GSEB Class 11 Maths Solutions Chapter 10 Straight Lines Exercise 10.2

In questions 1 to 7, find the equations of the line, which satisfy the given conditions:

Question 1. Write the equations for x and y-axis.
Answer: The equation of the x-axis is \( y = 0 \). The equation of the y-axis is \( x = 0 \).
In simple words: The x-axis is where the y-coordinate is always zero. The y-axis is where the x-coordinate is always zero.

Exam Tip: Remember these fundamental equations as they are basic building blocks for coordinate geometry problems.

Question 2. passing through (- 4, 3) with slope \( \frac{1}{2} \).
Answer: We use the formula for a line passing through \( (x_1, y_1) \) with slope \( m \), which is \( (y - y_1) = m(x - x_1) \).
Here, \( x_1 = -4 \), \( y_1 = 3 \), and \( m = \frac{1}{2} \).
Substitute these values into the formula:
\( (y - 3) = \frac{1}{2} (x + 4) \)
Multiply both sides by 2 to remove the fraction:
\( 2(y - 3) = 1(x + 4) \)
\( 2y - 6 = x + 4 \)
Rearrange the terms to get the standard form of the equation:
\( x - 2y + 4 + 6 = 0 \)
\( x - 2y + 10 = 0 \)
So, the required line's equation is \( x - 2y + 10 = 0 \).
In simple words: We used the point-slope form of a line. We put in the given point and slope, then simplified it to get the line's equation.

Exam Tip: Always remember the point-slope form \( (y - y_1) = m(x - x_1) \). It's very useful for finding a line's equation when you have a point and its slope. Simplify the equation to its general form \( Ax + By + C = 0 \).

Question 3. Passing through (0, 0) with slope m.
Answer: For a line passing through \( (x_1, y_1) \) with slope \( m \), the equation is \( (y - y_1) = m(x - x_1) \).
Here, the point is the origin \( (0, 0) \), so \( x_1 = 0 \) and \( y_1 = 0 \). The slope is \( m \).
Substitute these values into the formula:
\( (y - 0) = m(x - 0) \)
Simplify the equation:
\( y = mx \)
This means the equation of the required line is \( y = mx \).
In simple words: If a line goes through the origin and has a certain slope, its equation is simply y equals the slope times x.

Exam Tip: The equation \( y = mx \) represents any line that passes through the origin. This form is common in physics and other areas, so recognize it quickly.

Question 4. Passing through (2, \( 2\sqrt{3} \)) and inclined with the x-axis at an angle of 75°.
Answer: Here, the given point is \( (x_1, y_1) = (2, 2\sqrt{3}) \).
The angle of inclination with the x-axis is \( \theta = 75^\circ \).
First, find the slope \( m \) using the formula \( m = \tan \theta \):
\( m = \tan 75^\circ \)
We can write \( \tan 75^\circ \) as \( \tan (45^\circ + 30^\circ) \).
Using the tangent addition formula, \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \):
\( m = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \)
Substitute the known values \( \tan 45^\circ = 1 \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\( m = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} \)
\( m = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} \)
\( m = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)
To rationalize the denominator, multiply the numerator and denominator by the conjugate \( (\sqrt{3} + 1) \):
\( m = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)
\( m = \frac{(\sqrt{3})^2 + 2\sqrt{3} + 1^2}{(\sqrt{3})^2 - 1^2} \)
\( m = \frac{3 + 2\sqrt{3} + 1}{3 - 1} \)
\( m = \frac{4 + 2\sqrt{3}}{2} \)
\( m = 2 + \sqrt{3} \)
Now, use the point-slope form \( (y - y_1) = m(x - x_1) \):
\( y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2) \)
Expand the right side:
\( y - 2\sqrt{3} = (2 + \sqrt{3})x - 2(2 + \sqrt{3}) \)
\( y - 2\sqrt{3} = (2 + \sqrt{3})x - 4 - 2\sqrt{3} \)
Rearrange terms to get the equation:
\( (2 + \sqrt{3})x - y - 4 - 2\sqrt{3} + 2\sqrt{3} = 0 \)
\( (2 + \sqrt{3})x - y - 4 = 0 \)
This is the equation of the required line.
In simple words: First, we found the slope by using the tangent of the given angle. We rationalized the slope to simplify it. Then, we put the slope and the given point into the point-slope formula to find the line's equation.

Exam Tip: When given an angle of inclination, remember to use \( m = \tan \theta \) to find the slope. For common angles like \( 75^\circ \) or \( 15^\circ \), use trigonometric identities like \( \tan(A+B) \) or \( \tan(A-B) \). Always simplify the slope as much as possible before using it in the line equation.

