NCERT Class 11 Maths Chapter 10 Exercise 10.3 Solutions & Concepts

Overview of General Equation and Reductions

The general equation of a straight line is written as $Ax + By + C = 0$, where $A$, $B$, and $C$ are real constants, and $A$ and $B$ are not both zero. While this general form is universally applicable, it does not directly convey the line's geometric characteristics. To immediately find properties like the slope, intercepts, or normal distance, we reduce the general equation into standard forms:

1. Slope-Intercept Form ($y = mx + c$): By expressing $y$ in terms of $x$, we get $y = -\frac{A}{B}x - \frac{C}{B}$ (when $B \neq 0$). Here, the slope is $m = -\frac{A}{B}$ and the y-intercept is $c = -\frac{C}{B}$.

1. Intercept Form ($\frac{x}{a} + \frac{y}{b} = 1$): By rearranging the general equation to $Ax + By = -C$ and dividing by $-C$ (when $C \neq 0$), we get $\frac{x}{-C/A} + \frac{y}{-C/B} = 1$. The x-intercept is $a = -\frac{C}{A}$ and the y-intercept is $b = -\frac{C}{B}$.

1. Normal Form ($x\cos\omega + y\sin\omega = p$): Here, $p$ is the perpendicular distance from the origin to the line ($p > 0$), and $\omega$ is the angle made by this normal with the positive x-axis. To reduce to this form, we divide the entire equation by $\pm\sqrt{A^2 + B^2}$. The sign of the denominator is chosen to make the constant term on the right-hand side positive.

Key Distance Formulas in Exercise 10.3

Distance of a Point from a Line

The perpendicular distance $d$ from a point $P(x_1, y_1)$ to a line $Ax + By + C = 0$ is calculated using the formula: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

Distance Between Parallel Lines

The shortest distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula: $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.

Step-by-Step Reduction to Normal Form

1. Isolate the Constant — Rearrange the general equation $Ax + By + C = 0$ by shifting the constant term to the right-hand side: $Ax + By = -C$.
2. Ensure Positive Right-Hand Side — Since the distance $p$ must always be positive, the term on the right-hand side must be positive. If $-C$ is negative, multiply the entire equation by $-1$ so the right-hand side constant becomes positive.
3. Calculate the Normalizing Factor — Find the value of $\sqrt{A^2 + B^2}$ using the coefficients of $x$ and $y$ from the current equation.
4. Divide and Identify Parameters — Divide every term in the equation by $\sqrt{A^2 + B^2}$. Compare the resulting equation with $x\cos\omega + y\sin\omega = p$ to determine $\cos\omega$, $\sin\omega$, and $p$. Determine the correct quadrant of $\omega$ based on the signs of $\sin\omega$ and $\cos\omega$.

Crucial Exam Traps & Tips

• The Positive distance rule: In the normal form $x\cos\omega + y\sin\omega = p$, $p$ represents distance and MUST be positive. If you get a negative value, you forgot to adjust the sign of the constant term before dividing.
• Matching Coefficients for Parallel Lines: Before applying the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$, make sure the coefficients of $x$ and $y$ are identical in both equations. For example, if the lines are $3x + 4y - 6 = 0$ and $6x + 8y + 9 = 0$, divide the second equation by 2 first to get $3x + 4y + 4.5 = 0$ so that $A = 3$ and $B = 4$ match.

Practice Questions with Solutions

• Q: Reduce the equation $x - \sqrt{3}y + 8 = 0$ into normal form. Find the values of $p$ and $\omega$. A: Step 1: Write down the given equation: $x - \sqrt{3}y + 8 = 0$. Step 2: Move the constant term to the right-hand side: $x - \sqrt{3}y = -8$. Step 3: Since the RHS must be positive, multiply the entire equation by $-1$: $-x + \sqrt{3}y = 8$. Step 4: Divide both sides by $\sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. This gives: $-\frac{1}{2}x + \frac{\sqrt{3}}{2}y = 4$. Step 5: Compare this with $x\cos\omega + y\sin\omega = p$. We get $\cos\omega = -\frac{1}{2}$, $\sin\omega = \frac{\sqrt{3}}{2}$, and $p = 4$. Step 6: Since $\cos\omega$ is negative and $\sin\omega$ is positive, $\omega$ lies in the second quadrant. Therefore, $\omega = 180^\circ - 60^\circ = 120^\circ$ (or $\frac{2\pi}{3}$ radians). Final answer: The normal form is $x\cos 120^\circ + y\sin 120^\circ = 4$, where $p = 4$ and $\omega = 120^\circ$.
• Q: Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$. A: Step 1: Simplify the given equation of the line into general form: $12x + 72 = 5y - 10 \implies 12x - 5y + 82 = 0$. Step 2: Identify the coefficients: $A = 12$, $B = -5$, $C = 82$, and the point $(x_1, y_1) = (-1, 1)$. Step 3: Use the perpendicular distance formula: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Step 4: Substitute the values: $d = \frac{|12(-1) - 5(1) + 82|}{\sqrt{12^2 + (-5)^2}} = \frac{|-12 - 5 + 82|}{\sqrt{144 + 25}} = \frac{|65|}{\sqrt{169}} = \frac{65}{13} = 5$. Final answer: The distance of the point from the line is 5 units.
• Q: Find the distance between the parallel lines $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$. A: Step 1: Identify the coefficients. Here, $A = 15$, $B = 8$, $C_1 = -34$, and $C_2 = 31$. Step 2: Use the distance formula between parallel lines: $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Step 3: Substitute the coefficients: $d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}} = \frac{|-65|}{\sqrt{225 + 64}} = \frac{65}{\sqrt{289}} = \frac{65}{17}$. Final answer: The distance between the parallel lines is $\frac{65}{17}$ units.
• Q: Find the equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through the point $(-2, 3)$. A: Step 1: Any line parallel to the line $3x - 4y + 2 = 0$ can be represented as $3x - 4y + k = 0$, where $k$ is a constant. Step 2: Since this line passes through the point $(-2, 3)$, substitute $x = -2$ and $y = 3$ into the equation: $3(-2) - 4(3) + k = 0$. Step 3: Solve for $k$: $-6 - 12 + k = 0 \implies -18 + k = 0 \implies k = 18$. Step 4: Substitute the value of $k$ back into the parallel line equation: $3x - 4y + 18 = 0$. Final answer: The required equation of the line is $3x - 4y + 18 = 0$.

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