OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Exercise 16 (D)

Exercise 16(D)

Question 1. Write down the slopes of the following lines :
(i) \( 2x + 3y + 1 = 0 \)
(ii) \( 7x – 5y + 8 = 0 \)
(iii) \( – 6y – 11x = 0 \)
(iv) \( xx_1 + yy_1 = a^2 \)
(v) \( 3x + 4y - 2(x + x_1) - 5(y + y_1) + 2 = 0 \)
Answer:
(i) The given equation of the line is \( 2x + 3y + 1 = 0 \).
The slope of a line in the form \( Ax + By + C = 0 \) is given by \( m = -\frac{\text{coefficient of } x}{\text{coefficient of } y} \).
So, the slope of the given line is \( -\frac{2}{3} \). This is a fundamental formula for finding the slope.
(ii) The given equation of the line is \( 7x – 5y + 8 = 0 \).
Using the same formula, the slope is \( -\frac{7}{-5} = \frac{7}{5} \). The signs cancel out to give a positive slope.
(iii) The given equation of the line is \( – 6y – 11x = 0 \).
We can rewrite this as \( -11x - 6y = 0 \).
Using the formula, the slope is \( -\frac{-11}{-6} = -\frac{11}{6} \). A negative slope means the line goes downwards from left to right.
(iv) The given equation of the line is \( xx_1 + yy_1 = a^2 \).
Here, \( x_1 \) and \( y_1 \) are constants. So the coefficient of \( x \) is \( x_1 \) and the coefficient of \( y \) is \( y_1 \).
The slope is \( -\frac{x_1}{y_1} \). This represents the slope of a tangent line to a circle or similar curve.
(v) The given equation of the line is \( 3x + 4y - 2(x + x_1) - 5(y + y_1) + 2 = 0 \).
First, we need to simplify this equation:
\( 3x + 4y - 2x - 2x_1 - 5y - 5y_1 + 2 = 0 \)
Combine the \( x \) and \( y \) terms:
\( (3x - 2x) + (4y - 5y) - 2x_1 - 5y_1 + 2 = 0 \)
\( x - y - 2x_1 - 5y_1 + 2 = 0 \)
Now, the coefficient of \( x \) is 1 and the coefficient of \( y \) is -1.
The slope is \( -\frac{1}{-1} = 1 \). This simplified form helps in easily finding the slope.
In simple words: To find the slope of a line written as \( Ax + By + C = 0 \), you use the rule \( -\frac{A}{B} \). This tells you how steep the line is and in which direction it moves.

🎯 Exam Tip: Always rearrange the equation into the form \( Ax + By + C = 0 \) before applying the slope formula \( -\frac{A}{B} \). Remember to pay close attention to the signs of A and B.

Question 2. Find the value of k such that the line \( (k – 2) x+ (k + 3) y − 5 = 0 \) is
(i) parallel to the line \( 2x – y + 7 = 0 \)
(ii) perpendicular to it.
Answer:
Given equation of the first line is \( (k – 2)x + (k + 3)y – 5 = 0 \). We can call this line (1).
The slope of line (1), let's call it \( m_1 \), is \( -\frac{(k-2)}{(k+3)} \).
The given second line is \( 2x – y + 7 = 0 \). Let's call this line (2).
The slope of line (2), \( m_2 \), is \( -\frac{2}{-1} = 2 \). This provides a known value to compare with.

(i) If both lines are parallel, then their slopes must be equal.
So, \( m_1 = m_2 \)
\( -\frac{(k-2)}{(k+3)} = 2 \)
Multiply both sides by \( (k+3) \):
\( -(k-2) = 2(k+3) \)
\( -k + 2 = 2k + 6 \)
Now, bring all k terms to one side and constants to the other:
\( 2 - 6 = 2k + k \)
\( -4 = 3k \)
\( k = -\frac{4}{3} \). This value makes the lines run in the same direction.