Question 5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope – 2.
Answer: The line intersects the x-axis at a distance of 3 units to the left of the origin. This means the x-intercept is \( -3 \). So, the point on the x-axis is \( (-3, 0) \).
The slope of the line is given as \( m = -2 \).
Now, use the point-slope form of the equation of a line: \( (y - y_1) = m(x - x_1) \).
Substitute \( x_1 = -3 \), \( y_1 = 0 \), and \( m = -2 \):
\( y - 0 = -2(x - (-3)) \)
\( y = -2(x + 3) \)
\( y = -2x - 6 \)
Rearrange the terms to get the standard form of the equation:
\( 2x + y + 6 = 0 \)
This is the equation of the required line.
In simple words: We used the given information to find a point on the line (the x-intercept). Then, with that point and the slope, we put the values into the line equation formula and simplified it.

Exam Tip: An x-intercept \( c \) means the line passes through the point \( (c, 0) \). A y-intercept \( d \) means the line passes through \( (0, d) \). Clearly identify the given point from the intercept information before applying the point-slope formula.

Question 6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis.
Answer: The line intersects the y-axis at a distance of 2 units above the origin. This means the y-intercept is \( 2 \). So, the point on the y-axis is \( (0, 2) \).
The angle made with the positive direction of the x-axis is \( \theta = 30^\circ \).
First, find the slope \( m \) using the formula \( m = \tan \theta \):
\( m = \tan 30^\circ = \frac{1}{\sqrt{3}} \)
Now, use the point-slope form of the equation of a line: \( (y - y_1) = m(x - x_1) \).
Substitute \( x_1 = 0 \), \( y_1 = 2 \), and \( m = \frac{1}{\sqrt{3}} \):
\( y - 2 = \frac{1}{\sqrt{3}}(x - 0) \)
\( y - 2 = \frac{x}{\sqrt{3}} \)
Multiply both sides by \( \sqrt{3} \):
\( \sqrt{3}(y - 2) = x \)
\( \sqrt{3}y - 2\sqrt{3} = x \)
Rearrange the terms to get the standard form of the equation:
\( x - \sqrt{3}y + 2\sqrt{3} = 0 \)
This is the equation of the required line.
In simple words: We found the point where the line crosses the y-axis. Then, we used the angle to find the slope. Finally, we used the point and slope to write the line's equation.

Exam Tip: Remember that a line intersecting the y-axis at a certain distance from the origin gives you the y-intercept, which is a point \( (0, y_{intercept}) \). The slope is derived from the angle of inclination using \( \tan \theta \).

Question 7. Passing through the points (- 1, 1) and (2, – 4).
Answer: For a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the equation is \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \).
Here, the points are \( A(-1, 1) \) and \( B(2, -4) \). So, \( x_1 = -1 \), \( y_1 = 1 \), \( x_2 = 2 \), and \( y_2 = -4 \).
First, calculate the slope \( m = \frac{y_2 - y_1}{x_2 - x_1} \):
\( m = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{2 + 1} = \frac{-5}{3} \)
Now, substitute the slope and point \( (x_1, y_1) \) into the point-slope formula:
\( y - 1 = \frac{-5}{3}(x - (-1)) \)
\( y - 1 = \frac{-5}{3}(x + 1) \)
Multiply both sides by 3:
\( 3(y - 1) = -5(x + 1) \)
\( 3y - 3 = -5x - 5 \)
Rearrange the terms to get the standard form of the equation:
\( 5x + 3y - 3 + 5 = 0 \)
\( 5x + 3y + 2 = 0 \)
This is the equation of the line AB.
In simple words: We first found the slope of the line using the coordinates of the two given points. Then, we used one of the points and the calculated slope in the point-slope form to write the line's equation and simplified it.

Exam Tip: When given two points, always calculate the slope first using \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Then, use either of the given points with the calculated slope in the point-slope form to find the line's equation. Be careful with signs, especially when subtracting negative numbers.

