Q1. Find the values of k for which the line \((k–3) x – (4 – k^2) y + k^2 –7k + 6 = 0\) is
(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

Solution

The given equation of the line is \[ (k-3)x-(4-k^{2})y+k^{2}-7k+6=0 \]

For a line to be parallel to the x-axis, the coefficient of \(y\) must be zero. Hence, \[ -4+k^{2}=0 \] \[ k^{2}=4 \] \[ k=\pm 2 \] Therefore, the given line is parallel to the x-axis for \(k=2\) or \(k=-2\).

For a line to be parallel to the y-axis, the coefficient of \(x\) must be zero. Thus, \[ k-3=0 \] \[ k=3 \] Hence, the line is parallel to the y-axis when \(k=3\).

For the line to pass through the origin, the point \((0,0)\) must satisfy the given equation. Substituting \(x=0\) and \(y=0\), we get \[ k^{2}-7k+6=0 \] \[ k^{2}-6k-k+6=0 \] \[ k(k-6)-1(k-6)=0 \] \[ (k-6)(k-1)=0 \] \[ k=1 \text{ or } k=6 \] Thus, the line passes through the origin for \(k=1\) or \(k=6\).

Q2. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

Solution

Let the intercepts cut off by the required line on the x-axis and y-axis be \(a\) and \(b\), respectively. According to the given conditions, the sum of the intercepts is \(1\) and their product is \(-6\). Thus, \[ a+b=1 \] \[ ab=-6 \]

From the relation \(ab=-6\), we can write \[ b=-\dfrac{6}{a} \] Substituting this value of \(b\) in \(a+b=1\), we get \[ a+\left(-\dfrac{6}{a}\right)=1 \] \[ a^{2}-6=a \] \[ a^{2}-a-6=0 \] \[ a^{2}-3a+2a-6=0 \] \[ a(a-3)+2(a-3)=0 \] \[ (a+2)(a-3)=0 \]

Hence, the possible values of \(a\) are \[ a=3 \quad \text{or} \quad a=-2 \] Using \(a+b=1\), when \(a=3\), \[ b=1-3=-2 \] and when \(a=-2\), \[ b=1-(-2)=3 \] Thus, the pairs of intercepts are \((3,-2)\) and \((-2,3)\).

The equation of a line in intercept form is \(\dfrac{x}{a}+\dfrac{y}{b}=1\). For the intercepts \(a=3\) and \(b=-2\), the equation of the line is \[ \dfrac{x}{3}+\dfrac{y}{-2}=1 \] or equivalently \[ \dfrac{x}{3}-\dfrac{y}{2}=1 \]

For the intercepts \(a=-2\) and \(b=3\), the equation of the line is \[ \dfrac{x}{-2}+\dfrac{y}{3}=1 \] or equivalently \[ \dfrac{y}{3}-\dfrac{x}{2}=1 \]

Q3. What are the points on the y-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}=1\) is 4 units.

Solution

Let the required point on the y-axis be \((0,b)\), since any point on the y-axis has x-coordinate equal to zero.

The given equation of the line is \[ \dfrac{x}{3}+\dfrac{y}{4}=1 \] Multiplying throughout by \(12\), we get \[ 4x+3y-12=0 \]

The distance of a point \((x_1,y_1)\) from the line \(Ax+By+C=0\) is given by \[ \dfrac{|Ax_1+By_1+C|}{\sqrt{A^{2}+B^{2}}} \] Here, \(A=4\), \(B=3\), \(C=-12\), and the point is \((0,b)\). Since the distance is given as \(4\) units, we have \[ 4=\dfrac{|4\cdot 0+3b-12|}{\sqrt{4^{2}+3^{2}}} \] \[ 4=\dfrac{|3b-12|}{\sqrt{16+9}} \] \[ 4=\dfrac{|3b-12|}{5} \]

Multiplying both sides by \(5\), we get \[ |3b-12|=20 \] This gives two cases. First, \[ 3b-12=20 \] \[ 3b=32 \] \[ b=\dfrac{32}{3} \] Second, \[ 3b-12=-20 \] \[ 3b=-8 \] \[ b=\dfrac{-8}{3} \]

Hence, the required points on the y-axis are \[ \left(0,\dfrac{32}{3}\right) \text{ and } \left(0,\dfrac{-8}{3}\right) \]

Q4. Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ) and (cos φ, sin φ).