(ii) If both given lines are perpendicular, the product of their slopes must be -1.
So, \( m_1 m_2 = -1 \)
\( -\frac{(k-2)}{(k+3)} \times 2 = -1 \)
\( -\frac{2(k-2)}{(k+3)} = -1 \)
Multiply both sides by \( -(k+3) \):
\( 2(k-2) = (k+3) \)
\( 2k - 4 = k + 3 \)
Subtract k from both sides and add 4 to both sides:
\( 2k - k = 3 + 4 \)
\( k = 7 \). This value makes the lines cross at a 90-degree angle.
In simple words: When two lines are parallel, they have the same slope. When they are perpendicular, their slopes multiply to give -1. We use these rules to find the unknown value 'k' in the line's equation.

🎯 Exam Tip: Remember the conditions for parallel lines (\( m_1 = m_2 \)) and perpendicular lines (\( m_1 m_2 = -1 \)). Make sure to simplify the algebraic expressions carefully to avoid calculation errors.

Question 3. Prove that the lines
(i) \( 3x + 4y − 7 = 0 \) and \( 28x – 21y + 50 = 0 \) are mutually perpendicular;
(ii) \( px + qy − r = 0 \) and \( – 4px – 4qy + 5s = 0 \) are parallel.
Answer:
(i) First, let's find the slopes of the given lines.
Line (1): \( 3x + 4y – 7 = 0 \)
The slope of line (1), \( m_1 \), is \( -\frac{3}{4} \).
Line (2): \( 28x – 21y + 50 = 0 \)
The slope of line (2), \( m_2 \), is \( -\frac{28}{-21} = \frac{28}{21} = \frac{4}{3} \).
Now, let's multiply their slopes:
\( m_1 m_2 = \left(-\frac{3}{4}\right) \times \left(\frac{4}{3}\right) = -1 \).
Since the product of their slopes is -1, both lines (1) and (2) are mutually perpendicular. This is the definition of perpendicular lines.

(ii) Next, let's find the slopes of these two lines.
Line (1): \( px + qy − r = 0 \)
The slope of line (1), \( m_1 \), is \( -\frac{p}{q} \).
Line (2): \( – 4px – 4qy + 5s = 0 \)
The slope of line (2), \( m_2 \), is \( -\frac{-4p}{-4q} = -\frac{4p}{4q} = -\frac{p}{q} \).
Here, we see that \( m_1 = m_2 \).
Thus, both given lines are parallel. This means they will never intersect.
In simple words: To check if lines are perpendicular, multiply their slopes; if the answer is -1, they are. To check if they are parallel, just compare their slopes; if they are the same, the lines are parallel.

🎯 Exam Tip: Always simplify the slope fractions to their lowest terms before comparing or multiplying them. This helps in easily identifying parallel or perpendicular relationships.

Question 4. Find the slope of the line which is perpendicular to the line \( 7x + 11y – 2 = 0 \).
Answer:
The given equation of the line is \( 7x + 11y – 2 = 0 \). Let's call this line (1).
The slope of line (1), let's call it \( m \), is \( -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{7}{11} \).
We need to find the slope of a line that is perpendicular to line (1). Let this be \( m_{\perp} \).
For perpendicular lines, the product of their slopes is -1, i.e., \( m \times m_{\perp} = -1 \).
So, \( m_{\perp} = -\frac{1}{m} \).
\( m_{\perp} = -\frac{1}{-\frac{7}{11}} \)
\( \implies m_{\perp} = \frac{11}{7} \). This is the slope of any line forming a 90-degree angle with the given line.
In simple words: If you know the slope of one line, you can find the slope of a line perpendicular to it by flipping the fraction and changing its sign. For example, if a slope is \( \frac{a}{b} \), the perpendicular slope is \( -\frac{b}{a} \).

🎯 Exam Tip: The negative reciprocal rule \( m_2 = -1/m_1 \) is essential for finding the slope of a perpendicular line. Be careful with fractions and negative signs during calculation.