Question 8. Find the equation of a line whose perpendicular distance from the origin is 5 units and the angle made by the perpendicular with positive x-axis is 30°.
Answer: We need to find the equation of a line given its perpendicular distance from the origin and the angle the perpendicular makes with the x-axis. This is the normal form of a line's equation.
The normal form of the equation of a line is \( x \cos \omega + y \sin \omega = p \).
Here, \( p \) is the perpendicular distance from the origin, which is given as \( p = 5 \) units.
The angle \( \omega \) made by the perpendicular with the positive x-axis is given as \( \omega = 30^\circ \).
Substitute these values into the normal form equation:
\( x \cos 30^\circ + y \sin 30^\circ = 5 \)
Now, substitute the values of \( \cos 30^\circ \) and \( \sin 30^\circ \):
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 30^\circ = \frac{1}{2} \)
So, the equation becomes:
\( x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 5 \)
To eliminate the denominators, multiply the entire equation by 2:
\( \sqrt{3}x + y = 10 \)
Rearrange into the general form:
\( \sqrt{3}x + y - 10 = 0 \)
This is the equation of the required line.
In simple words: We used the normal form of a line's equation, which uses the distance from the origin to the line and the angle of the perpendicular. We put in the given distance and angle, then calculated the cosine and sine values, and finally simplified the equation.

Exam Tip: Recognize when to use the normal form \( x \cos \omega + y \sin \omega = p \). This form is used when the perpendicular distance from the origin to the line (\( p \)) and the angle (\( \omega \)) that this perpendicular makes with the positive x-axis are known. Make sure to correctly recall the values of sine and cosine for common angles.

Question 9. The vertices of a triangle PQR are P(2, 1), Q(- 2, 3) and R(4, 5). Find the equation of the median through the vertex R.
Answer: A median of a triangle connects a vertex to the midpoint of the opposite side. We need to find the equation of the median from vertex R to the midpoint of side PQ.
The vertices are \( P(2, 1) \), \( Q(-2, 3) \), and \( R(4, 5) \).
First, find the midpoint S of the side PQ. Use the midpoint formula: \( M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \).
For P(2, 1) and Q(-2, 3):
\( S = \left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) \)
\( S = \left(\frac{0}{2}, \frac{4}{2}\right) \)
\( S = (0, 2) \)
Now we have two points for the median RS: \( R(4, 5) \) and \( S(0, 2) \).
Next, find the equation of the line passing through R(4, 5) and S(0, 2) using the two-point form: \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \).
Let \( (x_1, y_1) = (4, 5) \) and \( (x_2, y_2) = (0, 2) \).
Calculate the slope \( m = \frac{2 - 5}{0 - 4} = \frac{-3}{-4} = \frac{3}{4} \).
Substitute the slope and point R(4, 5) into the formula:
\( y - 5 = \frac{3}{4}(x - 4) \)
Multiply both sides by 4:
\( 4(y - 5) = 3(x - 4) \)
\( 4y - 20 = 3x - 12 \)
Rearrange the terms to get the standard form of the equation:
\( 3x - 4y + 20 - 12 = 0 \)
\( 3x - 4y + 8 = 0 \)
This is the equation of the median RS.
In simple words: We first found the middle point of side PQ. Then, we used that middle point and vertex R to find the line's equation, as that line is the median we were looking for.

Exam Tip: To find the equation of a median, always start by finding the midpoint of the side opposite the given vertex using the midpoint formula. Then, use this midpoint and the vertex itself to find the equation of the line using the two-point form.

Question 10. Find the equation of the line passing through the point (- 3, 5) and perpendicular to the line through the points (2, 5) and (- 3, 6).
Answer: We need to find the equation of a line that passes through \( (-3, 5) \) and is perpendicular to another line passing through \( (2, 5) \) and \( (-3, 6) \).
First, find the slope of the line joining points \( A(2, 5) \) and \( B(-3, 6) \). Let this slope be \( m_2 \).
Using the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \):
\( m_2 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5} \)
Now, the required line is perpendicular to this line. If two lines are perpendicular, the product of their slopes is \( -1 \). Let the slope of the required line be \( m_1 \).
\( m_1 \cdot m_2 = -1 \)
\( m_1 \cdot \left(-\frac{1}{5}\right) = -1 \)
\( m_1 = 5 \)
So, the required line has a slope of 5 and passes through the point \( (-3, 5) \).
Use the point-slope form: \( (y - y_1) = m(x - x_1) \).
Substitute \( x_1 = -3 \), \( y_1 = 5 \), and \( m_1 = 5 \):
\( y - 5 = 5(x - (-3)) \)
\( y - 5 = 5(x + 3) \)
\( y - 5 = 5x + 15 \)
Rearrange the terms to get the standard form of the equation:
\( 5x - y + 15 + 5 = 0 \)
\( 5x - y + 20 = 0 \)
This is the equation of the required line.
In simple words: We first found the slope of the reference line. Since our line is perpendicular, we found its slope by taking the negative reciprocal. Then, we used this new slope and the given point to find the equation of our line.