Solution

Let the given points be \(A(\cos\theta,\sin\theta)\) and \(B(\cos\phi,\sin\phi)\). The perpendicular distance from the origin to the line joining two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \[ \text{Distance}=\dfrac{|x_1y_2-y_1x_2|}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}} \]

Substituting the coordinates of \(A\) and \(B\), we obtain \[ \text{Distance}=\dfrac{|\cos\theta\sin\phi-\sin\theta\cos\phi|}{\sqrt{(\cos\phi-\cos\theta)^2+(\sin\phi-\sin\theta)^2}} \]

Using the trigonometric identity \(\sin\phi\cos\theta-\cos\phi\sin\theta=\sin(\phi-\theta)\), the numerator becomes \[ |\sin(\phi-\theta)| \]

Now simplifying the denominator, \[ \begin{aligned} (\cos\phi-\cos\theta)^2+(\sin\phi-\sin\theta)^2 &=\cos^2\phi+\cos^2\theta-2\cos\phi\cos\theta \\ &\quad+\sin^2\phi+\sin^2\theta-2\sin\phi\sin\theta \\ &=2-2(\cos\phi\cos\theta+\sin\phi\sin\theta) \\ &=2-2\cos(\phi-\theta) \end{aligned} \]

Using the identity \(1-\cos\alpha=2\sin^2\frac{\alpha}{2}\), we get \[ 2-2\cos(\phi-\theta)=4\sin^2\frac{\phi-\theta}{2} \] Hence, the denominator becomes \[ 2\left|\sin\frac{\phi-\theta}{2}\right| \]

Therefore, the required perpendicular distance is \[ \begin{aligned} \text{Distance} &=\dfrac{|\sin(\phi-\theta)|}{2\left|\sin\frac{\phi-\theta}{2}\right|} \\ &=\left|\cos\frac{\phi-\theta}{2}\right| \end{aligned} \]

Hence, the perpendicular distance from the origin to the line joining the points \((\cos\theta,\sin\theta)\) and \((\cos\phi,\sin\phi)\) is \[ \boxed{\left|\cos\dfrac{\phi-\theta}{2}\right|} \]

Q5. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines \(x – 7y + 5 = 0\) and \(3x + y = 0\).

Solution

To find the required line, we first determine the point of intersection of the given lines \[ x-7y+5=0 \] and \[ 3x+y=0 \]

From the second equation, \[ y=-3x \] Substituting this value of \(y\) in the first equation, we get \[ x-7(-3x)+5=0 \] \[ x+21x+5=0 \] \[ 22x+5=0 \] \[ x=-\dfrac{5}{22} \]

Substituting \(x=-\dfrac{5}{22}\) in \(3x+y=0\), \[ 3\left(-\dfrac{5}{22}\right)+y=0 \] \[ -\dfrac{15}{22}+y=0 \] \[ y=\dfrac{15}{22} \] Hence, the point of intersection of the two lines is \(\left(-\dfrac{5}{22},\dfrac{15}{22}\right)\).

A line parallel to the y-axis has the equation \(x=\text{constant}\). Since the required line passes through the point \(\left(-\dfrac{5}{22},\dfrac{15}{22}\right)\), its equation is \[ x=-\dfrac{5}{22} \]

Q6. Find the equation of a line drawn perpendicular to the line \(\frac{x}{4} + \frac{y}{6} = 1\) through the point, where it meets the y-axis.

Solution

The given line is \[ \dfrac{x}{4}+\dfrac{y}{6}=1 \] Putting \(x=0\), the line meets the y-axis at the point \((0,6)\).

To find the slope of the given line, we rewrite its equation in slope–intercept form \[ 6x+4y=24 \] \[ 4y=-6x+24 \] \[ y=-\dfrac{3}{2}x+6 \] Hence, the slope of the given line is \(-\dfrac{3}{2}\).

If a line is perpendicular to another line, the product of their slopes is \(-1\). Let the slope of the required line be \(m\). Then \[ \begin{aligned} m\left(-\dfrac{3}{2}\right)=-1 \\ m=\dfrac{2}{3} \end{aligned} \]

The required line passes through the point \((0,6)\) and has slope \(\dfrac{2}{3}\). Using the point–slope form of the equation of a line, \[ \begin{aligned} y-y_{0}=m(x-x_{0}) \\ y-6=\dfrac{2}{3}(x-0) \\ 3y-18=2x \\ 3y-2x-18=0 \end{aligned} \]

Q7. Find the area of the triangle formed by the lines \(y – x = 0,\; x + y = 0\) and \(x – k = 0\).

