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

Question 1. Find the distance of the point P from the line AB in the following cases :
(i) P(4, 2), AB is \( 5x - 12y - 9 = 0 \),
(ii) P(0, 0), AB is \( h(x + h) + k (y + k) = 0 \).
Answer:
(i) We need to find the distance of point P(4, 2) from the line AB, which is \( 5x - 12y - 9 = 0 \). We use the formula for the perpendicular distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \), which is \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
Here, \( (x_1, y_1) = (4, 2) \), \( A = 5 \), \( B = -12 \), \( C = -9 \).
So, the distance \( d = \frac{|5(4) + (-12)(2) - 9|}{\sqrt{5^2 + (-12)^2}} \)
\( \implies d = \frac{|20 - 24 - 9|}{\sqrt{25 + 144}} \)
\( \implies d = \frac{|-13|}{\sqrt{169}} \)
\( \implies d = \frac{13}{13} \)
\( \implies d = 1 \) unit.

(ii) We need the distance of point P(0, 0) from the line AB, which is \( h(x + h) + k(y + k) = 0 \). First, let's simplify the equation of line AB:
\( hx + h^2 + ky + k^2 = 0 \)
\( \implies hx + ky + (h^2 + k^2) = 0 \)
Here, \( (x_1, y_1) = (0, 0) \), \( A = h \), \( B = k \), \( C = (h^2 + k^2) \).
Using the distance formula:
\( d = \frac{|h(0) + k(0) + (h^2 + k^2)|}{\sqrt{h^2 + k^2}} \)
\( \implies d = \frac{|h^2 + k^2|}{\sqrt{h^2 + k^2}} \)
Since \( h^2 + k^2 \) is always non-negative, \( |h^2 + k^2| = h^2 + k^2 \).
\( \implies d = \frac{h^2 + k^2}{\sqrt{h^2 + k^2}} \)
We can write \( h^2 + k^2 \) as \( (\sqrt{h^2 + k^2})^2 \).
\( \implies d = \frac{(\sqrt{h^2 + k^2})^2}{\sqrt{h^2 + k^2}} \)
\( \implies d = \sqrt{h^2 + k^2} \) units.
In simple words: To find how far a point is from a line, we use a special formula. For the first part, we plug in the numbers and get a distance of 1. For the second part, we simplify the line's equation and then use the same formula. The distance turns out to be the square root of \( h^2 \) plus \( k^2 \).

🎯 Exam Tip: Remember the formula for the perpendicular distance from a point to a line. Pay close attention to signs when substituting values and simplifying the numerator and denominator.

Question 2. Calculate the length of the perpendicular from (7, 0) to the straight line \( 5x + 12y - 9 = 0 \) and show that it is twice the length of the perpendicular from (2, 1).
Answer:
First, let's find the length of the perpendicular from point P(7, 0) to the line \( 5x + 12y - 9 = 0 \).
Using the distance formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \):
Here, \( (x_1, y_1) = (7, 0) \), \( A = 5 \), \( B = 12 \), \( C = -9 \).
Distance from P(7, 0) \( = \frac{|5(7) + 12(0) - 9|}{\sqrt{5^2 + 12^2}} \)
\( \implies = \frac{|35 + 0 - 9|}{\sqrt{25 + 144}} \)
\( \implies = \frac{|26|}{\sqrt{169}} \)
\( \implies = \frac{26}{13} \)
\( \implies = 2 \) units.

Next, let's find the length of the perpendicular from point Q(2, 1) to the same line \( 5x + 12y - 9 = 0 \).
Here, \( (x_1, y_1) = (2, 1) \), \( A = 5 \), \( B = 12 \), \( C = -9 \).
Distance from Q(2, 1) \( = \frac{|5(2) + 12(1) - 9|}{\sqrt{5^2 + 12^2}} \)
\( \implies = \frac{|10 + 12 - 9|}{\sqrt{25 + 144}} \)
\( \implies = \frac{|13|}{\sqrt{169}} \)
\( \implies = \frac{13}{13} \)
\( \implies = 1 \) unit.

We found that the distance from P(7, 0) is 2 units and the distance from Q(2, 1) is 1 unit. This clearly shows that the length of the perpendicular from P(7, 0) is twice the length of the perpendicular from Q(2, 1) to the same line.
In simple words: We calculated how far two different points are from the same straight line. The first point was 2 units away. The second point was 1 unit away. This proves that the first distance is double the second distance.

🎯 Exam Tip: When asked to show a relationship between two distances, calculate both distances separately and then compare their values at the end.

Question 3. The point A(0, 0), B(1, 7), C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from A to BC and hence the area of the \( \triangle ABC \).
Answer:
First, we need the equation of the line BC. We use the two-point form \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \) with B(1, 7) and C(5, 1).
\( y - 7 = \frac{1 - 7}{5 - 1}(x - 1) \)
\( \implies y - 7 = \frac{-6}{4}(x - 1) \)
\( \implies y - 7 = -\frac{3}{2}(x - 1) \)
Now, we cross-multiply to get rid of the fraction:
\( 2(y - 7) = -3(x - 1) \)
\( \implies 2y - 14 = -3x + 3 \)
Rearranging into the standard form \( Ax + By + C = 0 \):
\( 3x + 2y - 17 = 0 \). This is the equation of line BC.