Exam Tip: When dealing with perpendicular lines, remember that their slopes \( m_1 \) and \( m_2 \) satisfy \( m_1 m_2 = -1 \) (unless one line is vertical and the other is horizontal). For parallel lines, their slopes are equal (\( m_1 = m_2 \)). Carefully calculate the first slope and then apply the perpendicular condition to get the second slope.

Question 11. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.
Answer: We need to find the equation of a line (let's call it CD) that is perpendicular to the line segment AB (joining A(1, 0) and B(2, 3)) and divides AB in the ratio 1:n.
First, find the slope of the line segment AB. Let \( (x_1, y_1) = (1, 0) \) and \( (x_2, y_2) = (2, 3) \).
Slope of AB, \( m_{AB} = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3 \).
The required line CD is perpendicular to AB. So, its slope \( m_{CD} \) must satisfy \( m_{CD} \cdot m_{AB} = -1 \).
\( m_{CD} \cdot 3 = -1 \)
\( m_{CD} = -\frac{1}{3} \)
Next, find the coordinates of point P where the line CD divides AB in the ratio 1:n. Use the section formula: \( \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \).
Here, the ratio is \( 1:n \). So \( m=1 \) and \( n=n \). Points are \( A(1, 0) \) and \( B(2, 3) \).
\( P_x = \frac{1 \cdot 2 + n \cdot 1}{1 + n} = \frac{2 + n}{1 + n} \)
\( P_y = \frac{1 \cdot 3 + n \cdot 0}{1 + n} = \frac{3}{1 + n} \)
So, point \( P = \left(\frac{n+2}{n+1}, \frac{3}{n+1}\right) \).
Now, we have the slope of line CD (\( m_{CD} = -\frac{1}{3} \)) and a point it passes through (\( P \)). Use the point-slope form: \( y - y_1 = m(x - x_1) \).
\( y - \frac{3}{n+1} = -\frac{1}{3}\left(x - \frac{n+2}{n+1}\right) \)
Multiply both sides by \( 3(n+1) \) to clear denominators:
\( 3(n+1)\left(y - \frac{3}{n+1}\right) = (n+1)\left(-\frac{1}{3}\right) \cdot 3\left(x - \frac{n+2}{n+1}\right) \)
\( 3(n+1)y - 3 \cdot 3 = -(n+1)\left(x - \frac{n+2}{n+1}\right) \)
\( 3(n+1)y - 9 = -(n+1)x + (n+1) \cdot \frac{n+2}{n+1} \)
\( 3(n+1)y - 9 = -(n+1)x + (n+2) \)
Rearrange the terms to get the standard form:
\( (n+1)x + 3(n+1)y = n+2+9 \)
\( (n+1)x + 3(n+1)y = n+11 \)
This is the equation of the required line.
In simple words: First, we calculated the slope of the given line segment. Since our line is perpendicular, we found its slope. Then, we used the section formula to find the exact point where our line crosses the segment. Finally, with this point and the perpendicular slope, we wrote the line's equation and simplified it.

Exam Tip: This problem combines several concepts: slope of a line, perpendicular lines, and section formula. Break down the problem into smaller, manageable steps. First find the slope of the given segment, then the perpendicular slope, then the point of division using the section formula, and finally the equation of the line using the point-slope form.

Question 12. Find the equation of a line that cuts off equal intercepts on the co-ordinate axes and passes through (2, 3).
Answer: We need to find the equation of a line that makes equal intercepts on the coordinate axes and passes through the point \( (2, 3) \).
If a line cuts off equal intercepts on the coordinate axes, let the intercepts be \( a \). So, the x-intercept is \( a \) and the y-intercept is also \( a \).
The intercept form of the equation of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Since the intercepts are equal, \( b = a \). Substitute this into the formula:
\( \frac{x}{a} + \frac{y}{a} = 1 \)
Multiply the entire equation by \( a \):
\( x + y = a \)
The line passes through the point \( (2, 3) \). Substitute these coordinates into the equation to find \( a \):
\( 2 + 3 = a \)
\( a = 5 \)
Now, substitute the value of \( a \) back into the equation \( x + y = a \):
\( x + y = 5 \)
Rearrange into the standard form:
\( x + y - 5 = 0 \)
This is the equation of the required line.
Alternatively, if the intercepts are equal, it implies that the slope of the line is either \( -1 \) (for positive intercepts) or \( 1 \) (for negative intercepts).
Case 1: Slope \( m = -1 \).
Using the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (2, 3) \) and \( m = -1 \):
\( y - 3 = -1(x - 2) \)
\( y - 3 = -x + 2 \)
\( x + y - 3 - 2 = 0 \)
\( x + y - 5 = 0 \)
This matches our previous result. This is the correct line for positive intercepts.
Case 2: Slope \( m = 1 \).
Using the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (2, 3) \) and \( m = 1 \):
\( y - 3 = 1(x - 2) \)
\( y - 3 = x - 2 \)
\( x - y - 2 + 3 = 0 \)
\( x - y + 1 = 0 \)
This line has intercepts \( (-1, 0) \) and \( (0, 1) \), which are not equal in value, only in magnitude. For "equal intercepts", usually it implies equal values. If the question implies equal magnitude intercepts, then both lines are possible, but typically "equal intercepts" means \( a=b \). The first answer \( x+y-5=0 \) gives \( x \)-intercept \( 5 \) and \( y \)-intercept \( 5 \). This is correct.
In simple words: A line with equal x and y-intercepts can be written as \( x+y=a \). We used the given point to find the value of \( a \), and then put it back into the equation to get the final line.