Solution

The given lines are \[ y-x=0 \] \[ x+y=0 \] \[ x-k=0 \] The area of the triangle formed by these lines can be found after determining their points of intersection.

First, we find the point of intersection of the lines \(y-x=0\) and \(x+y=0\). Solving these simultaneously, \[ \begin{aligned} y-x=0 \\ x+y=0 \\ 2y=0 \\ y=0 \\ x=0 \end{aligned} \] Hence, one vertex of the triangle is \((0,0)\).

Next, we find the point of intersection of the lines \(x+y=0\) and \(x-k=0\). From \(x-k=0\), we get \(x=k\). Substituting this in \(x+y=0\), \[ \begin{aligned} k+y=0 \\ y=-k \end{aligned} \] Thus, the second vertex is \((k,-k)\).

Now, we find the point of intersection of the lines \(y-x=0\) and \(x-k=0\). From \(x-k=0\), we have \(x=k\). Substituting this in \(y-x=0\), \[ \begin{aligned} y-k=0 \\ y=k \end{aligned} \] Hence, the third vertex of the triangle is \((k,k)\).

The vertices of the triangle are therefore \((0,0)\), \((k,-k)\), and \((k,k)\). Using the shoelace formula for the area of a triangle, \[ \begin{aligned} \Delta &=\dfrac{1}{2}\left| 0(-k-k)+k(k-0)+k(0-(-k)) \right| \\ &=\dfrac{1}{2}\left| k^{2}+k^{2}\right| \\ &=\left|k^{2}\right| \end{aligned} \]

Hence, the area of the triangle formed by the given lines is \[ \boxed{\;k^{2}\;} \]

Q8. Find the value of p so that the three lines \(3x + y – 2 = 0,\; px + 2 y – 3 = 0\) and \(2x – y – 3 = 0\) may intersect at one point.

Solution

For the three given lines to intersect at one point, they must be concurrent, that is, they must pass through a common point. We first find the point of intersection of the lines \[ 3x+y-2=0 \] and \[ 2x-y-3=0 \]

Solving these two equations simultaneously, \[ \begin{aligned} 3x+y-2=0 \\ 2x-y-3=0 \\ 5x-5=0 \\ x=1 \end{aligned} \] Substituting \(x=1\) in \(2x-y-3=0\), \[ \begin{aligned} 2(1)-y-3=0 \\ -y-1=0 \\ y=-1 \end{aligned} \] Hence, the point of intersection of the first two lines is \((1,-1)\).

For all three lines to intersect at one point, this point must also satisfy the third line \[ px+2y-3=0 \] Substituting \(x=1\) and \(y=-1\), \[ \begin{aligned} p(1)+2(-1)-3=0 \\ p-2-3=0 \\ p=5 \end{aligned} \]

Therefore, the value of \(p\) for which the three lines intersect at one point is \[ \boxed{5} \]

Q9. If three lines whose equations are \(y = m_1 x + c_1 ,\; y = m_2 x + c_2\) and \(y = m_3 x + c_3\) are concurrent, then show that \(m_1 (c_2 – c_3 ) + m_2 (c_3 – c_1 ) + m_3 (c_1 – c_2 ) = 0\).

Solution

Since the three given lines are concurrent, they pass through one common point. Let their point of concurrency be \((x_0,y_0)\).

Because the point \((x_0,y_0)\) lies on each line, it must satisfy all three equations \[ \begin{aligned} y_0&=m_1x_0+c_1 \\ y_0&=m_2x_0+c_2 \\ y_0&=m_3x_0+c_3 \end{aligned} \]

Subtracting the second equation from the first, we get \[ \begin{aligned} m_1x_0+c_1&=m_2x_0+c_2 \\ (m_1-m_2)x_0&=c_2-c_1 \end{aligned} \] Similarly, subtracting the third equation from the second, we obtain \[ \begin{aligned} m_2x_0+c_2&=m_3x_0+c_3 \\ (m_2-m_3)x_0&=c_3-c_2 \end{aligned} \]