Next, we find the length of the perpendicular from point A(0, 0) to line BC (\( 3x + 2y - 17 = 0 \)). Let this length be \( d \).
Using the distance formula:
\( d = \frac{|3(0) + 2(0) - 17|}{\sqrt{3^2 + 2^2}} \)
\( \implies d = \frac{|-17|}{\sqrt{9 + 4}} \)
\( \implies d = \frac{17}{\sqrt{13}} \) units. This is the altitude from A to BC.

Now, we need the length of the base BC. We use the distance formula between two points \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
BC \( = \sqrt{(5 - 1)^2 + (1 - 7)^2} \)
\( \implies BC = \sqrt{(4)^2 + (-6)^2} \)
\( \implies BC = \sqrt{16 + 36} \)
\( \implies BC = \sqrt{52} \)
\( \implies BC = \sqrt{4 \times 13} \)
\( \implies BC = 2\sqrt{13} \) units.

Finally, we calculate the area of \( \triangle ABC \). The formula is \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \).
Area \( = \frac{1}{2} \times BC \times d \)
\( \implies \text{Area} = \frac{1}{2} \times 2\sqrt{13} \times \frac{17}{\sqrt{13}} \)
The \( \sqrt{13} \) terms cancel out.
\( \implies \text{Area} = \frac{1}{2} \times 2 \times 17 \)
\( \implies \text{Area} = 17 \) square units.
In simple words: First, we find the straight line equation that connects points B and C. Then, we find the distance from point A to that line, which is the triangle's height. After that, we calculate the length of the line BC, which is the base. Lastly, we use the base and height to find the total area of the triangle.

🎯 Exam Tip: When finding the area of a triangle given vertices, first determine the equation of one side, then the perpendicular distance from the opposite vertex to that side, and finally calculate the length of that side to use in the area formula.

Question 4. Find the lengths of altitudes of the triangle whose sides are given by \( 3x - 4y = 5 \), \( 4x + 3y = 5 \) and \( x + y = 1 \).
Answer:
Let the three given equations of the sides be:
(1) \( 3x - 4y = 5 \)
(2) \( 4x + 3y = 5 \)
(3) \( x + y = 1 \)

To find the vertices of the triangle, we solve these equations in pairs.

1. **Intersection of (1) and (2):**
Multiply (1) by 3: \( 9x - 12y = 15 \)
Multiply (2) by 4: \( 16x + 12y = 20 \)
Add the two equations: \( (9x - 12y) + (16x + 12y) = 15 + 20 \)
\( \implies 25x = 35 \)
\( \implies x = \frac{35}{25} = \frac{7}{5} \)
Substitute \( x = \frac{7}{5} \) into (3) for simplicity: \( \frac{7}{5} + y = 1 \)
\( \implies y = 1 - \frac{7}{5} \)
\( \implies y = \frac{5 - 7}{5} = -\frac{2}{5} \)
Let this vertex be A: \( \left(\frac{7}{5}, -\frac{2}{5}\right) \)

2. **Intersection of (2) and (3):**
From (3), \( y = 1 - x \). Substitute into (2):
\( 4x + 3(1 - x) = 5 \)
\( \implies 4x + 3 - 3x = 5 \)
\( \implies x + 3 = 5 \)
\( \implies x = 2 \)
Substitute \( x = 2 \) into \( y = 1 - x \): \( y = 1 - 2 = -1 \)
Let this vertex be B: \( (2, -1) \)

3. **Intersection of (1) and (3):**
From (3), \( y = 1 - x \). Substitute into (1):
\( 3x - 4(1 - x) = 5 \)
\( \implies 3x - 4 + 4x = 5 \)
\( \implies 7x - 4 = 5 \)
\( \implies 7x = 9 \)
\( \implies x = \frac{9}{7} \)
Substitute \( x = \frac{9}{7} \) into \( y = 1 - x \): \( y = 1 - \frac{9}{7} \)
\( \implies y = \frac{7 - 9}{7} = -\frac{2}{7} \)
Let this vertex be C: \( \left(\frac{9}{7}, -\frac{2}{7}\right) \)

Now we find the length of each altitude.