Exam Tip: "Equal intercepts" means the x-intercept and y-intercept have the same value, not just the same magnitude. The general form for such a line is \( x+y=a \). Remember this shortcut, as it greatly simplifies finding the equation.

Question 13. Find the equations of the lines passing through the point (2, 2), such that the sum of their intercepts on the axes is 9.
Answer: We need to find the equation(s) of line(s) passing through \( (2, 2) \) where the sum of their x-intercept and y-intercept is 9.
Let the x-intercept be \( a \) and the y-intercept be \( b \).
According to the problem, \( a + b = 9 \).
From this, we can express \( b \) in terms of \( a \): \( b = 9 - a \).
The intercept form of the equation of a line is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Substitute \( b = 9 - a \) into the intercept form:
\( \frac{x}{a} + \frac{y}{9 - a} = 1 \)
Since the line passes through the point \( (2, 2) \), substitute \( x = 2 \) and \( y = 2 \) into the equation:
\( \frac{2}{a} + \frac{2}{9 - a} = 1 \)
To solve for \( a \), find a common denominator, which is \( a(9 - a) \):
\( \frac{2(9 - a) + 2a}{a(9 - a)} = 1 \)
\( 18 - 2a + 2a = a(9 - a) \)
\( 18 = 9a - a^2 \)
Rearrange this into a quadratic equation:
\( a^2 - 9a + 18 = 0 \)
Factor the quadratic equation:
\( (a - 3)(a - 6) = 0 \)
This gives two possible values for \( a \):
Case 1: \( a = 3 \)
If \( a = 3 \), then \( b = 9 - a = 9 - 3 = 6 \).
Substitute \( a = 3 \) and \( b = 6 \) into the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \):
\( \frac{x}{3} + \frac{y}{6} = 1 \)
Multiply by the least common multiple (6) to clear denominators:
\( 2x + y = 6 \)
Case 2: \( a = 6 \)
If \( a = 6 \), then \( b = 9 - a = 9 - 6 = 3 \).
Substitute \( a = 6 \) and \( b = 3 \) into the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \):
\( \frac{x}{6} + \frac{y}{3} = 1 \)
Multiply by the least common multiple (6) to clear denominators:
\( x + 2y = 6 \)
Therefore, there are two possible equations for the lines: \( 2x + y = 6 \) and \( x + 2y = 6 \).
In simple words: We used the intercept form of a line and the fact that the sum of intercepts is 9. We put the given point into the equation, which gave us a quadratic equation for one intercept. Solving it gave us two possible values for the intercepts, leading to two different line equations.

Exam Tip: Problems involving intercepts often lead to quadratic equations. Ensure you solve the quadratic correctly to find all possible values for the intercepts. Each pair of (a, b) values will give you one valid line equation.

Question 14. Find the equation of the line through the point (0, 2) making an angle \( \frac{2\pi}{3} \) with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Answer: We have two parts to this question.
**Part 1: Equation of the first line (AC)**
The line AC passes through \( (0, 2) \) and makes an angle of \( \theta = \frac{2\pi}{3} \) with the positive x-axis.
First, find the slope \( m_1 \) of this line:
\( m_1 = \tan \left(\frac{2\pi}{3}\right) = \tan (120^\circ) \)
Since \( 120^\circ \) is in the second quadrant, \( \tan 120^\circ = \tan (180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \).
So, \( m_1 = -\sqrt{3} \).
Using the point-slope form \( y - y_1 = m(x - x_1) \) with \( (x_1, y_1) = (0, 2) \) and \( m_1 = -\sqrt{3} \):
\( y - 2 = -\sqrt{3}(x - 0) \)
\( y - 2 = -\sqrt{3}x \)
Rearrange into the standard form:
\( \sqrt{3}x + y - 2 = 0 \)
This is the equation of the line AC.