Multiplying the first of these results by \((m_3-m_1)\) and the second by \((m_1-m_2)\), we have \[ \begin{aligned} (m_1-m_2)(m_3-m_1)x_0&=(c_2-c_1)(m_3-m_1) \\ (m_2-m_3)(m_1-m_2)x_0&=(c_3-c_2)(m_1-m_2) \end{aligned} \]

Adding the above two equations and simplifying, the terms involving \(x_0\) cancel, giving \[ \begin{aligned} m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0 \end{aligned} \]

Hence, if the three lines are concurrent, then \[ \boxed{m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0} \]

Q10. Find the equation of the lines through the point (3, 2) which make an angle of \(45^\circ\) with the line \(x – 2y = 3\).

Solution

The required line passes through the point \((3,2)\) and makes an angle of \(45^\circ\) with the line \[ x-2y=3 \]

First, we find the slope of the given line \[ \begin{aligned} x-2y=3 \\ 2y=x-3 \\ y=\dfrac{1}{2}x-\dfrac{3}{2} \end{aligned} \] Hence, the slope of the given line is \[ m_{2}=\dfrac{1}{2} \]

If \(m_{1}\) is the slope of the required line and the angle between the two lines is \(45^\circ\), then \[ \tan\theta=\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}} \] Since \(\theta=45^\circ\) and \(\tan45^\circ=1\), \[ \begin{aligned} 1=\dfrac{m_{1}-\frac{1}{2}}{1+\frac{m_{1}}{2}} \\ 1+\dfrac{m_{1}}{2}=m_{1}-\dfrac{1}{2} \\ \dfrac{m_{1}}{2}-m_{1}=-\dfrac{1}{2}-1 \\ -m_{1}=-3 \\ m_{1}=3 \end{aligned} \]

Using the point–slope form of the equation of a line through \((3,2)\), \[ \begin{aligned} y-y_{0}=m(x-x_{0}) \\ y-2=3(x-3) \\ y-2=3x-9 \\ y-3x+7=0 \end{aligned} \]

Since the angle between two lines can also be \(45^\circ\) in the other orientation, we take \[ \tan45^\circ=\dfrac{m_{2}-m_{1}}{1+m_{1}m_{2}} \] which gives \[ \begin{aligned} 1=\dfrac{\frac{1}{2}-m_{1}}{1+\frac{m_{1}}{2}} \\ 1+\dfrac{m_{1}}{2}=\dfrac{1}{2}-m_{1} \\ \dfrac{3m_{1}}{2}=-\dfrac{1}{2} \\ m_{1}=-\dfrac{1}{3} \end{aligned} \]

The equation of the second required line through \((3,2)\) is \[ \begin{aligned} y-2=-\dfrac{1}{3}(x-3) \\ 3y-6=-x+3 \\ x+3y-9=0 \end{aligned} \]

Hence, the required equations of the lines are \[ y-3x+7=0 \quad \text{and} \quad x+3y-9=0 \]

Q11. Find the equation of the line passing through the point of intersection of the lines \(4x + 7y – 3 = 0\) and \(2x – 3y + 1 = 0\) that has equal intercepts on the axes.

Solution

First, we find the point of intersection of the given lines \[ 4x+7y-3=0 \] and \[ 2x-3y+1=0 \]

Solving these equations simultaneously, \[ \begin{aligned} 4x+7y&=3 \\ 2x-3y&=-1 \end{aligned} \] Multiplying the second equation by \(2\), \[ \begin{aligned} 4x-6y&=-2 \end{aligned} \] Subtracting this from the first equation, \[ \begin{aligned} 13y&=5 \\ y&=\dfrac{5}{13} \end{aligned} \] Substituting \(y=\dfrac{5}{13}\) in \(2x-3y=-1\), \[ \begin{aligned} 2x-\dfrac{15}{13}&=-1 \\ 2x&=\dfrac{2}{13} \\ x&=\dfrac{1}{13} \end{aligned} \] Hence, the point of intersection is \(\left(\dfrac{1}{13},\dfrac{5}{13}\right)\).