Question 5. Determine the angle between the lines whose equations are
(i) \( 3x + y − 7 = 0 \) and \( x + 2y + 9 = 0 \),
(ii) \( 2x - y + 3 = 0 \) and \( x + y - 2 = 0 \).
Answer:
(i) First, find the slopes of both lines.
For line (1): \( 3x + y − 7 = 0 \)
The slope, \( m_1 \), is \( -\frac{3}{1} = -3 \).
For line (2): \( x + 2y + 9 = 0 \)
The slope, \( m_2 \), is \( -\frac{1}{2} \).
Let \( \theta \) be the acute angle between the given lines.
The formula for the tangent of the angle between two lines is \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \). This formula helps us find the angle based on their steepness.
\( \tan \theta = \left| \frac{-3 - \left(-\frac{1}{2}\right)}{1 + (-3)\left(-\frac{1}{2}\right)} \right| \)
\( \implies \tan \theta = \left| \frac{-3 + \frac{1}{2}}{1 + \frac{3}{2}} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{-6+1}{2}}{\frac{2+3}{2}} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{-5}{2}}{\frac{5}{2}} \right| \)
\( \implies \tan \theta = |-1| \)
\( \implies \tan \theta = 1 \).
Therefore, \( \theta = \tan^{-1}(1) = 45^\circ \). The acute angle is 45 degrees.

(ii) Again, find the slopes of both lines.
For line (1): \( 2x - y + 3 = 0 \)
The slope, \( m_1 \), is \( -\frac{2}{-1} = 2 \).
For line (2): \( x + y - 2 = 0 \)
The slope, \( m_2 \), is \( -\frac{1}{1} = -1 \).
Let \( \theta \) be the acute angle between these lines.
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \)
\( \implies \tan \theta = \left| \frac{2 - (-1)}{1 + (2)(-1)} \right| \)
\( \implies \tan \theta = \left| \frac{2 + 1}{1 - 2} \right| \)
\( \implies \tan \theta = \left| \frac{3}{-1} \right| \)
\( \implies \tan \theta = |-3| \)
\( \implies \tan \theta = 3 \).
Therefore, \( \theta = \tan^{-1}(3) \). Using a calculator or tables, this is approximately \( 71^\circ 34' \). This means the lines are quite steep relative to each other.
In simple words: To find the angle between two lines, first find their individual slopes. Then, use the special formula for \( \tan \theta \) with these slopes. The absolute value ensures you always get the acute angle.

🎯 Exam Tip: Remember to use the absolute value in the \( \tan \theta \) formula to find the acute angle. Carefully handle negative signs when substituting slope values into the formula to avoid calculation errors.

Question 6. Use tables to find the acute angle between the lines \( 2y + x = 0 \) and \( \frac {x}{1} + \frac { y }{ 2 } = 2 \).
Answer:
First, let's find the slopes of both lines.
For line (1): \( 2y + x = 0 \), which can be written as \( x + 2y = 0 \).
The slope, \( m_1 \), is \( -\frac{1}{2} \).
For line (2): \( \frac {x}{1} + \frac { y }{ 2 } = 2 \). To find its slope easily, we can rewrite it as \( x + \frac{y}{2} = 2 \). Multiply by 2 to clear the fraction: \( 2x + y = 4 \).
The slope, \( m_2 \), is \( -\frac{2}{1} = -2 \). This makes it easier to use the slope formula.
Let \( \theta \) be the acute angle between the given lines.
Using the formula \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \):
\( \tan \theta = \left| \frac{-\frac{1}{2} - (-2)}{1 + \left(-\frac{1}{2}\right)(-2)} \right| \)
\( \implies \tan \theta = \left| \frac{-\frac{1}{2} + 2}{1 + 1} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{-1+4}{2}}{2} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{3}{2}}{2} \right| \)
\( \implies \tan \theta = \left| \frac{3}{4} \right| \)
\( \implies \tan \theta = \frac{3}{4} \).
Therefore, \( \theta = \tan^{-1}\left(\frac{3}{4}\right) \). Using trigonometric tables, \( \theta \) is approximately \( 36^\circ 52' \). This angle describes the relative orientation of the two lines.
In simple words: To find the acute angle, get the slope for each line. Then, plug these slopes into the tangent formula, and use a table or calculator to find the angle itself.