1. **Altitude from A \( \left(\frac{7}{5}, -\frac{2}{5}\right) \) to side BC (line \( x + y = 1 \), or \( x + y - 1 = 0 \)):**
Using the distance formula:
\( d_A = \frac{|1(\frac{7}{5}) + 1(-\frac{2}{5}) - 1|}{\sqrt{1^2 + 1^2}} \)
\( \implies d_A = \frac{|\frac{7}{5} - \frac{2}{5} - \frac{5}{5}|}{\sqrt{1 + 1}} \)
\( \implies d_A = \frac{|\frac{7 - 2 - 5}{5}|}{\sqrt{2}} \)
\( \implies d_A = \frac{|\frac{0}{5}|}{\sqrt{2}} \)
\( \implies d_A = 0 \) units. (This indicates that point A actually lies on the line \( x+y=1 \), which contradicts it being a vertex of a triangle formed by these lines if it's supposed to be an altitude. Let's recheck the calculation of vertices for Q4 from the source. The source has `x = \frac{7}{5}; y =\frac{-1}{5}` for intersection of (1) and (2) and `x = \frac{9}{7} and y = \frac{-2}{7}` for intersection of (1) and (3). Let's follow the source's calculated vertices to avoid discrepancies, as per Iron Rule 6. Source Vertices: A (from (1) and (2)): \( \left(\frac{7}{5}, -\frac{1}{5}\right) \) B (from (2) and (3)): \( (2, -1) \) C (from (1) and (3)): \( \left(\frac{9}{7}, -\frac{2}{7}\right) \) Let's recalculate the altitudes with the source's vertex coordinates. Side equations: (1) \( 3x - 4y - 5 = 0 \) (2) \( 4x + 3y - 5 = 0 \) (3) \( x + y - 1 = 0 \) Vertices: A \( \left(\frac{7}{5}, -\frac{1}{5}\right) \) (from lines (1) and (2)) B \( (2, -1) \) (from lines (2) and (3)) C \( \left(\frac{9}{7}, -\frac{2}{7}\right) \) (from lines (1) and (3))

Now calculate altitudes:

1. **Altitude from A \( \left(\frac{7}{5}, -\frac{1}{5}\right) \) to side BC (line \( x + y - 1 = 0 \)):**
\( d_A = \frac{|1(\frac{7}{5}) + 1(-\frac{1}{5}) - 1|}{\sqrt{1^2 + 1^2}} \)
\( \implies d_A = \frac{|\frac{7}{5} - \frac{1}{5} - \frac{5}{5}|}{\sqrt{2}} \)
\( \implies d_A = \frac{|\frac{7 - 1 - 5}{5}|}{\sqrt{2}} \)
\( \implies d_A = \frac{|\frac{1}{5}|}{\sqrt{2}} \)
\( \implies d_A = \frac{1}{5\sqrt{2}} \) units.

2. **Altitude from B \( (2, -1) \) to side AC (line \( 3x - 4y - 5 = 0 \)):**
\( d_B = \frac{|3(2) - 4(-1) - 5|}{\sqrt{3^2 + (-4)^2}} \)
\( \implies d_B = \frac{|6 + 4 - 5|}{\sqrt{9 + 16}} \)
\( \implies d_B = \frac{|5|}{\sqrt{25}} \)
\( \implies d_B = \frac{5}{5} \)
\( \implies d_B = 1 \) unit.

3. **Altitude from C \( \left(\frac{9}{7}, -\frac{2}{7}\right) \) to side AB (line \( 4x + 3y - 5 = 0 \)):**
\( d_C = \frac{|4(\frac{9}{7}) + 3(-\frac{2}{7}) - 5|}{\sqrt{4^2 + 3^2}} \)
\( \implies d_C = \frac{|\frac{36}{7} - \frac{6}{7} - \frac{35}{7}|}{\sqrt{16 + 9}} \)
\( \implies d_C = \frac{|\frac{36 - 6 - 35}{7}|}{\sqrt{25}} \)
\( \implies d_C = \frac{|-\frac{5}{7}|}{5} \)
\( \implies d_C = \frac{\frac{5}{7}}{5} \)
\( \implies d_C = \frac{5}{7 \times 5} \)
\( \implies d_C = \frac{1}{7} \) unit.
In simple words: First, we find the corner points (vertices) of the triangle by solving the equations for the sides two at a time. Then, for each corner point, we calculate the shortest distance from that point to the line forming the opposite side. These distances are the lengths of the altitudes. A triangle has three altitudes, and we found all three lengths.

🎯 Exam Tip: When finding altitudes of a triangle given its sides, always remember to first find the coordinates of all three vertices by solving the side equations in pairs. Then, use the distance formula from each vertex to its corresponding opposite side.

Question 5. If P is the perpendicular distance of the origin from the line whose intercepts on the axes are a and b, show that \( \frac{1}{p^2} = \frac{1}{a^2} = \frac{1}{b^2} \)
Answer:
The equation of a line with intercepts \( a \) and \( b \) on the axes is given by \( \frac{x}{a} + \frac{y}{b} = 1 \).
We can rewrite this equation in the standard form \( Ax + By + C = 0 \):
Multiply by \( ab \) to clear fractions: \( bx + ay = ab \)
\( \implies bx + ay - ab = 0 \).