**Part 2: Equation of the second line (BD)**
The second line BD is parallel to line AC. This means its slope \( m_2 \) is equal to \( m_1 \).
So, \( m_2 = -\sqrt{3} \).
This line crosses the y-axis at a distance of 2 units below the origin. This means its y-intercept is \( -2 \). So, it passes through the point \( (0, -2) \).
Using the point-slope form \( y - y_1 = m(x - x_1) \) with \( (x_1, y_1) = (0, -2) \) and \( m_2 = -\sqrt{3} \):
\( y - (-2) = -\sqrt{3}(x - 0) \)
\( y + 2 = -\sqrt{3}x \)
Rearrange into the standard form:
\( \sqrt{3}x + y + 2 = 0 \)
This is the equation of the line BD.
Thus, the equations of the required lines are \( \sqrt{3}x + y - 2 = 0 \) and \( \sqrt{3}x + y + 2 = 0 \).
In simple words: First, we found the slope of the first line using its angle and point. Then, for the second line, we knew it was parallel, so it had the same slope. We used its y-intercept as a point and its slope to find its equation.

Exam Tip: Pay close attention to radian measures for angles and convert them to degrees if needed for trigonometric values. Remember that parallel lines have equal slopes, and the y-intercept provides a direct point \( (0, y_{intercept}) \) for finding the equation.

Question 15. The perpendicular from the origin to a line meets at the point (- 2, 9). Find the equation of the line.
Answer: We need to find the equation of a line (let's call it AB) such that the perpendicular from the origin O to this line meets AB at point N(-2, 9).
First, find the slope of the line ON, which connects the origin \( (0, 0) \) to point \( N(-2, 9) \). Let this slope be \( m_{ON} \).
\( m_{ON} = \frac{9 - 0}{-2 - 0} = \frac{9}{-2} = -\frac{9}{2} \)
The required line AB is perpendicular to ON. Let the slope of line AB be \( m_{AB} \).
Since AB is perpendicular to ON, the product of their slopes must be \( -1 \): \( m_{AB} \cdot m_{ON} = -1 \).
\( m_{AB} \cdot \left(-\frac{9}{2}\right) = -1 \)
\( m_{AB} = \frac{2}{9} \)
Now, we have the slope of line AB (\( m_{AB} = \frac{2}{9} \)) and a point it passes through, which is \( N(-2, 9) \).
Use the point-slope form of the equation of a line: \( y - y_1 = m(x - x_1) \).
Substitute \( x_1 = -2 \), \( y_1 = 9 \), and \( m_{AB} = \frac{2}{9} \):
\( y - 9 = \frac{2}{9}(x - (-2)) \)
\( y - 9 = \frac{2}{9}(x + 2) \)
Multiply both sides by 9:
\( 9(y - 9) = 2(x + 2) \)
\( 9y - 81 = 2x + 4 \)
Rearrange the terms to get the standard form of the equation:
\( 2x - 9y + 4 + 81 = 0 \)
\( 2x - 9y + 85 = 0 \)
This is the equation of the required line.
In simple words: We found the slope of the line from the origin to the given point. Since the line we need to find is perpendicular to this one, we calculated its slope. Then, using that slope and the given point, we found the line's equation.

Exam Tip: When a perpendicular from the origin meets a line at a specific point, that point lies on both the perpendicular and the main line. Use the origin and the given point to find the slope of the perpendicular, then use the negative reciprocal for the slope of the required line. Finally, use the given point and the required slope in the point-slope formula.

