# Grade 10 Optional Mathematics: Angle Between Two Lines
## Unit: Coordinate Geometry | Chapter: Angle Between Two Lines | Complete Solutions

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## 🎯 Chapter Introduction

Welcome to the comprehensive study guide for finding the **Angle Between Two Straight Lines**. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to all 29 unique problems from your materials. By the end of this guide, you will have mastered the concepts of slope, parallel and perpendicular lines, homogeneous equations, and finding the equations of lines based on various conditions. 📐

### Key Learning Areas
1.  **Slope (Gradient):** Calculating the slope of a line from two points or from its equation.
2.  **Angle Formula:** Using the slopes of two lines to find the angle between them.
3.  **Parallel & Perpendicular Lines:** Understanding and applying the conditions for parallelism and perpendicularity.
4.  **Equations of Lines:** Finding the equation of a line that is parallel or perpendicular to another line and passes through a given point or satisfies other conditions.
5.  **Homogeneous Equations:** Finding the angle between the pair of lines represented by a second-degree homogeneous equation.

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## 🔑 Key Concepts & Formulas

Before we solve the problems, let's understand the main tools we'll be using.

### 1. The Slope of a Straight Line (m)

The slope represents the steepness of a line.

* **From Two Points $(x_1, y_1)$ and $(x_2, y_2)$:**
    $$m = \frac{y_2 - y_1}{x_2 - x_1}$$

* **From the General Equation $Ax + By + C = 0$:**
    $$m = -\frac{\text{Coefficient of x}}{\text{Coefficient of y}} = -\frac{A}{B}$$

* **From the Slope-Intercept Form $y = mx + c$:**
    The slope is simply `m`.

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### 2. Angle Between Two Lines



If two lines have slopes $m_1$ and $m_2$, the angle ($\theta$) between them is given by:

* **General Formula:**
    $$\tan\theta = \pm \left( \frac{m_1 - m_2}{1 + m_1m_2} \right)$$
    * **Acute Angle:** For the smaller angle (less than 90°), we take the positive value: $\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$.
    * **Obtuse Angle:** For the larger angle (greater than 90°), we take the negative value.

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### 3. Conditions for Parallel and Perpendicular Lines

* **Parallel Lines:** Two lines are parallel if and only if their slopes are equal.
    $$m_1 = m_2$$

* **Perpendicular Lines:** Two lines are perpendicular if and only if the product of their slopes is -1.
    $$m_1 \cdot m_2 = -1$$

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### 4. Special Equations of Lines

* **Equation of a line parallel to $Ax + By + C = 0$:**
    The equation will be of the form $Ax + By + k = 0$, where `k` is a new constant found using another condition (like a point on the line).

* **Equation of a line perpendicular to $Ax + By + C = 0$:**
    The equation will be of the form $Bx - Ay + k = 0$. (Note the swapped coefficients of x and y, and the sign change for the y-coefficient).

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### 5. Pair of Straight Lines (Homogeneous Equation)

A second-degree homogeneous equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines passing through the origin.

* **Angle ($\theta$) between the pair of lines:**
    $$\tan\theta = \frac{\pm 2\sqrt{h^2 - ab}}{a+b}$$

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## 📚 Exercise Solutions

### Short Questions

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#### Question 1: Finding the Slope of a Perpendicular Line
**Find the slope of the line perpendicular to the straight line 3x - 4y = 10.**

**Solution:**
First, find the slope of the given line $3x - 4y = 10$.
$$m_1 = -\frac{\text{Coefficient of x}}{\text{Coefficient of y}} = -\frac{3}{-4} = \frac{3}{4}$$
Let the slope of the perpendicular line be $m_2$. The condition for perpendicularity is $m_1 \cdot m_2 = -1$.
$$
\begin{aligned}
\frac{3}{4} \cdot m_2 &= -1 \\
m_2 &= -\frac{4}{3}
\end{aligned}
$$
The slope of the perpendicular line is **$-\frac{4}{3}$**.