🎯 Exam Tip: Always convert fractional equations of lines into the standard \( Ax + By + C = 0 \) form before finding their slopes. This reduces errors in calculation and makes the process straightforward.

Question 7. Reduce the following equations to the normal form and find the values of p and \( \alpha \).
(i) \( \sqrt{3}x − y + 2 = 0 \).
(ii) \( 3x + 4y + 10 = 0 \) (Use tables).
Answer:
(i) The given equation is \( \sqrt{3}x − y + 2 = 0 \).
To convert to normal form \( x \cos \alpha + y \sin \alpha = p \), we need the constant term on the right side to be positive.
\( \sqrt{3}x - y = -2 \)
Multiply by -1 to make the right side positive:
\( -\sqrt{3}x + y = 2 \).
Now, divide the entire equation by \( \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \). This step normalizes the coefficients.
\( -\frac{\sqrt{3}}{2}x + \frac{1}{2}y = 1 \).
Comparing this with \( x \cos \alpha + y \sin \alpha = p \):
\( \cos \alpha = -\frac{\sqrt{3}}{2} \)
\( \sin \alpha = \frac{1}{2} \)
\( p = 1 \).
Since \( \sin \alpha \) is positive and \( \cos \alpha \) is negative, \( \alpha \) lies in the second quadrant. In the second quadrant, \( \alpha = 180^\circ - \text{reference angle} \). The reference angle for \( \sin \alpha = \frac{1}{2} \) and \( \cos \alpha = \frac{\sqrt{3}}{2} \) is \( 30^\circ \) or \( \frac{\pi}{6} \).
So, \( \alpha = 180^\circ - 30^\circ = 150^\circ \) or \( \frac{5\pi}{6} \). The normal form gives the perpendicular distance from the origin and its angle.

(ii) The given equation is \( 3x + 4y + 10 = 0 \).
Rewrite it as \( 3x + 4y = -10 \).
Multiply by -1 to make the right side positive:
\( -3x - 4y = 10 \).
Divide by \( \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). This is the length of the normal vector.
\( -\frac{3}{5}x - \frac{4}{5}y = \frac{10}{5} \)
\( -\frac{3}{5}x - \frac{4}{5}y = 2 \).
Comparing this with \( x \cos \alpha + y \sin \alpha = p \):
\( \cos \alpha = -\frac{3}{5} \)
\( \sin \alpha = -\frac{4}{5} \)
\( p = 2 \).
Since both \( \sin \alpha \) and \( \cos \alpha \) are negative, \( \alpha \) lies in the third quadrant. In the third quadrant, \( \alpha = 180^\circ + \text{reference angle} \).
The reference angle \( \alpha' = \tan^{-1}\left(\left|\frac{\sin \alpha}{\cos \alpha}\right|\right) = \tan^{-1}\left(\left|\frac{-4/5}{-3/5}\right|\right) = \tan^{-1}\left(\frac{4}{3}\right) \).
Using tables, \( \tan^{-1}\left(\frac{4}{3}\right) \) is approximately \( 53^\circ 8' \).
So, \( \alpha = 180^\circ + 53^\circ 8' = 233^\circ 8' \). The angle is measured from the positive x-axis counter-clockwise.
In simple words: To change a line's equation into normal form, make sure the number on the right side is positive. Then, divide the whole equation by the square root of the sum of the squares of the x and y coefficients. This gives you \( \cos \alpha \), \( \sin \alpha \), and the perpendicular distance \( p \).

🎯 Exam Tip: When converting to normal form, ensure the constant 'p' is always positive. The quadrant of \( \alpha \) is determined by the signs of \( \cos \alpha \) and \( \sin \alpha \), which helps in finding the correct angle value.

