easier to see.

Question 4. Find the equation of the line which is equidistant from the parallel lines \(9x + 6y - 7 = 0\) and \(3x + 2y + 6 = 0\).
Answer: First, let's write down the equations of the given lines. We can divide the first equation by 3 to make its coefficients match the second line:
\(l_1: 3x + 2y - \frac{7}{3} = 0\) ............(i)
And \(l_3: 3x + 2y + 6 = 0\) ............. (ii)
The slope of both lines is \(-\frac{3}{2}\), as determined by the formula \(m = \frac{-\text{coefficient of } x}{\text{coefficient of } y}\):
\(m = \frac{-3}{2}\)
Since the required line \(l_2\) lies exactly in between and parallel to these two lines, its slope must also be \(-\frac{3}{2}\). We can assume the equation of this parallel line \(l_2\) to be:
\(y = mx + c\)
\(\Rightarrow y = -\frac{3}{2}x + c\)
\(\Rightarrow 2y = -3x + 2c\)
\(\Rightarrow 3x + 2y - 2c = 0\) (\(l_2\)) ........... (iii)
We are given that \(l_2\) is equidistant from \(l_1\) and \(l_3\). Thus, the distance from \(l_1\) to \(l_2\) must equal the distance from \(l_2\) to \(l_3\):
\(\frac{\left|-\frac{7}{3} + 2c\right|}{\sqrt{3^2 + 2^2}} = \frac{|6 + 2c|}{\sqrt{3^2 + 2^2}}\) .......... (using the parallel distance formula)
\(\Rightarrow \left|-\frac{7}{3} + 2c\right| = |6 + 2c|\)
This yields two possibilities:
\(-\frac{7}{3} + 2c = \pm(6 + 2c)\)
The positive case gives no valid value for \(c\) as the terms cancel out:
\(-\frac{7}{3} + 2c = 6 + 2c\)
The negative case gives:
\(-\frac{7}{3} + 2c = -6 - 2c\)
\(\Rightarrow 4c = -6 + \frac{7}{3}\)
\(\Rightarrow 4c = \frac{-11}{3} \Rightarrow c = \frac{-11}{12}\)
Substituting this value of \(c\) back into our assumed equation (iii) for \(l_2\), we obtain:
\(3x + 2y - 2\left(\frac{-11}{12}\right) = 0\)
\(\Rightarrow 3x + 2y + \frac{11}{6} = 0\)
\(\Rightarrow 18x + 12y + 11 = 0\) ans.
In simple words: To find a line that is exactly halfway between two parallel lines, we rewrite them with the same \(x\) and \(y\) coefficients. Then, the constant term of our halfway line is simply the average of the constant terms of the two parallel lines.

Exam Tip: When dealing with parallel lines, always normalize their \(x\) and \(y\) coefficients to be identical first. The constant term of the equidistant line is always the arithmetic mean of the two normalized constant terms, which serves as a great mental check!

Question 5. A line is such that its segment between the lines \(5x - y + 4 = 0\) and \(3x + 4y = 4\) is bisected at the point \((1,5)\) obtain its equation.
Answer: Let the line intersect \(5x - y + 4 = 0\) at point \(P(x_1, y_1)\). This point must satisfy the line's equation:
\(5x_1 - y_1 = -4\) ............. (i)
Similarly, let the line intersect the second line \(3x + 4y = 4\) at point \(Q(x_2, y_2)\):
\(3x_2 + 4y_2 = 4\) ............ (ii)
Since the segment \(PQ\) is bisected at \(R(1,5)\), this point is the midpoint of \(P\) and \(Q\):
\(1 = \frac{x_1 + x_2}{2}\) and \(5 = \frac{y_1 + y_2}{2}\)
\(\Rightarrow x_1 + x_2 = 2\) and \(y_1 + y_2 = 10\)
\(\Rightarrow x_2 = 2 - x_1\) and \(y_2 = 10 - y_1\)
Substituting these expressions for \(x_2\) and \(y_2\) into the second line's equation (ii), we get:
\(3(2 - x_1) + 4(10 - y_1) = 4\)
\(\Rightarrow 6 - 3x_1 + 40 - 4y_1 = 4\)
\(\Rightarrow 3x_1 + 4y_1 = 42\) ............(iii)
By solving the system of equations (i) and (iii) simultaneously, we find the coordinates of \(P\):
\(x = \frac{26}{23}\) and \(y = \frac{222}{23}\)
\(\therefore P\left(\frac{26}{23}, \frac{222}{23}\right)\)
Now, we can find the equation of the required line passing through the midpoint \((1,5)\) and the point \(P \left(\frac{26}{23}, \frac{222}{23}\right)\) using the two-point formula:
\(y - 5 = \left(\frac{\frac{222}{23} - 5}{\frac{26}{23} - 1}\right) (x - 1)\)
\(\Rightarrow y - 5 = \frac{107}{3} (x - 1)\)
\(\Rightarrow 107x - 3y = 92\) ans.