***

#### Question 2: Proving Lines are Parallel
**Prove that the straight line passing through the points (3, -4) and (-2, 6) and the straight line with the equation $2x + y + 3 = 0$ are parallel.**

**Solution:**
**Step 1: Find the slope of the line from the equation.**
For the line $2x + y + 3 = 0$:
$$m_1 = -\frac{\text{Coefficient of x}}{\text{Coefficient of y}} = -\frac{2}{1} = -2$$
**Step 2: Find the slope of the line from the two points.**
For the points $(3, -4)$ and $(-2, 6)$:
$$m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-4)}{-2 - 3} = \frac{10}{-5} = -2$$
**Step 3: Compare the slopes.**
Since $m_1 = m_2 = -2$, the two lines are parallel. **(Proved)**

***

#### Question 3: Finding an Unknown for Parallel Lines
**If the lines $3x + my = 5$ and $\frac{x}{2} + \frac{y}{3} = 1$ are parallel to each other, find the value of m.**

**Solution:**
**Step 1: Find the slope of the first line.**
For $3x + my = 5$, the slope is $m_1 = -\frac{3}{m}$.

**Step 2: Find the slope of the second line.**
First, convert $\frac{x}{2} + \frac{y}{3} = 1$ to the general form by taking the LCM (6):
$$3x + 2y = 6 \implies 3x + 2y - 6 = 0$$
The slope is $m_2 = -\frac{3}{2}$.

**Step 3: Equate the slopes and solve for m.**
For parallel lines, $m_1 = m_2$.
$$
\begin{aligned}
-\frac{3}{m} &= -\frac{3}{2} \\
\frac{3}{m} &= \frac{3}{2} \\
m &= 2
\end{aligned}
$$
Thus, **$m = 2$**.

***

#### Question 4: Finding an Unknown for Perpendicular Lines
**If the straight line having equation 3x + 4y = 12 and 2x - py = 5 are perpendicular to each other, find the value of p.**

**Solution:**
**Step 1: Find the slopes of both lines.**
For $3x + 4y = 12$, the slope is $m_1 = -\frac{3}{4}$.
For $2x - py = 5$, the slope is $m_2 = -\frac{2}{-p} = \frac{2}{p}$.

**Step 2: Use the perpendicularity condition and solve for p.**
For perpendicular lines, $m_1 \cdot m_2 = -1$.
$$
\begin{aligned}
\left(-\frac{3}{4}\right) \cdot \left(\frac{2}{p}\right) &= -1 \\
-\frac{6}{4p} &= -1 \\
6 &= 4p \\
p &= \frac{6}{4} = \frac{3}{2}
\end{aligned}
$$
Thus, **$p = \frac{3}{2}$**.

***

#### Question 5: Finding an Unknown for Perpendicular Lines
**If the line passing through points (3, -4) and (-2, k) is perpendicular to the line having equation 5x + y = -3, find the value of k.**

**Solution:**
**Step 1: Find the slope of the line from the two points.**
$$m_1 = \frac{k - (-4)}{-2 - 3} = \frac{k+4}{-5}$$
**Step 2: Find the slope of the line from the equation.**
For $5x + y = -3$, the slope is $m_2 = -\frac{5}{1} = -5$.

**Step 3: Use the perpendicularity condition and solve for k.**
$$
\begin{aligned}
m_1 \cdot m_2 &= -1 \\
\left(\frac{k+4}{-5}\right) \cdot (-5) &= -1 \\
k+4 &= -1 \\
k &= -5
\end{aligned}
$$
Thus, **$k=-5$**.

***

#### Question 6: Acute Angle Between Lines
**Find the acute angle between the lines having slope $-\sqrt{3}$ and $\sqrt{3}$ respectively.**

**Solution:**
Let $m_1 = -\sqrt{3}$ and $m_2 = \sqrt{3}$. Let $\theta$ be the acute angle.
$$
\begin{aligned}
\tan\theta &= \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| \\
&= \left| \frac{-\sqrt{3} - \sqrt{3}}{1 + (-\sqrt{3})(\sqrt{3})} \right| \\
&= \left| \frac{-2\sqrt{3}}{1 - 3} \right| \\
&= \left| \frac{-2\sqrt{3}}{-2} \right| \\
\tan\theta &= \sqrt{3}
\end{aligned}
$$
Since $\tan(60^\circ) = \sqrt{3}$, the acute angle is **60°**.