Now, we need to find the perpendicular distance, \( p \), from the origin (0, 0) to this line.
Using the distance formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \):
Here, \( (x_1, y_1) = (0, 0) \), \( A = b \), \( B = a \), \( C = -ab \).
So, \( p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} \)
\( \implies p = \frac{|-ab|}{\sqrt{a^2 + b^2}} \)
Since \( |-ab| = |ab| \), and assuming \( a, b \) are real numbers, \( |ab| \) is just \( ab \) if \( a \) and \( b \) have the same sign, or \( -ab \) if they have different signs, but for distance, it's always positive. So, we'll use \( ab \).
\( \implies p = \frac{ab}{\sqrt{a^2 + b^2}} \)

To get to the desired form, we square both sides:
\( p^2 = \left(\frac{ab}{\sqrt{a^2 + b^2}}\right)^2 \)
\( \implies p^2 = \frac{a^2 b^2}{a^2 + b^2} \)
Now, we take the reciprocal of both sides:
\( \frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} \)
We can split the fraction on the right side:
\( \frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} \)
\( \implies \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2} \)
Thus, we have shown that \( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \). This relationship is a key result in coordinate geometry for lines intersecting the axes.
In simple words: We start with the line equation where it cuts the x and y axes at points 'a' and 'b'. Then, we find the straight-line distance from the center point (origin) to this line using a formula. After that, we square this distance and flip it upside down. This shows a special connection: one divided by the square of the distance is equal to one divided by the square of 'a' plus one divided by the square of 'b'.

🎯 Exam Tip: Remember the intercept form of a line \( \frac{x}{a} + \frac{y}{b} = 1 \) and how to convert it to general form. The distance formula from a point to a line is crucial here, and algebraic manipulation (squaring and taking reciprocals) is key to reaching the final proof.

Question 6. Find the perpendicular distance between the lines
(i) \( 3x + 4y + 5 = 0 \), \( 3x + 4y + 17 = 0 \).
(ii) \( 9x + 40y - 20 = 0 \), \( 9x + 40y + 21 = 0 \).
(iii) \( y = mx + c \), \( y = mx + d \).
Answer:
To find the distance between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \), the formula is \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \). This formula works directly when the coefficients of \( x \) and \( y \) are the same in both equations.

(i) Given lines are: \( 3x + 4y + 5 = 0 \) and \( 3x + 4y + 17 = 0 \).
Here, \( A = 3 \), \( B = 4 \), \( C_1 = 5 \), \( C_2 = 17 \).
Distance \( d = \frac{|5 - 17|}{\sqrt{3^2 + 4^2}} \)
\( \implies d = \frac{|-12|}{\sqrt{9 + 16}} \)
\( \implies d = \frac{12}{\sqrt{25}} \)
\( \implies d = \frac{12}{5} \) units.

(ii) Given lines are: \( 9x + 40y - 20 = 0 \) and \( 9x + 40y + 21 = 0 \).
Here, \( A = 9 \), \( B = 40 \), \( C_1 = -20 \), \( C_2 = 21 \).
Distance \( d = \frac{|-20 - 21|}{\sqrt{9^2 + 40^2}} \)
\( \implies d = \frac{|-41|}{\sqrt{81 + 1600}} \)
\( \implies d = \frac{41}{\sqrt{1681}} \)
\( \implies d = \frac{41}{41} \)
\( \implies d = 1 \) unit.

(iii) Given lines are: \( y = mx + c \) and \( y = mx + d \).
First, rewrite them in the standard form \( Ax + By + C = 0 \):
Line 1: \( mx - y + c = 0 \)
Line 2: \( mx - y + d = 0 \)
Here, \( A = m \), \( B = -1 \), \( C_1 = c \), \( C_2 = d \).
Distance \( d = \frac{|c - d|}{\sqrt{m^2 + (-1)^2}} \)
\( \implies d = \frac{|c - d|}{\sqrt{m^2 + 1}} \) units. This general formula helps find the distance between any two parallel lines if their equations are known.
In simple words: When two lines are parallel, they never meet, and the distance between them is always the same. We use a special formula that takes the difference between the constant parts of their equations and divides it by the square root of the sum of squares of the x and y coefficients. We apply this for all three given pairs of lines.

🎯 Exam Tip: Always check if the coefficients of x and y are identical for both parallel lines. If not, divide or multiply one equation by a constant to make them identical before applying the distance formula \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \).

Question 7. Find the equations of two straight lines which are parallel to the straight line \( x + 7y + 2 = 0 \), and a unit distance from the point (2, – 1).
Answer:
The given straight line is \( x + 7y + 2 = 0 \) (Equation 1).
Any line parallel to Equation 1 will have the form \( x + 7y + k = 0 \) (Equation 2), where \( k \) is a constant.

We are given that the perpendicular distance from the point (2, –1) to this parallel line (Equation 2) is 1 unit. We use the distance formula:
\( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)
Here, \( (x_1, y_1) = (2, -1) \), \( A = 1 \), \( B = 7 \), \( C = k \), and \( d = 1 \).
So, \( 1 = \frac{|1(2) + 7(-1) + k|}{\sqrt{1^2 + 7^2}} \)
\( \implies 1 = \frac{|2 - 7 + k|}{\sqrt{1 + 49}} \)
\( \implies 1 = \frac{|k - 5|}{\sqrt{50}} \)
Multiply both sides by \( \sqrt{50} \):
\( |k - 5| = \sqrt{50} \)
We know that \( \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \).
So, \( |k - 5| = 5\sqrt{2} \)
This means \( k - 5 = 5\sqrt{2} \) or \( k - 5 = -5\sqrt{2} \).
Therefore, \( k = 5 + 5\sqrt{2} \) or \( k = 5 - 5\sqrt{2} \).
These are two possible values for \( k \). We need to substitute them back into Equation 2 to find the two equations of the lines.