A line having equal intercepts on the coordinate axes has the intercept form \[ \dfrac{x}{a}+\dfrac{y}{a}=1 \] which simplifies to \[ x+y=a \]

Since the required line passes through the point \(\left(\dfrac{1}{13},\dfrac{5}{13}\right)\), we substitute these coordinates in \(x+y=a\), \[ \begin{aligned} \dfrac{1}{13}+\dfrac{5}{13}&=a \\ a&=\dfrac{6}{13} \end{aligned} \]

Therefore, the equation of the required line is \[ x+y=\dfrac{6}{13} \] or equivalently \[ 13x+13y-6=0 \]

Q12. Show that the equation of the line passing through the origin and making an angle \(\theta\) with the line \(y=mc+c\) is \(\frac{y}{x}=\frac{m\pm \tan \theta}{1\mp m\tan \theta}\)

Solution

Let the given line be \[ y=mx+c \] Its slope is \(m\). Suppose the required line passes through the origin and makes an angle \(\theta\) with this line. Let the slope of the required line be \(m_1\).

Since the required line passes through the origin, its equation is \[ y=m_1x \] or equivalently \[ \dfrac{y}{x}=m_1 \]

The angle \(\theta\) between two lines having slopes \(m_1\) and \(m\) is given by \[ \tan\theta=\left|\dfrac{m_1-m}{1+m_1m}\right| \] Removing the modulus sign to include both possible orientations of the angle, we write \[ \tan\theta=\dfrac{m_1-m}{1+m_1m} \] or \[ \tan\theta=\dfrac{m-m_1}{1+m_1m} \]

From \[ \tan\theta=\dfrac{m_1-m}{1+m_1m} \] we get \[ \begin{aligned} \tan\theta(1+m_1m)&=m_1-m \\ \tan\theta+m_1m\tan\theta&=m_1-m \\ m_1(1-m\tan\theta)&=m+\tan\theta \\ m_1&=\dfrac{m+\tan\theta}{1-m\tan\theta} \end{aligned} \]

Similarly, from \[ \tan\theta=\dfrac{m-m_1}{1+m_1m} \] we obtain \[ \begin{aligned} \tan\theta(1+m_1m)&=m-m_1 \\ \tan\theta+m_1m\tan\theta&=m-m_1 \\ m_1(1+m\tan\theta)&=m-\tan\theta \\ m_1&=\dfrac{m-\tan\theta}{1+m\tan\theta} \end{aligned} \]

Since \(\dfrac{y}{x}=m_1\), the equation of the required line is \[ \dfrac{y}{x}=\dfrac{m\pm\tan\theta}{1\mp m\tan\theta} \]

Hence proved.

Q13. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Solution

First, we find the equation of the line passing through the points \((-1,1)\) and \((5,7)\). The slope of this line is \[ \begin{aligned} m&=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ &=\dfrac{7-1}{5-(-1)} \\ &=\dfrac{6}{6} \\ &=1 \end{aligned} \]

Using the point–slope form of the equation of a line through \((-1,1)\), \[ \begin{aligned} y-y_{0}&=m(x-x_{0}) \\ y-1&=1(x+1) \\ y-x-2&=0 \end{aligned} \]

Now we find the point of intersection of the lines \[ y-x-2=0 \] and \[ x+y-4=0 \] Solving these simultaneously, \[ \begin{aligned} y&=x+2 \\ x+(x+2)-4&=0 \\ 2x-2&=0 \\ x&=1 \\ y&=1+2=3 \end{aligned} \] Thus, the point of intersection is \((1,3)\).

Let the line joining \((-1,1)\) and \((5,7)\) be divided by the line \(x+y=4\) in the ratio \(m:n\). Using the section formula for the x-coordinate, \[ \begin{aligned} 1&=\dfrac{m\cdot 5+n(-1)}{m+n} \\ m+n&=5m-n \\ 2n&=4m \\ n&=2m \end{aligned} \]

Hence, the ratio in which the line is divided is \[ m:n=1:2 \]

Q15. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line \(x + y = 4\) may be at a distance of 3 units from this point.