Question 16. The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942, when C = 20 and L = 125.134, when C = 110, express L in terms of C.
Answer: The length L is a linear function of Celsius temperature C. This means it can be expressed in the form \( L = a + bC \), where \( a \) and \( b \) are constants.
We are given two sets of values:
1. When \( L = 124.942 \), \( C = 20 \).
Substitute these values into the equation:
\( 124.942 = a + 20b \) ... (1)
2. When \( L = 125.134 \), \( C = 110 \).
Substitute these values into the equation:
\( 125.134 = a + 110b \) ... (2)
Now, we have a system of two linear equations with two variables \( a \) and \( b \). Subtract equation (1) from equation (2) to eliminate \( a \):
\( (125.134 - 124.942) = (a + 110b) - (a + 20b) \)
\( 0.192 = 90b \)
Solve for \( b \):
\( b = \frac{0.192}{90} \)
\( b = 0.0021333... \)
Let's use \( b \approx 0.00213 \).
Now, substitute the value of \( b \) back into equation (1) to solve for \( a \):
\( 124.942 = a + 20(0.00213) \)
\( 124.942 = a + 0.0426 \)
\( a = 124.942 - 0.0426 \)
\( a = 124.8994 \)
Finally, express L in terms of C by substituting the values of \( a \) and \( b \) into the linear function equation \( L = a + bC \):
\( L = 124.8994 + 0.00213C \)
This is the required expression for L in terms of C.
In simple words: We know the length and temperature have a straight-line relationship. We used the two given examples to create two math problems. We solved these problems to find the starting length (a) and how much length changes per degree (b). Then, we wrote the final formula using these numbers.

Exam Tip: When a relationship is described as a "linear function," immediately think of the form \( y = mx + c \) or \( L = a + bC \). Use the given data points to set up a system of linear equations and solve for the unknown constants. Always present the final answer in the requested functional form.

Question 17. The owner of a milk store finds that he can sell 980 litres of milk each week at ₹ 14/litre and 1220 litres of milk each week at ₹ 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹ 17/litre.
Answer: We are given a linear relationship between selling price (per litre) and the quantity of milk sold. Let \( x \) be the selling price per litre (in Rs.) and \( y \) be the quantity of milk sold (in litres). The relationship is a straight line, so it can be expressed as \( y = a + bx \).
We have two data points:
1. When \( x_1 = \text{Rs. } 14/\text{litre} \), \( y_1 = 980 \text{ litres} \).
2. When \( x_2 = \text{Rs. } 16/\text{litre} \), \( y_2 = 1220 \text{ litres} \).
First, find the slope \( b \) of this linear relationship:
\( b = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1220 - 980}{16 - 14} \)
\( b = \frac{240}{2} = 120 \)
Now use the point-slope form to find the equation of the line. Using point \( (x_1, y_1) = (14, 980) \) and slope \( b = 120 \):
\( y - y_1 = b(x - x_1) \)
\( y - 980 = 120(x - 14) \)
The question asks how many litres could be sold weekly at \( x = \text{Rs. } 17/\text{litre} \). Substitute \( x = 17 \) into the equation:
\( y - 980 = 120(17 - 14) \)
\( y - 980 = 120(3) \)
\( y - 980 = 360 \)
\( y = 980 + 360 \)
\( y = 1340 \)
Therefore, 1340 litres of milk could be sold weekly at Rs. 17/litre.
In simple words: We treated the price and litres sold as points on a line. We found how steep the line was (its slope). Then, we used that slope and one of the given points to make a formula. Finally, we put the new price into our formula to find out how many litres would be sold.

Exam Tip: When given two data points that follow a linear relationship, you can find the equation of the line (demand/supply curve in economics context). Calculate the slope using the two points, then use the point-slope form to get the equation. Finally, use the equation to predict values for new inputs.

Question 18. P(a, b) is the mid-point of a line segment between axes. Show that the equation of the corresponding line is \( \frac{x}{a} + \frac{y}{b} = 2 \).
Answer: Let the line segment be AB, where A is the x-intercept and B is the y-intercept. This means the line makes intercepts on the x-axis and y-axis.
Let the x-intercept be \( p \) and the y-intercept be \( q \).
So, point A is \( (p, 0) \) and point B is \( (0, q) \).
The midpoint of this line segment AB is given as \( P(a, b) \).
Using the midpoint formula: \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \).
\( a = \frac{p + 0}{2} \implies p = 2a \)
\( b = \frac{0 + q}{2} \implies q = 2b \)
Now, the equation of the line in intercept form is \( \frac{x}{p} + \frac{y}{q} = 1 \).
Substitute the values of \( p = 2a \) and \( q = 2b \) into the intercept form equation:
\( \frac{x}{2a} + \frac{y}{2b} = 1 \)
To remove the denominators, multiply the entire equation by 2:
\( 2 \left(\frac{x}{2a} + \frac{y}{2b}\right) = 2 \cdot 1 \)
\( \frac{2x}{2a} + \frac{2y}{2b} = 2 \)
\( \frac{x}{a} + \frac{y}{b} = 2 \)
This is the required equation of the line.
In simple words: We first set up the x and y-intercepts as points. Then, we used the midpoint formula to relate these intercepts to the given midpoint coordinates (a,b). Finally, we put the intercepts into the standard intercept form of a line and simplified it to get the desired equation.