1. For \( k = 5 + 5\sqrt{2} \):
The equation is \( x + 7y + (5 + 5\sqrt{2}) = 0 \).
2. For \( k = 5 - 5\sqrt{2} \):
The equation is \( x + 7y + (5 - 5\sqrt{2}) = 0 \).
These are the two required equations of the straight lines. They represent lines equally distant from the given point.
In simple words: We want to find two lines that run next to a given line and are a fixed distance away from a specific point. We first write the general form of lines parallel to the given line, including an unknown number \( k \). Then, we use the distance formula to say that the distance from the specific point to our new line must be 1. Solving this gives us two possible values for \( k \), which then gives us the two equations for the parallel lines.

🎯 Exam Tip: When finding equations of lines parallel to a given line, use the general form \( Ax + By + k = 0 \). Remember that the absolute value in the distance formula leads to two possible solutions for \( k \) (positive and negative cases), giving two possible lines.

Question 8. Find the equations of the two straight lines drawn through the point (0, 1) on which the perpendiculars dropped from the point (2, 2) are each of unit length.
Answer:
Let the equation of the straight line passing through the point (0, 1) with slope \( m \) be given by the point-slope form \( y - y_1 = m(x - x_1) \).
\( y - 1 = m(x - 0) \)
\( \implies y - 1 = mx \)
Rearranging into the standard form \( Ax + By + C = 0 \):
\( mx - y + 1 = 0 \) (Equation 1).

We are given that the length of the perpendicular from the point (2, 2) to this line (Equation 1) is 1 unit. We use the distance formula:
\( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)
Here, \( (x_1, y_1) = (2, 2) \), \( A = m \), \( B = -1 \), \( C = 1 \), and \( d = 1 \).
So, \( 1 = \frac{|m(2) - 1(2) + 1|}{\sqrt{m^2 + (-1)^2}} \)
\( \implies 1 = \frac{|2m - 2 + 1|}{\sqrt{m^2 + 1}} \)
\( \implies 1 = \frac{|2m - 1|}{\sqrt{m^2 + 1}} \)

To solve for \( m \), we square both sides:
\( 1^2 = \frac{(2m - 1)^2}{(\sqrt{m^2 + 1})^2} \)
\( \implies 1 = \frac{4m^2 - 4m + 1}{m^2 + 1} \)
Multiply both sides by \( (m^2 + 1) \):
\( m^2 + 1 = 4m^2 - 4m + 1 \)
Now, bring all terms to one side:
\( 0 = 4m^2 - m^2 - 4m + 1 - 1 \)
\( \implies 3m^2 - 4m = 0 \)
Factor out \( m \):
\( m(3m - 4) = 0 \)
This gives two possible values for \( m \):
\( m = 0 \) or \( 3m - 4 = 0 \implies m = \frac{4}{3} \).

Now, substitute these values of \( m \) back into Equation 1 (\( mx - y + 1 = 0 \)) to find the equations of the lines:

1. For \( m = 0 \):
\( 0(x) - y + 1 = 0 \)
\( \implies -y + 1 = 0 \)
\( \implies y = 1 \).
2. For \( m = \frac{4}{3} \):
\( \frac{4}{3}x - y + 1 = 0 \)
Multiply by 3 to clear the fraction:
\( 4x - 3y + 3 = 0 \).
These are the two required equations of the straight lines. Finding the slope allows us to determine the precise path of each line.
In simple words: We are looking for lines that pass through one point and are a certain distance away from another point. We start by writing the general equation for a line through the first point, using 'm' for its slope. Then, we use the distance formula, setting the distance to 1. This gives us an equation to solve for 'm'. We find two possible values for 'm', which then give us the two equations for the lines.

🎯 Exam Tip: Remember to express the line equation in the general form \( Ax + By + C = 0 \) before applying the distance formula. Squaring both sides of the distance equation is a common step that may introduce multiple solutions for the slope.

Question 9. A straight line is parallel to the lines \( 3x - y - 3 = 0 \) and \( 3x - y + 5 = 0 \), and lies between them. Find its equation if its distances from these lines are in the ratio 3 : 5.
Answer:
Let the given parallel lines be:
Line 1: \( 3x - y - 3 = 0 \)
Line 2: \( 3x - y + 5 = 0 \)

A line parallel to these two lines will have the form \( 3x - y + k = 0 \) (Line 3). This line lies between them.

The distance from Line 3 to Line 1 is \( d_1 = \frac{|k - (-3)|}{\sqrt{3^2 + (-1)^2}} = \frac{|k + 3|}{\sqrt{10}} \).
The distance from Line 3 to Line 2 is \( d_2 = \frac{|k - 5|}{\sqrt{3^2 + (-1)^2}} = \frac{|k - 5|}{\sqrt{10}} \).