***

#### Question 7: Acute Angle Between Lines
**If the slopes of the lines AB and CD are 3 and -2 respectively, find the acute angle between them.**

**Solution:**
Given slopes, $m_1 = 3$ and $m_2 = -2$.
Let $\theta$ be the acute angle.
$$
\begin{aligned}
\tan\theta &= \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| \\
&= \left| \frac{3 - (-2)}{1 + (3)(-2)} \right| \\
&= \left| \frac{5}{1 - 6} \right| \\
&= \left| \frac{5}{-5} \right| = 1
\end{aligned}
$$
Since $\tan(45^\circ) = 1$, the acute angle is **45°**.

***

#### Question 8: Finding an Unknown for Parallel Lines
**If the straight lines Px + 3y - 12 = 0 and 4y - 3x + 7 = 0 are parallel to each other, find the value of P.**

**Solution:**
**Step 1: Find the slopes.**
For $Px + 3y - 12 = 0$, the slope is $m_1 = -\frac{P}{3}$.
For $-3x + 4y + 7 = 0$, the slope is $m_2 = -\frac{-3}{4} = \frac{3}{4}$.

**Step 2: Equate the slopes.**
For parallel lines, $m_1 = m_2$.
$$
\begin{aligned}
-\frac{P}{3} &= \frac{3}{4} \\
-4P &= 9 \\
P &= -\frac{9}{4}
\end{aligned}
$$
Thus, **$P = -9/4$**.

***

#### Question 9: Finding an Unknown for Parallel Lines
**If the lines 2x - 8y + 6 = 0 and px - 12y - 4 = 3 are parallel to each other, find the value of p.**

**Solution:**
**Step 1: Find the slopes.**
For $2x - 8y + 6 = 0$, the slope is $m_1 = -\frac{2}{-8} = \frac{1}{4}$.
For $px - 12y - 7 = 0$, the slope is $m_2 = -\frac{p}{-12} = \frac{p}{12}$.

**Step 2: Equate the slopes.**
$$
\begin{aligned}
m_1 &= m_2 \\
\frac{1}{4} &= \frac{p}{12} \\
12 &= 4p \\
p &= 3
\end{aligned}
$$
Thus, **$p=3$**.

***

#### Question 10: Prove Lines are Perpendicular
**Show that the lines $y + \sqrt{3}x + 4 = 0$ and $x - \sqrt{3}y = 5$ are perpendicular to each other.**

**Solution:**
**Step 1: Find the slopes.**
For $\sqrt{3}x + y + 4 = 0$, the slope is $m_1 = -\frac{\sqrt{3}}{1} = -\sqrt{3}$.
For $x - \sqrt{3}y - 5 = 0$, the slope is $m_2 = -\frac{1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}$.

**Step 2: Check the perpendicularity condition.**
$$m_1 \cdot m_2 = (-\sqrt{3}) \cdot \left(\frac{1}{\sqrt{3}}\right) = -1$$
Since the product of the slopes is -1, the lines are perpendicular. **(Proved)**

***

#### Question 11: Prove Lines are Perpendicular
**Show that the lines $ax + by + c = 0$ and $bx - ay + c = 0$ are perpendicular to each other.**

**Solution:**
**Step 1: Find the slopes.**
For $ax + by + c = 0$, the slope is $m_1 = -\frac{a}{b}$.
For $bx - ay + c = 0$, the slope is $m_2 = -\frac{b}{-a} = \frac{b}{a}$.