Exam Tip: When a point is described as the midpoint of a line segment between the axes, it means the x-intercept is \( (2 \times \text{midpoint x-coord}) \) and the y-intercept is \( (2 \times \text{midpoint y-coord}) \). Use these values directly in the intercept form of the line equation.

Question 19. Point (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the corresponding line.
Answer: Let the line AB cut the x-axis at A and the y-axis at B. Let the x-intercept be \( a_0 \) and the y-intercept be \( b_0 \). So, the coordinates of A are \( (a_0, 0) \) and B are \( (0, b_0) \).
The point R(h, k) divides the line segment AB in the ratio 1:2.
Using the section formula: \( \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \).
Here, the ratio is \( m:n = 1:2 \). Points are \( A(a_0, 0) \) and \( B(0, b_0) \).
For the x-coordinate \( h \):
\( h = \frac{1 \cdot 0 + 2 \cdot a_0}{1 + 2} = \frac{2a_0}{3} \)
From this, we get \( a_0 = \frac{3h}{2} \).
For the y-coordinate \( k \):
\( k = \frac{1 \cdot b_0 + 2 \cdot 0}{1 + 2} = \frac{b_0}{3} \)
From this, we get \( b_0 = 3k \).
Now, the equation of the line in intercept form is \( \frac{x}{a_0} + \frac{y}{b_0} = 1 \).
Substitute the values of \( a_0 = \frac{3h}{2} \) and \( b_0 = 3k \) into the intercept form:
\( \frac{x}{\frac{3h}{2}} + \frac{y}{3k} = 1 \)
Simplify the fractions:
\( \frac{2x}{3h} + \frac{y}{3k} = 1 \)
To clear the denominators, multiply the entire equation by \( 3hk \):
\( 3hk \left(\frac{2x}{3h}\right) + 3hk \left(\frac{y}{3k}\right) = 3hk \cdot 1 \)
\( 2kx + hy = 3hk \)
This is the equation of the corresponding line.
In simple words: We used the section formula to find the x-intercept and y-intercept in terms of \( h \) and \( k \). Then, we put these new intercepts into the standard intercept form of a line and simplified it to get the final equation.

Exam Tip: This problem involves the section formula and the intercept form of a line. Be careful with algebraic manipulation when substituting the expressions for intercepts back into the line equation. Always aim to clear denominators for a cleaner final equation.

Question 20. By using the concept of equation of a line, prove that the three points (3, 0), (- 2, – 2) and (8, 2) are collinear.
Answer: To prove that three points are collinear, we can find the equation of the line passing through any two of the points and then check if the third point satisfies this equation.
Let the three points be \( A(3, 0) \), \( B(-2, -2) \), and \( C(8, 2) \).
We will find the equation of the line passing through points A(3, 0) and B(-2, -2).
Using the two-point form of the equation of a line: \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \).
Let \( (x_1, y_1) = (3, 0) \) and \( (x_2, y_2) = (-2, -2) \).
First, calculate the slope \( m_{AB} \):
\( m_{AB} = \frac{-2 - 0}{-2 - 3} = \frac{-2}{-5} = \frac{2}{5} \)
Now, substitute the slope and point A(3, 0) into the point-slope formula:
\( y - 0 = \frac{2}{5}(x - 3) \)
\( y = \frac{2}{5}(x - 3) \)
Multiply both sides by 5:
\( 5y = 2(x - 3) \)
\( 5y = 2x - 6 \)
Rearrange into the standard form:
\( 2x - 5y - 6 = 0 \)
Now, check if the third point C(8, 2) lies on this line. Substitute \( x = 8 \) and \( y = 2 \) into the equation \( 2x - 5y - 6 = 0 \):
\( 2(8) - 5(2) - 6 \)
\( 16 - 10 - 6 \)
\( 6 - 6 = 0 \)
Since substituting the coordinates of point C into the equation results in 0, the point C(8, 2) satisfies the equation of the line.
Therefore, the three points (3, 0), (-2, -2), and (8, 2) are collinear.
In simple words: To show the points are in a straight line, we found the equation for the line connecting two of them. Then, we checked if the third point perfectly fits into that same line's equation. If it did, all three points are in a straight line.

Exam Tip: A common method to prove collinearity of three points is to find the equation of the line passing through any two points and then verify if the third point satisfies this equation. Alternatively, you can calculate the slope between the first two points and the slope between the second and third points; if the slopes are equal, the points are collinear.