In simple words: We find two points on the given lines such that their average (midpoint) is exactly \((1,5)\). Once we find these points, we simply write the equation of the line passing through them.

Exam Tip: Using parametric coordinates or expressing one point's coordinates in terms of the other using the midpoint formula is the cleanest way to solve segment-bisection problems without introducing too many variables.

Question 6. If the lines \(2x + y = 3\), \(5x + ky - 3 = 0\) and \(3x - y - 2 = 0\) are concurrent (intersect at one point). Find the value of \(k\).
Answer: Let's list the equations of the given lines:
\(2x + y = 3\) ........ (i)
\(5x + ky = 3\) ......... (ii)
\(3x - y = 2\) ......... (iii)
By solving the first and third equations simultaneously, we find their point of intersection:
Adding (i) and (iii):
\(5x = 5 \Rightarrow x = 1\)
Substituting \(x = 1\) into (i):
\(2(1) + y = 3 \Rightarrow y = 1\)
So, the intersection point is \((1,1)\). Since all three lines meet at a single point (concurrency), this intersection point must also satisfy the second equation:
\(5(1) + k(1) - 3 = 0\)
\(\Rightarrow 5 + k - 3 = 0\)
\(\Rightarrow k = -2\) ans.
In simple words: If three lines meet at the same point, we can find the meeting point using the two fully known lines. Then, we plug this point into the third line's equation to find the missing number \(k\).

Exam Tip: For three lines \(a_1x + b_1y + c_1 = 0\), \(a_2x + b_2y + c_2 = 0\), and \(a_3x + b_3y + c_3 = 0\) to be concurrent, the determinant of their coefficients must be zero. This determinant method is a very fast way to solve objective-type questions!

Question 7. If the lines \(y = 3x + 1\) and \(2y = x + 3\) are equally inclined to the line \(y = mx + 4\). Find the value of \(m\).
Answer: Let's first list the slopes of the given lines:
\(l_1: y = 3x + 1 \Rightarrow m_1 = 3\)
\(l_3: 2y = x + 3 \Rightarrow y = \frac{1}{2}x + \frac{3}{2} \Rightarrow m_3 = \frac{1}{2}\)
Let the slope of our target line \(l_2\) be \(m\). The angle \(\theta\) between line \(l_1\) and line \(l_2\) is given by:
\(\tan\theta = \left|\frac{3 - m}{1 + 3m}\right|\) .......... (i)
Similarly, the angle \(\theta\) between line \(l_2\) and line \(l_3\) is calculated as:
\(\tan\theta = \left|\frac{\frac{1}{2} - m}{1 + \frac{1}{2}m}\right| = \left|\frac{1 - 2m}{2 + m}\right|\) .......... (ii)
Equating the two expressions for \(\tan\theta\), we get:
\(\left|\frac{3 - m}{1 + 3m}\right| = \left|\frac{1 - 2m}{2 + m}\right|\)
\(\Rightarrow \frac{3 - m}{1 + 3m} = \pm \left(\frac{1 - 2m}{2 + m}\right)\)
Case 1: \(\frac{3 - m}{1 + 3m} = \frac{1 - 2m}{2 + m}\)
\(\Rightarrow 6 - 2m + 3m - m^2 = 1 - 2m + 3m - 6m^2\)
\(\Rightarrow 5m^2 = -5\)
\(\Rightarrow m^2 = -1 \Rightarrow m = \pm i\) (discarded, no real value of \(m\))
Case 2: \(\frac{3 - m}{1 + 3m} = -\left(\frac{1 - 2m}{2 + m}\right)\)
\(\Rightarrow 6 - 2m + 3m - m^2 = -1 + 2m - 3m + 6m^2\)
\(\Rightarrow 7m^2 - 2m - 7 = 0\)
Applying the quadratic formula to solve for \(m\):
\(m = \frac{2 \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)}\)
\(m = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14}\)
\(m = \frac{2 \pm 10\sqrt{2}}{14}\)
\(\therefore m = \frac{1 \pm 5\sqrt{2}}{7}\) ans.

In simple words: If a line makes equal angles with two other lines, the tangent of the angle between the first and second line must equal the tangent of the angle between the second and third line. We solve this equation to find the slope \(m\).

Exam Tip: Do not forget the \(\pm\) sign when removing absolute value bars. One sign will lead to imaginary slopes (which are rejected), while the other will yield the correct real slopes.