**Step 2: Check the perpendicularity condition.**
$$m_1 \cdot m_2 = \left(-\frac{a}{b}\right) \cdot \left(\frac{b}{a}\right) = -1$$
Since the product of the slopes is -1, the lines are perpendicular. **(Proved)**

***

#### Question 12: Finding an Unknown for Perpendicular Lines
**If the straight lines 2x + 3y + 6 = 0 and ax - 5y + 20 = 0 are perpendicular to each other, find the value of a.**

**Solution:**
**Step 1: Find the slopes.**
For $2x + 3y + 6 = 0$, the slope is $m_1 = -\frac{2}{3}$.
For $ax - 5y + 20 = 0$, the slope is $m_2 = -\frac{a}{-5} = \frac{a}{5}$.

**Step 2: Use the perpendicularity condition.**
$$
\begin{aligned}
m_1 \cdot m_2 &= -1 \\
\left(-\frac{2}{3}\right) \cdot \left(\frac{a}{5}\right) &= -1 \\
-\frac{2a}{15} &= -1 \\
2a &= 15 \\
a &= \frac{15}{2}
\end{aligned}
$$
Thus, **$a = 15/2$**.

***

#### Question 13: Finding an Unknown for Parallel Lines
**If a line passing through the points (4, -p) and (-2, 6) is parallel to the line having equation 2y + 3x = -3, find the value of p.**

**Solution:**
**Step 1: Find the slopes.**
Slope of the line passing through points (4, -p) and (-2, 6):
$$m_1 = \frac{6 - (-p)}{-2 - 4} = \frac{6+p}{-6}$$
Slope of the line $3x + 2y + 3 = 0$:
$$m_2 = -\frac{3}{2}$$
**Step 2: Equate the slopes.**
$$
\begin{aligned}
\frac{6+p}{-6} &= -\frac{3}{2} \\
2(6+p) &= (-3)(-6) \\
12 + 2p &= 18 \\
2p &= 6 \\
p &= 3
\end{aligned}
$$
Thus, **$p=3$**.

***

### Long Questions

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#### Question 14: Angle between a Pair of Lines
**If an angle between the pair of lines represented by the equation $2x^2 + kxy + 3y^2 = 0$ is 45°, find the positive value of k and also find the separate equations of the lines.**

**Solution:**
**Step 1: Find k.**
Comparing $2x^2 + kxy + 3y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$, we get:
$a=2$, $b=3$, $2h=k \implies h=k/2$.
Given $\theta = 45^\circ$.
$$
\begin{aligned}
\tan\theta &= \frac{\pm 2\sqrt{h^2 - ab}}{a+b} \\
\tan45^\circ &= \frac{\pm 2\sqrt{(k/2)^2 - (2)(3)}}{2+3} \\
1 &= \frac{\pm 2\sqrt{k^2/4 - 6}}{5} \\
5 &= \pm \sqrt{k^2 - 24} \quad (\text{since } \sqrt{4}=2)
\end{aligned}
$$Squaring both sides:$$
\begin{aligned}
25 &= k^2 - 24 \\
k^2 &= 49 \\
k &= \pm 7
\end{aligned}
$$
The positive value is **$k=7$**.

**Step 2: Find the separate equations.**
Substitute $k=7$ into the original equation:
$$
\begin{aligned}
2x^2 + 7xy + 3y^2 &= 0 \\
2x^2 + 6xy + xy + 3y^2 &= 0 \\
2x(x+3y) + y(x+3y) &= 0 \\
(2x+y)(x+3y) &= 0
\end{aligned}
$$
The separate equations are **$2x+y=0$** and **$x+3y=0$**.

***

#### Question 15: Equation of a Line through Intersection
**Find the equation of a straight line passing through the point of intersection of the lines $3x + 4y = 7$ and $5x - 2y = 3$ and perpendicular to the line $2x + 3y = 5$.**

**Solution:**
**Step 1: Find the point of intersection.**
From $3x + 4y = 7$, we get $x = \frac{7-4y}{3}$.
Substitute into the second equation:
$$
\begin{aligned}
5\left(\frac{7-4y}{3}\right) - 2y &= 3 \\
35 - 20y - 6y &= 9 \\
35 - 26y &= 9 \\
26 &= 26y \\
y &= 1
\end{aligned}
$$
Now find x: $x = \frac{7-4(1)}{3} = \frac{3}{3} = 1$. The intersection point is (1, 1).