We are given that the ratio of these distances is 3 : 5, so \( \frac{d_1}{d_2} = \frac{3}{5} \).
\( \frac{\frac{|k + 3|}{\sqrt{10}}}{\frac{|k - 5|}{\sqrt{10}}} = \frac{3}{5} \)
\( \implies \frac{|k + 3|}{|k - 5|} = \frac{3}{5} \)
This means \( 5|k + 3| = 3|k - 5| \).
Squaring both sides to remove the absolute value signs:
\( (5(k + 3))^2 = (3(k - 5))^2 \)
\( \implies 25(k + 3)^2 = 9(k - 5)^2 \)
\( \implies 25(k^2 + 6k + 9) = 9(k^2 - 10k + 25) \)
\( \implies 25k^2 + 150k + 225 = 9k^2 - 90k + 225 \)
Bring all terms to one side:
\( 25k^2 - 9k^2 + 150k + 90k + 225 - 225 = 0 \)
\( \implies 16k^2 + 240k = 0 \)
Factor out \( 16k \):
\( 16k(k + 15) = 0 \)
This gives two possible values for \( k \):
\( k = 0 \) or \( k = -15 \).

Since the line lies between the two given lines, \( k \) must be between -3 and 5. Both 0 and -15 are between -3 and 5 if we consider the interpretation of the formula, but mathematically, \( k \) should lie in the interval (-3, 5) for the line to be *strictly* between them. The values of C are -3 and 5. For \( k \) to be between these, \( |k - (-3)| \) and \( |k - 5| \) must be considered. Let's re-examine the distances. The values of C are \( C_1 = -3 \) and \( C_2 = 5 \). For a line \( 3x - y + k = 0 \) to be between \( 3x - y - 3 = 0 \) and \( 3x - y + 5 = 0 \), the value of \( k \) must be between -3 and 5. If \( k = 0 \), then the line is \( 3x - y = 0 \). This value 0 is between -3 and 5. If \( k = -15 \), then the line is \( 3x - y - 15 = 0 \). This value -15 is not between -3 and 5. It would be outside the given range. Therefore, \( k = -15 \) is not a valid solution for a line *between* the given lines in this context, although it satisfies the ratio. We reject \( k = -15 \).

So, the only valid value for \( k \) is 0.

Substituting \( k = 0 \) into \( 3x - y + k = 0 \):
The required equation of the line is \( 3x - y + 0 = 0 \implies 3x - y = 0 \).
In simple words: We have two parallel lines. We are looking for a third line, also parallel, that sits exactly between them and is closer to one line than the other in a 3:5 ratio. We use a general equation for this new line with an unknown number \( k \). We calculate its distance to both original lines and set up a ratio. Solving this gives us possible values for \( k \). We pick the value of \( k \) that makes the line actually sit in between the first two. This gives us the final line equation.

🎯 Exam Tip: When a problem states that a line lies "between" two other parallel lines, ensure that the constant term \( k \) in the resulting line's equation falls numerically between the constant terms of the given lines. This often helps in rejecting extraneous solutions obtained from squaring absolute values.

Question 10. Find the equation of the locus of a point P which is equidistant from the st. line \( 3x - 4y + 2 = 0 \) and the origin.
Answer:
Let P be a point with coordinates \( (x, y) \).
The locus of P is such that its distance from the line \( 3x - 4y + 2 = 0 \) is equal to its distance from the origin (0, 0).

1. **Distance from P(x, y) to the origin (0, 0):**
Using the distance formula between two points: \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Distance \( d_1 = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \).

2. **Distance from P(x, y) to the line \( 3x - 4y + 2 = 0 \):**
Using the perpendicular distance formula \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)
Here, \( (x_1, y_1) = (x, y) \), \( A = 3 \), \( B = -4 \), \( C = 2 \).
Distance \( d_2 = \frac{|3x - 4y + 2|}{\sqrt{3^2 + (-4)^2}} \)
\( \implies d_2 = \frac{|3x - 4y + 2|}{\sqrt{9 + 16}} \)
\( \implies d_2 = \frac{|3x - 4y + 2|}{\sqrt{25}} \)
\( \implies d_2 = \frac{|3x - 4y + 2|}{5} \).

Since the point P is equidistant from the line and the origin, we set \( d_1 = d_2 \):
\( \sqrt{x^2 + y^2} = \frac{|3x - 4y + 2|}{5} \)

To remove the square root and absolute value, we square both sides:
\( (\sqrt{x^2 + y^2})^2 = \left(\frac{|3x - 4y + 2|}{5}\right)^2 \)
\( \implies x^2 + y^2 = \frac{(3x - 4y + 2)^2}{25} \)
Multiply both sides by 25:
\( 25(x^2 + y^2) = (3x - 4y + 2)^2 \)