Question 8. Find the values of \(\alpha\) and \(p\) if the equation \(x \cos \alpha + y \sin \alpha = p\) is the normal form of \(\sqrt{3}x + y + 2 = 0\).
Answer: The given equation of the line is:
\(\sqrt{3}x + y + 2 = 0\)
\(\Rightarrow \sqrt{3}x + y = -2\)
We multiply the entire equation by \(-1\) to ensure the constant on the right-hand side is positive:
\(\Rightarrow -\sqrt{3}x - y = 2\) - (making the RHS positive)
Comparing this with \(ax + by = d\), we have \(a = -\sqrt{3}\) and \(b = -1\).
Now, we divide every term by \(\sqrt{a^2 + b^2} = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\):
\(\Rightarrow -\frac{\sqrt{3}}{2}x + \left(-\frac{1}{2}\right)y = \frac{2}{2}\)
\(\Rightarrow -\frac{\sqrt{3}}{2}x - \frac{1}{2}y = 1\)
By comparing this resulting equation with the standard normal form \(x \cos\alpha + y \sin\alpha = p\), we find:
\(\cos\alpha = -\frac{\sqrt{3}}{2}\) and \(\sin\alpha = -\frac{1}{2}\)
Since both sine and cosine are negative, \(\alpha\) lies in the third quadrant:
\(\alpha = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\)
Comparing the constants, we get:
\(p = 1\)
Thus, \(\alpha = \frac{7\pi}{6}\) & \(p = 1\) ans.
In simple words: To convert any line to its normal form, we first move the constant to the right side and make it positive. Then, we divide every term by the square root of the sum of the squares of the \(x\) and \(y\) coefficients.

Exam Tip: The perpendicular distance \(p\) from the origin must always be positive. If your constant on the right side of the equation is negative, multiply the whole equation by \(-1\) before dividing by \(\sqrt{A^2 + B^2}\).

Question 9. Show that the path of a moving point such that its distance from two lines \(3x - 2y = 5\) and \(3x + 2y = 5\) are equal is a straight line.
Answer: First, let's write the given lines in general form:
\(3x - 2y - 5 = 0\) ....... (i)
and \(3x + 2y - 5 = 0\) ....... (ii)
Let \(P(x,y)\) be any variable point on the path. According to the problem, the perpendicular distances from \(P\) to both lines are equal. Applying the point-to-line distance formula:
\(\frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}}\)
\(\Rightarrow \frac{|3x - 2y - 5|}{\sqrt{13}} = \frac{|3x + 2y - 5|}{\sqrt{13}}\)
\(\Rightarrow |3x - 2y - 5| = |3x + 2y - 5|\)
Removing the absolute value signs gives two possible linear equations:
\(3x - 2y - 5 = \pm(3x + 2y - 5)\)
Case A (positive sign):
\(3x - 2y - 5 = 3x + 2y - 5\)
\(\Rightarrow -4y = 0 \Rightarrow y = 0\)
Case B (negative sign):
\(3x - 2y - 5 = -(3x + 2y - 5)\)
\(\Rightarrow 3x - 2y - 5 = -3x - 2y + 5\)
\(\Rightarrow 6x = 10 \Rightarrow x = \frac{5}{3}\)
Both equations represent straight lines (specifically, the angle bisectors of the given lines). Thus, the path of the point \(P\) is indeed a straight line.

In simple words: The set of points that are at equal distances from two intersecting lines forms the angle bisectors of those lines. Since these bisectors are straight lines, the path is always a straight line.

Exam Tip: When an absolute value equation \(|A| = |B|\) is solved, it splits into \(A = B\) and \(A = -B\). Both paths represent straight lines which bisect the angles between the original lines.

Question 10. If sum of the perpendicular distances of a variable point \(P(x, y)\) from the lines \(x + y - 5 = 0\) and \(3x - 2y + 7 = 0\) is always 10. Show that \(P\) must move on a line.
Answer: We are given that the sum of the two perpendicular distances is \(10\):
\(\frac{|x + y - 5|}{\sqrt{1^2 + 1^2}} + \frac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = 10\)
\(\Rightarrow \frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10\)
Multiplying the entire equation by \(\sqrt{26}\):
\(\Rightarrow \sqrt{13} |x + y - 5| + \sqrt{2} |3x - 2y + 7| = 10\sqrt{26}\)
Since there are absolute value signs, the coordinates can lie in four different regions, giving four cases based on the signs of the terms inside the absolute values: \((+, +)\), \((+, -)\), \((-, +)\), \((-, -)\).
Let's examine the first case where both expressions inside the absolute values are positive:
\(\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}\)
Rearranging the terms:
\(\Rightarrow x(\sqrt{13} + 3\sqrt{2}) + y(\sqrt{13} - 2\sqrt{2}) - 5\sqrt{13} + 7\sqrt{2} - 10\sqrt{26} = 0\)
This simplifies to a linear equation of the form \(Ax + By + C = 0\), which is the standard equation of a straight line. Hence, the locus of point \(P\) is a straight line. By applying similar sign combinations, the other three cases also yield equations of straight lines.

In simple words: The sum of the distances from a point to two lines is written using absolute values. When we remove these absolute value signs for any of the four possible regions, we get a first-degree equation in \(x\) and \(y\), which always represents a straight line.

Exam Tip: Always state that removing absolute values yields a linear equation of the form \(Ax + By + C = 0\). Mentioning that any first-degree polynomial in two variables represents a straight line is key to scoring full explanation marks.

Click on link below to download CBSE Class 11 Mathematics Straight Lines Worksheet (4).