Question 8. Put the equation \( 12y = 5x + 65 \) in the form \( x \cos \theta + y \sin \theta = p \) and indicate clearly in a rough diagram the position of the straight line and the meaning of the constant \( \theta \) and \( p \).
Answer:
The given equation of the line is \( 12y = 5x + 65 \).
First, rearrange it into the standard form \( Ax + By + C = 0 \):
\( 5x - 12y + 65 = 0 \).
To convert to normal form \( x \cos \theta + y \sin \theta = p \), we need the constant term on the right side to be positive.
\( 5x - 12y = -65 \).
Multiply by -1 to make the right side positive:
\( -5x + 12y = 65 \).
Now, divide the entire equation by \( \sqrt{(-5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \). This normalization step is crucial.
\( -\frac{5}{13}x + \frac{12}{13}y = \frac{65}{13} \)
\( -\frac{5}{13}x + \frac{12}{13}y = 5 \).
Comparing this with \( x \cos \theta + y \sin \theta = p \):
\( \cos \theta = -\frac{5}{13} \)
\( \sin \theta = \frac{12}{13} \)
\( p = 5 \).
Since \( \cos \theta \) is negative and \( \sin \theta \) is positive, \( \theta \) lies in the second quadrant. In the second quadrant, \( \theta = 180^\circ - \text{reference angle} \).
The reference angle is \( \alpha = \tan^{-1}\left(\left|\frac{12/13}{-5/13}\right|\right) = \tan^{-1}\left(\frac{12}{5}\right) \).
Using tables, \( \tan^{-1}\left(\frac{12}{5}\right) \) is approximately \( 67^\circ 23' \).
So, \( \theta = 180^\circ - 67^\circ 23' = 112^\circ 37' \).
The normal form of the equation is \( x \cos 112^\circ 37' + y \sin 112^\circ 37' = 5 \). Here, \( p \) represents the perpendicular distance from the origin to the line (which is 5 units), and \( \theta \) represents the angle that the normal (perpendicular) from the origin to the line makes with the positive x-axis (which is \( 112^\circ 37' \)).
In simple words: First, rewrite the line equation in a special form where the constant is positive. Then, divide by a certain number to get \( \cos \theta \) and \( \sin \theta \). The 'p' tells you how far the line is from the center (origin), and 'θ' tells you the angle of the line that goes from the center straight to your line.

🎯 Exam Tip: The constant \( p \) is always the perpendicular distance from the origin to the line and must be positive. The angle \( \theta \) is the angle the normal to the line makes with the positive x-axis, measured counter-clockwise.

Question 9. If \( Ax + By = C \) and \( x \cos \alpha + y \sin \alpha = p \) represent the same line, find \( p \) in terms of A, B, C.
Answer:
We are given two equations that represent the same straight line:
Equation (1): \( Ax + By = C \)
Equation (2): \( x \cos \alpha + y \sin \alpha = p \)
If two equations represent the same line, their corresponding coefficients must be proportional.
So, we can write:
\( \frac{A}{\cos \alpha} = \frac{B}{\sin \alpha} = \frac{C}{p} \). This is a key property of identical lines.
From these equalities, we can derive expressions for \( \cos \alpha \) and \( \sin \alpha \):
\( \cos \alpha = \frac{Ap}{C} \)
\( \sin \alpha = \frac{Bp}{C} \)
We know the fundamental trigonometric identity: \( \cos^2 \alpha + \sin^2 \alpha = 1 \). This identity is always true.
Substitute the expressions for \( \cos \alpha \) and \( \sin \alpha \) into the identity:
\( \left(\frac{Ap}{C}\right)^2 + \left(\frac{Bp}{C}\right)^2 = 1 \)
\( \frac{A^2 p^2}{C^2} + \frac{B^2 p^2}{C^2} = 1 \)
Multiply both sides by \( C^2 \):
\( A^2 p^2 + B^2 p^2 = C^2 \)
Factor out \( p^2 \):
\( p^2 (A^2 + B^2) = C^2 \)
Solve for \( p^2 \):
\( p^2 = \frac{C^2}{A^2 + B^2} \)
Take the square root of both sides. Since \( p \) represents a distance in the normal form, it must be positive.
\( p = \frac{|C|}{\sqrt{A^2 + B^2}} \). This gives the perpendicular distance from the origin to the line.
In simple words: When two equations describe the exact same line, their parts are proportional. By using a basic angle rule (\( \cos^2 \alpha + \sin^2 \alpha = 1 \)), we can find 'p', which is the distance from the origin to the line, using the 'A', 'B', and 'C' values from the first equation.