Expand the right side. Remember \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \).
Here, \( a = 3x \), \( b = -4y \), \( c = 2 \).
\( 25x^2 + 25y^2 = (3x)^2 + (-4y)^2 + (2)^2 + 2(3x)(-4y) + 2(-4y)(2) + 2(2)(3x) \)
\( \implies 25x^2 + 25y^2 = 9x^2 + 16y^2 + 4 - 24xy - 16y + 12x \)

Now, move all terms to the left side and combine like terms:
\( (25x^2 - 9x^2) + (25y^2 - 16y^2) + 24xy + 16y - 12x - 4 = 0 \)
\( \implies 16x^2 + 9y^2 + 24xy + 16y - 12x - 4 = 0 \).
This is the required equation of the locus of point P. This curved path is a parabola, defined by points equidistant from a point (focus) and a line (directrix).
In simple words: We are looking for all points that are the same distance from a given straight line and from the center point (origin). We write down the distance from a general point \( (x, y) \) to the origin, and then the distance from \( (x, y) \) to the line. We set these two distances equal to each other. Then, we do some algebra, including squaring both sides, to get a final equation that describes all such points.

🎯 Exam Tip: The locus of a point equidistant from a fixed point and a fixed line is a parabola. When solving, ensure you square both sides of the distance equation carefully, expanding binomials and trinomials correctly, and collecting all terms to one side for the final equation.

Question 11. A point P is such that the sum of the squares of its distances from the two axes of co-ordinates is equal to the square of its distance from the line \( x - y = 1 \). Find the equation of the locus of P.
Answer:
Let the coordinates of point P be \( (x, y) \).
The two coordinate axes are: the x-axis (equation \( y = 0 \)) and the y-axis (equation \( x = 0 \)).
The given line is \( x - y = 1 \), which can be written as \( x - y - 1 = 0 \).

1. **Distance from P(x, y) to the x-axis (y=0):**
\( d_x = \frac{|0x + 1y + 0|}{\sqrt{0^2 + 1^2}} = \frac{|y|}{1} = |y| \).
The square of this distance is \( d_x^2 = y^2 \).

2. **Distance from P(x, y) to the y-axis (x=0):**
\( d_y = \frac{|1x + 0y + 0|}{\sqrt{1^2 + 0^2}} = \frac{|x|}{1} = |x| \).
The square of this distance is \( d_y^2 = x^2 \).

3. **Distance from P(x, y) to the line \( x - y - 1 = 0 \):**
\( d_{line} = \frac{|1x - 1y - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y - 1|}{\sqrt{1 + 1}} = \frac{|x - y - 1|}{\sqrt{2}} \).
The square of this distance is \( d_{line}^2 = \left(\frac{|x - y - 1|}{\sqrt{2}}\right)^2 = \frac{(x - y - 1)^2}{2} \).

According to the problem statement, the sum of the squares of its distances from the two axes is equal to the square of its distance from the line \( x - y = 1 \).
So, \( d_x^2 + d_y^2 = d_{line}^2 \)
\( \implies y^2 + x^2 = \frac{(x - y - 1)^2}{2} \)

Multiply both sides by 2:
\( 2(x^2 + y^2) = (x - y - 1)^2 \)

Expand the right side using \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \).
Here, \( a = x \), \( b = -y \), \( c = -1 \).
\( 2x^2 + 2y^2 = x^2 + (-y)^2 + (-1)^2 + 2(x)(-y) + 2(-y)(-1) + 2(-1)(x) \)
\( \implies 2x^2 + 2y^2 = x^2 + y^2 + 1 - 2xy + 2y - 2x \)

Move all terms to the left side and simplify:
\( (2x^2 - x^2) + (2y^2 - y^2) + 2xy - 2y + 2x - 1 = 0 \)
\( \implies x^2 + y^2 + 2xy - 2y + 2x - 1 = 0 \).
This is the equation of the locus of point P. The conditions lead to a specific type of curve in the coordinate plane. The distance to a line and perpendicular axes determine the form of the equation.
In simple words: We find a point \( P(x, y) \) that has a special property: if you add up the square of its distance to the x-axis and the square of its distance to the y-axis, that sum should be equal to the square of its distance to another specific line. We write down each distance, square them, and set up the equation as given. Then, we simplify this equation to find the path (locus) of point P.

🎯 Exam Tip: Remember that the distance from \( (x,y) \) to the x-axis is \( |y| \) and to the y-axis is \( |x| \). Carefully expand squared trinomials and combine like terms to avoid errors when finding the locus equation.

Question 12. Show that the equation to the parallel line mid-way between the parallel lines \( ax + by + c_1 = 0 \) and \( ax + by + c_2 = 0 \) is \( ax + by + \frac{c_1+c_2}{2} = 0 \).
Answer:
Let the two given parallel lines be:
Line 1: \( ax + by + c_1 = 0 \)
Line 2: \( ax + by + c_2 = 0 \)

Let the equation of the line that is mid-way between them be \( ax + by + c = 0 \) (Line 3). Since it's mid-way, it must be equidistant from Line 1 and Line 2.

To find a point on Line 3, let's set \( x = 0 \). Then \( by + c = 0 \implies y = -\frac{c}{b} \).
So, a point on Line 3 is \( \left(0, -\frac{c}{b}\right) \).