Solution

Let the required line pass through the point \((-1,2)\) and make an angle such that its slope is \(m\). The equation of the line passing through \((-1,2)\) is \[ \begin{aligned} y-y_{0}&=m(x-x_{0}) \\ y-2&=m(x+1) \\ y-mx-m-2&=0 \\ y&=m(x+1)+2 \end{aligned} \]

To find the point of intersection of this line with the line \(x+y=4\), we substitute the value of \(y\) from the above equation into \(x+y=4\), \[ \begin{aligned} x+m(x+1)+2&=4 \\ x+mx+m&=2 \\ x(m+1)+m&=2 \\ x&=\dfrac{2-m}{m+1} \end{aligned} \]

Using \(y=4-x\), we get \[ \begin{aligned} y&=4-\dfrac{2-m}{m+1} \\ y&=\dfrac{4(m+1)-(2-m)}{m+1} \\ y&=\dfrac{5m+2}{m+1} \end{aligned} \] Hence, the point of intersection is \[ \left(\dfrac{2-m}{m+1},\dfrac{5m+2}{m+1}\right) \]

The distance between this point and the point \((-1,2)\) is given to be \(3\) units. Using the distance formula, \[ \begin{aligned} 3&=\sqrt{\left(-1-\dfrac{2-m}{m+1}\right)^2+\left(2-\dfrac{5m+2}{m+1}\right)^2} \\ 9&=\left(\dfrac{-m-1-2+m}{m+1}\right)^2+\left(\dfrac{2m+2-5m-2}{m+1}\right)^2 \\ 9&=\left(\dfrac{-3}{m+1}\right)^2+\left(\dfrac{-3m}{m+1}\right)^2 \\ 9&=\dfrac{9+9m^2}{(m+1)^2} \end{aligned} \]

Multiplying both sides by \((m+1)^2\), \[ \begin{aligned} 9(m+1)^2&=9+9m^2 \\ 9m^2+18m+9&=9+9m^2 \\ 18m&=0 \\ m&=0 \end{aligned} \]

Thus, the required line has slope \(m=0\), and hence it is parallel to the x-axis. Therefore, the direction in which the line must be drawn is parallel to the x-axis.

Q16. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

Solution

Since the legs of the right angled triangle are parallel to the coordinate axes, the right angle must occur at a point whose x-coordinate is the same as one end of the hypotenuse and whose y-coordinate is the same as the other end. The given ends of the hypotenuse are \((1,3)\) and \((-4,1)\).

Taking the x-coordinate from \((1,3)\) and the y-coordinate from \((-4,1)\), the right angle is at the point \((1,1)\). Hence, the legs of the triangle are the lines through this point parallel to the coordinate axes.

The equations of these perpendicular sides are \[ \begin{aligned} x&=1 \\ y&=1 \end{aligned} \]

Alternatively, taking the x-coordinate from \((-4,1)\) and the y-coordinate from \((1,3)\), the right angle can also be at the point \((-4,3)\). In this case, the legs of the triangle are again parallel to the coordinate axes.

The equations of these perpendicular sides are \[ \begin{aligned} x&=-4 \\ y&=3 \end{aligned} \]

Thus, the required equations of the legs of the right angled triangle, which are parallel to the axes, are either \(x=1,\; y=1\) or \(x=-4,\; y=3\).

Q17. Find the image of the point (3, 8) with respect to the line \(x +3y = 7\) assuming the line to be a plane mirror.

Solution

The given point is \((3,8)\). Since the line \(x+3y=7\) acts as a plane mirror, the image of the point will lie on the perpendicular drawn from \((3,8)\) to the given line and at the same distance on the other side.

First, we find the equation of the line perpendicular to \(x+3y=7\) and passing through \((3,8)\). Rewriting the given line, \[ \begin{aligned} x+3y&=7 \\ 3y&=-x+7 \\ y&=-\dfrac{1}{3}x+\dfrac{7}{3} \end{aligned} \] Hence, the slope of the given line is \(-\dfrac{1}{3}\), so the slope of the perpendicular line is \(3\).

Using the point–slope form for the perpendicular line through \((3,8)\), \[ \begin{aligned} y-y_{0}&=m_{2}(x-x_{0}) \\ y-8&=3(x-3) \\ y-8&=3x-9 \\ y-3x+1&=0 \end{aligned} \]

Now we find the point of intersection of the given line and this perpendicular line. Solving \[ \begin{aligned} x+3y&=7 \\ y&=3x-1 \end{aligned} \] Substituting for \(y\), \[ \begin{aligned} x+3(3x-1)&=7 \\ x+9x-3&=7 \\ 10x&=10 \\ x&=1 \end{aligned} \] Substituting \(x=1\) in \(x+3y=7\), \[ \begin{aligned} 1+3y&=7 \\ 3y&=6 \\ y&=2 \end{aligned} \] Thus, the foot of the perpendicular is \((1,2)\).