**Step 2: Find the equation of the required line.**
The slope of $2x + 3y = 5$ is $m_1 = -2/3$.
The required line is perpendicular, so its slope is $m_2 = -1/m_1 = 3/2$.
The line passes through (1, 1) with slope 3/2. Using point-slope form:
$$
\begin{aligned}
y - 1 &= \frac{3}{2}(x - 1) \\
2y - 2 &= 3x - 3 \\
3x - 2y - 1 &= 0
\end{aligned}
$$
The required equation is **$3x - 2y - 1 = 0$**.

***

#### Question 16: Equation of a Perpendicular Bisector
**Find the equation of the perpendicular bisector of the line segment joining the points A(3, -7) and B(-5, 3).**

**Solution:**
**Step 1: Find the midpoint of AB.**
$$Midpoint = \left( \frac{3 + (-5)}{2}, \frac{-7+3}{2} \right) = (-1, -2)$$
**Step 2: Find the slope of line AB.**
$$m_{AB} = \frac{3 - (-7)}{-5 - 3} = \frac{10}{-8} = -\frac{5}{4}$$
**Step 3: Find the slope of the perpendicular bisector.**
$$m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-5/4} = \frac{4}{5}$$
**Step 4: Find the equation of the bisector.**
The line passes through $(-1, -2)$ with slope $4/5$.
$$
\begin{aligned}
y - (-2) &= \frac{4}{5}(x - (-1)) \\
5(y+2) &= 4(x+1) \\
5y + 10 &= 4x + 4 \\
4x - 5y - 6 &= 0
\end{aligned}
$$
The required equation is **$4x - 5y - 6 = 0$**.

***

#### Question 17: Equation of a Line through a Centroid
**A(3, 2), B(1, -1) and C(5, -5) are the vertices of a triangle ABC. Find the equation of a straight line passing through the centroid of triangle ABC and parallel to the side BC.**

**Solution:**
**Step 1: Find the centroid of triangle ABC.**
$$Centroid = \left( \frac{3+1+5}{3}, \frac{2+(-1)+(-5)}{3} \right) = \left( \frac{9}{3}, \frac{-4}{3} \right) = (3, -4/3)$$
**Step 2: Find the slope of the side BC.**
$$m_{BC} = \frac{-5 - (-1)}{5 - 1} = \frac{-4}{4} = -1$$
**Step 3: Find the equation of the required line.**
The required line is parallel to BC, so its slope is also $m = -1$.
It passes through the centroid $(3, -4/3)$.
$$
\begin{aligned}
y - (-\frac{4}{3}) &= -1(x - 3) \\
y + \frac{4}{3} &= -x + 3 \\
3y + 4 &= -3x + 9 \\
3x + 3y - 5 &= 0
\end{aligned}
$$
The required equation is **$3x + 3y - 5 = 0$**.

***

#### Question 18: Equation of a Line passing through Centroid
**Find the equation of the line passing through the centroid of triangle PQR with vertices P(3, 3), Q(-2, -6) and R(5, -3) and parallel to the line QR.**

**Solution:**
**Step 1: Find the centroid of triangle PQR.**
$$Centroid = \left( \frac{3+(-2)+5}{3}, \frac{3+(-6)+(-3)}{3} \right) = \left( \frac{6}{3}, \frac{-6}{3} \right) = (2, -2)$$
**Step 2: Find the slope of the side QR.**
$$m_{QR} = \frac{-3 - (-6)}{5 - (-2)} = \frac{3}{7}$$
**Step 3: Find the equation of the required line.**
The required line is parallel to QR, so its slope is also $m = 3/7$.
It passes through the centroid $(2, -2)$.
$$
\begin{aligned}
y - (-2) &= \frac{3}{7}(x - 2) \\
7(y+2) &= 3(x-2) \\
7y + 14 &= 3x - 6 \\
3x - 7y - 20 &= 0
\end{aligned}
$$
The required equation is **$3x - 7y - 20 = 0$**.