🎯 Exam Tip: When lines are identical, their coefficients are proportional. Always remember to take the absolute value of C and the positive square root for \( \sqrt{A^2 + B^2} \) since \( p \) represents a positive distance.

Question 10. Show that \( (2, – 1) \) and \( (1, 1) \) are on opposite sides of \( 3x + 4y = 6 \).
Answer:
The given equation of the line is \( 3x + 4y = 6 \).
To determine if points are on opposite sides, we bring all terms to one side to form a function \( f(x, y) = 3x + 4y - 6 \).
If the values of \( f(x, y) \) for two points have opposite signs, then the points lie on opposite sides of the line. This is a common method for point-line relationship.

First, let's test the point \( (2, -1) \):
Substitute \( x = 2 \) and \( y = -1 \) into \( f(x, y) \):
\( f(2, -1) = 3(2) + 4(-1) - 6 \)
\( = 6 - 4 - 6 \)
\( = -4 \).
Since \( f(2, -1) = -4 \), which is less than 0.

Next, let's test the point \( (1, 1) \):
Substitute \( x = 1 \) and \( y = 1 \) into \( f(x, y) \):
\( f(1, 1) = 3(1) + 4(1) - 6 \)
\( = 3 + 4 - 6 \)
\( = 7 - 6 \)
\( = 1 \).
Since \( f(1, 1) = 1 \), which is greater than 0.

Because the results \( -4 \) and \( 1 \) have opposite signs (one is negative and one is positive), the two points \( (2, -1) \) and \( (1, 1) \) lie on opposite sides of the line \( 3x + 4y = 6 \). This simple sign check efficiently determines their relative positions.
In simple words: To check if two points are on different sides of a line, put their x and y values into the line's equation (making sure one side is zero). If one point gives a positive answer and the other gives a negative answer, then they are on opposite sides.

🎯 Exam Tip: When checking if points are on opposite sides of a line, always rearrange the line equation so one side is zero (e.g., \( Ax + By - C = 0 \)). Then, substitute the coordinates of each point and observe the sign of the result; opposite signs confirm opposite sides.

Question 11. The sides of a straight triangles are given by the equations \( 3x + 4y = 10 \), \( 4x – 3y = 5 \), and \( 7x + y + 10 = 0 \); show that the origin lies within the triangle.
Answer:
Let the three given lines be:
Line (1): \( 3x + 4y - 10 = 0 \)
Line (2): \( 4x - 3y - 5 = 0 \)
Line (3): \( 7x + y + 10 = 0 \)
To show that the origin \( (0,0) \) lies within the triangle formed by these lines, we check the position of the origin relative to each line. If the origin is on the "interior" side of all three lines, it lies inside the triangle. This involves substituting the origin's coordinates into each equation and noting the sign of the result. It's important for the constant term to have a consistent sign for the "interior" side.

For Line (1): \( 3x + 4y - 10 = 0 \)
Substitute \( (0,0) \): \( 3(0) + 4(0) - 10 = -10 \). (Let's consider the side where the origin lies as negative).

For Line (2): \( 4x - 3y - 5 = 0 \)
Substitute \( (0,0) \): \( 4(0) - 3(0) - 5 = -5 \). (Origin is on the negative side).

For Line (3): \( 7x + y + 10 = 0 \)
Substitute \( (0,0) \): \( 7(0) + (0) + 10 = 10 \). (Origin is on the positive side).