Let the image of the point \((3,8)\) be \((x_{1},y_{1})\). Since the mirror line bisects the segment joining the point and its image at right angles, \((1,2)\) is the midpoint of \((3,8)\) and \((x_{1},y_{1})\). Using the midpoint formula, \[ \begin{aligned} 1&=\dfrac{x_{1}+3}{2} \\ 2&=\dfrac{y_{1}+8}{2} \end{aligned} \]

Solving these, \[ \begin{aligned} x_{1}+3&=2 \\ x_{1}&=-1 \\ y_{1}+8&=4 \\ y_{1}&=-4 \end{aligned} \]

Hence, the image of the point \((3,8)\) with respect to the line \(x+3y=7\) is \[ (-1,-4) \]

Q18. If the lines \(y = 3x +1\) and \(2y = x + 3\) are equally inclined to the line \(y = mx + 4\), find the value of \(m\).

Solution

The given lines are \[ y=3x+1 \] and \[ 2y=x+3 \quad \Rightarrow \quad y=\dfrac{1}{2}x+\dfrac{3}{2} \] Let the line \(y=mx+4\) make equal angles with both these lines.

If two lines are equally inclined to a third line, then the angles between the third line and each of the given lines are equal in magnitude but opposite in sense. Hence, using the formula for the tangent of the angle between two lines, \[ \tan\theta=\left|\dfrac{m_1-m}{1+m_1m}\right| \]

For the line \(y=3x+1\), the slope is \(3\). For the line \(y=\dfrac{1}{2}x+\dfrac{3}{2}\), the slope is \(\dfrac{1}{2}\). Thus, \[ \dfrac{3-m}{1+3m}=-\dfrac{\frac{1}{2}-m}{1+\frac{m}{2}} \]

Simplifying the right-hand side, \[ \dfrac{\frac{1}{2}-m}{1+\frac{m}{2}}=\dfrac{1-2m}{2+m} \] Hence, \[ \dfrac{3-m}{1+3m}=-\dfrac{1-2m}{2+m} \]

Cross-multiplying, \[ \begin{aligned} (3-m)(2+m)&=-(1+3m)(1-2m) \\ 6+3m-2m-m^2&=-\left(1-2m+3m-6m^2\right) \\ 6+m-m^2&=-1-m+6m^2 \end{aligned} \]

Bringing all terms to one side, \[ \begin{aligned} 6+m-m^2+1+m-6m^2&=0 \\ 7-7m^2+2m&=0 \\ 7m^2-2m-7&=0 \end{aligned} \]

Solving the quadratic equation, \[ \begin{aligned} m&=\dfrac{2\pm\sqrt{(-2)^2-4(7)(-7)}}{2\cdot7} \\ &=\dfrac{2\pm\sqrt{4+196}}{14} \\ &=\dfrac{2\pm\sqrt{200}}{14} \\ &=\dfrac{2\pm10\sqrt{2}}{14} \\ &=\dfrac{1\pm5\sqrt{2}}{7} \end{aligned} \]

Hence, the required values of \(m\) are \[ \boxed{\,m=\dfrac{1\pm5\sqrt{2}}{7}\,} \]

Q19. If sum of the perpendicular distances of a variable point \(P (x,\; y)\) from the lines \(x + y – 5 = 0\) and \(3x – 2y +7 = 0\) is always 10. Show that \(P\) must move on a line.

Solution

Let the variable point be \(P(x,y)\). The perpendicular distance of a point \((x_1,y_1)\) from the line \(Ax+By+C=0\) is \[ \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \]

The given lines are \[ x+y-5=0 \] and \[ 3x-2y+7=0 \]

The perpendicular distances of \(P(x,y)\) from these lines are \[ \dfrac{|x+y-5|}{\sqrt{1^2+1^2}}=\dfrac{|x+y-5|}{\sqrt{2}} \] and \[ \dfrac{|3x-2y+7|}{\sqrt{3^2+(-2)^2}}=\dfrac{|3x-2y+7|}{\sqrt{13}} \]

According to the given condition, the sum of these distances is always \(10\). Hence, \[ \begin{aligned} \dfrac{|x+y-5|}{\sqrt{2}}+\dfrac{|3x-2y+7|}{\sqrt{13}}&=10 \end{aligned} \]