The distance from this point \( \left(0, -\frac{c}{b}\right) \) to Line 1 \( (ax + by + c_1 = 0) \) is:
\( d_1 = \frac{|a(0) + b(-\frac{c}{b}) + c_1|}{\sqrt{a^2 + b^2}} = \frac{|-c + c_1|}{\sqrt{a^2 + b^2}} = \frac{|c_1 - c|}{\sqrt{a^2 + b^2}} \).

The distance from this point \( \left(0, -\frac{c}{b}\right) \) to Line 2 \( (ax + by + c_2 = 0) \) is:
\( d_2 = \frac{|a(0) + b(-\frac{c}{b}) + c_2|}{\sqrt{a^2 + b^2}} = \frac{|-c + c_2|}{\sqrt{a^2 + b^2}} = \frac{|c_2 - c|}{\sqrt{a^2 + b^2}} \).

Since Line 3 is mid-parallel, \( d_1 = d_2 \):
\( \frac{|c_1 - c|}{\sqrt{a^2 + b^2}} = \frac{|c_2 - c|}{\sqrt{a^2 + b^2}} \)
This means \( |c_1 - c| = |c_2 - c| \).
This equation implies two possibilities:
1. \( c_1 - c = c_2 - c \)
\( \implies c_1 = c_2 \). This would mean the two original lines are identical, which is not the case for two distinct parallel lines. 2. \( c_1 - c = -(c_2 - c) \)
\( \implies c_1 - c = -c_2 + c \)
\( \implies c_1 + c_2 = 2c \)
\( \implies c = \frac{c_1 + c_2}{2} \).

Substituting this value of \( c \) back into the equation of Line 3 \( (ax + by + c = 0) \):
The equation of the mid-parallel line is \( ax + by + \frac{c_1 + c_2}{2} = 0 \).
This shows how to find the exact middle line between any two parallel lines. The formula for the constant term is simply the average of the two original constant terms.
In simple words: We have two parallel lines and want to find a third line that runs exactly in the middle of them. We use a general equation for this middle line. For it to be in the middle, its distance to the first line must be the same as its distance to the second line. By setting these distances equal and doing some algebra, we find that the constant part of the middle line's equation is just the average of the constant parts of the first two lines.

🎯 Exam Tip: The key to this proof is recognizing that a mid-parallel line is equidistant from the two given parallel lines. Remember to handle the absolute value by considering both positive and negative cases; the case where the constant terms are equal is usually rejected as it implies identical lines.

Question 13. Prove that the line \( 12x - 5y - 3 = 0 \) is midparallel to the lines \( 12x - 5y + 7 = 0 \) and \( 12x - 5y - 13 = 0 \).
Answer:
Let the three given lines be:
Line (1): \( 12x - 5y - 3 = 0 \)
Line (2): \( 12x - 5y + 7 = 0 \)
Line (3): \( 12x - 5y - 13 = 0 \)

To prove that Line (1) is midparallel to Line (2) and Line (3), we need to show that the perpendicular distance from Line (1) to Line (2) is equal to the perpendicular distance from Line (1) to Line (3).

We use the formula for the distance between two parallel lines \( Ax + By + C_a = 0 \) and \( Ax + By + C_b = 0 \), which is \( d = \frac{|C_a - C_b|}{\sqrt{A^2 + B^2}} \).

For all three lines, \( A = 12 \) and \( B = -5 \). So, \( \sqrt{A^2 + B^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \).

1. **Distance between Line (1) and Line (2):**
Here, \( C_a = -3 \) (from Line 1) and \( C_b = 7 \) (from Line 2).
\( d_{12} = \frac{|-3 - 7|}{13} \)
\( \implies d_{12} = \frac{|-10|}{13} \)
\( \implies d_{12} = \frac{10}{13} \) units.

2. **Distance between Line (1) and Line (3):**
Here, \( C_a = -3 \) (from Line 1) and \( C_b = -13 \) (from Line 3).
\( d_{13} = \frac{|-3 - (-13)|}{13} \)
\( \implies d_{13} = \frac{|-3 + 13|}{13} \)
\( \implies d_{13} = \frac{|10|}{13} \)
\( \implies d_{13} = \frac{10}{13} \) units.

Since \( d_{12} = d_{13} = \frac{10}{13} \), we have proven that Line (1) is equidistant from Line (2) and Line (3). Therefore, Line (1) is midparallel to Line (2) and Line (3). The coefficients of x and y are identical, which ensures they are parallel, and the distances confirm it's exactly in the middle.
In simple words: We have three parallel lines. To show that the first line is exactly in the middle of the other two, we calculate how far it is from the second line. Then, we calculate how far it is from the third line. If these two distances are the same, it means the first line is indeed in the middle. We found both distances to be equal, so the proof is complete.

🎯 Exam Tip: For problems proving a line is "midparallel", ensure you use the correct formula for distance between parallel lines. Remember that the absolute value in the numerator ensures the distance is always positive, and be careful with the signs of the constant terms \( C_1 \) and \( C_2 \).