Now, we need to ensure that the origin is on the same side relative to *all* lines for it to be inside. We can re-orient the equations by multiplying by -1 if needed, so that the origin gives a consistent sign (e.g., all negative for inside).
Re-written lines such that the side of the origin has a consistent sign (e.g. positive interior for the triangle):
Line (1): \( -(3x + 4y - 10) = 0 \implies -3x - 4y + 10 = 0 \). For \( (0,0) \), result is \( 10 > 0 \).
Line (2): \( -(4x - 3y - 5) = 0 \implies -4x + 3y + 5 = 0 \). For \( (0,0) \), result is \( 5 > 0 \).
Line (3): \( 7x + y + 10 = 0 \). For \( (0,0) \), result is \( 10 > 0 \).
Since substituting the origin \( (0,0) \) into these re-oriented equations results in a positive value for all three lines, it means the origin lies on the interior side of each line and, therefore, lies within the triangle formed by these lines. This consistent sign confirms the origin's position.
In simple words: To check if the origin is inside a triangle made by three lines, put (0,0) into each line's equation. If you can change the equations (by multiplying by -1) so that all results from (0,0) have the same sign (all positive or all negative), then the origin is inside the triangle.

🎯 Exam Tip: The key to showing a point is inside a polygon is to ensure it lies on the same "side" (positive or negative result after substitution) relative to all bounding lines. Adjust the sign of the line equations if needed to achieve consistency.

Question 12. Find by calculation whether the points \( (13, 8) \), \( (26, – 4) \) lie in the same, adjacent, or opposite angles formed by the straight lines \( 5x + 6y – 12 = 0 \), and \( 10x + 11 y – 217 = 0 \).
Answer:
Let the two given straight lines be:
Line (1): \( L_1(x, y) = 5x + 6y - 12 = 0 \)
Line (2): \( L_2(x, y) = 10x + 11y - 217 = 0 \)
To determine the relative position of points with respect to two intersecting lines, we substitute the coordinates of each point into both line equations. The signs of the results indicate whether the points are in the same, adjacent, or opposite angles. This method is effective for identifying regions.

First, consider the point \( (13, 8) \):
Substitute into Line (1):
\( L_1(13, 8) = 5(13) + 6(8) - 12 \)
\( = 65 + 48 - 12 \)
\( = 113 - 12 = 101 \). (Positive value)
Substitute into Line (2):
\( L_2(13, 8) = 10(13) + 11(8) - 217 \)
\( = 130 + 88 - 217 \)
\( = 218 - 217 = 1 \). (Positive value)
For point \( (13, 8) \), both \( L_1 \) and \( L_2 \) yield positive results.

Next, consider the point \( (26, -4) \):
Substitute into Line (1):
\( L_1(26, -4) = 5(26) + 6(-4) - 12 \)
\( = 130 - 24 - 12 \)
\( = 106 - 12 = 94 \). (Positive value)
Substitute into Line (2):
\( L_2(26, -4) = 10(26) + 11(-4) - 217 \)
\( = 260 - 44 - 217 \)
\( = 216 - 217 = -1 \). (Negative value)
For point \( (26, -4) \), \( L_1 \) yields a positive result and \( L_2 \) yields a negative result.

Summary of signs:
Point \( (13, 8) \): \( L_1 > 0 \), \( L_2 > 0 \)
Point \( (26, -4) \): \( L_1 > 0 \), \( L_2 < 0 \)

If two points lie in the same angle, the signs of their results for both lines should be identical. If they lie in opposite angles, the signs should be opposite for both lines. If the signs are different for only one line, or partially match, they are in adjacent angles.
Here, the sign for \( L_1 \) is positive for both points, but the sign for \( L_2 \) is positive for \( (13, 8) \) and negative for \( (26, -4) \). This means the points are separated by line \( L_2 \), but not by line \( L_1 \). Thus, they lie in **adjacent angles** formed by the two lines. The change in sign for only one line indicates adjacency.
In simple words: To find if points are in the same, opposite, or adjacent areas made by two crossing lines, plug each point into both line equations. If the signs of the answers are completely the same (e.g., both plus, both minus for both lines), they are in the same area. If all signs are opposite, they are in opposite areas. If only some signs are different, they are in adjacent areas.

🎯 Exam Tip: To classify the region of points relative to two lines, create functions \( L_1(x,y) \) and \( L_2(x,y) \). Substitute each point into both functions. If the signs \((+,+)\) or \((-,-)\) match for both points, they are in the same angle. If \((+,+)\) and \((-,-)\) (or vice-versa), they are in opposite angles. Any other combination means adjacent angles.