Since the sum is constant, the point \(P\) must lie entirely on one fixed side of each line. Removing the modulus signs with appropriate choice of signs, we get \[ \begin{aligned} \dfrac{x+y-5}{\sqrt{2}}+\dfrac{3x-2y+7}{\sqrt{13}}&=10 \end{aligned} \]

Multiplying throughout by \(\sqrt{26}\), \[ \begin{aligned} \sqrt{13}(x+y-5)+\sqrt{2}(3x-2y+7)&=10\sqrt{26} \end{aligned} \]

This equation is linear in \(x\) and \(y\). Hence, it represents the equation of a straight line.

Therefore, the variable point \(P(x,y)\), whose sum of perpendicular distances from the given two lines is constant, must move on a straight line.

Q20. Find equation of the line which is equidistant from parallel lines \(9x + 6y – 7 = 0\) and \(3x + 2y + 6 = 0\).

Solution

Let an equidistant point be \((x_1, y_1)\). Since the given lines are parallel, rewrite the first line by dividing by 3:

\[ 9x + 6y - 7 = 0 \;\Rightarrow\; 3x + 2y - \frac{7}{3} = 0 \]

Distances of \((x_1, y_1)\) from the two lines are:

\[ \begin{aligned} d_1 &= \dfrac{3x_1 + 2y_1 - \dfrac{7}{3}}{\sqrt{3^2 + 2^2}} \\ d_2 &= \dfrac{3x_1 + 2y_1 + 6}{\sqrt{3^2 + 2^2}} \end{aligned} \]

Since the point is equidistant from both lines:

\[ \begin{aligned} \left| \dfrac{3x_1 + 2y_1 - \dfrac{7}{3}}{\sqrt{13}} \right| &= \left| \dfrac{3x_1 + 2y_1 + 6}{\sqrt{13}} \right| \\ \left| 3x_1 + 2y_1 - \frac{7}{3} \right| &= \left| 3x_1 + 2y_1 + 6 \right| \end{aligned} \]

Let \(3x_1 + 2y_1 = k\). Then:

\[ \begin{aligned} \left| k - \frac{7}{3} \right| &= \left| k + 6 \right| \\ \left( k - \frac{7}{3} \right)^2 &= \left( k + 6 \right)^2 \\ k^2 - \frac{14}{3}k + \frac{49}{9} &= k^2 + 12k + 36 \\ - \frac{14}{3}k - 12k &= 36 - \frac{49}{9} \\ - \frac{50}{3}k &= \frac{275}{9} \\ k &= -\frac{11}{6} \end{aligned} \]

Substituting back:

\[ 3x_1 + 2y_1 = -\frac{11}{6} \]

Multiplying throughout by 6, the required equation of the line equidistant from the given parallel lines is:

\[ \boxed{18x + 12y + 11 = 0} \]

Q21. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Solution

Let the ray of light pass through the point \(P(1, 2)\) and reflect on the x-axis at point \(A(x, 0)\). After reflection, the ray passes through the point \(Q(5, 3)\).

Reflection in the x-axis maps the point \(Q(5, 3)\) to \(Q'(5, -3)\). Hence, the reflected path can be treated as a straight line passing through \(P(1, 2)\) and \(Q'(5, -3)\).

The slope of line \(PQ'\) is:

\[ \begin{aligned} m &= \frac{-3 - 2}{5 - 1} \\ &= \frac{-5}{4} \end{aligned} \]

Equation of the line through \(P(1, 2)\) with slope \(-\frac{5}{4}\) is:

\[ \begin{aligned} y - 2 &= -\frac{5}{4}(x - 1) \end{aligned} \]

Since point \(A\) lies on the x-axis, substitute \(y = 0\):

\[ \begin{aligned} 0 - 2 &= -\frac{5}{4}(x - 1) \\ -2 &= -\frac{5}{4}(x - 1) \\ -8 &= -5(x - 1) \\ 8 &= 5(x - 1) \\ x - 1 &= \frac{8}{5} \\ x &= \frac{13}{5} \end{aligned} \]

Therefore, the coordinates of the point of reflection are:

\[ \boxed{A\left(\frac{13}{5},\,0\right)} \]

Q22. Prove that the product of the lengths of the perpendiculars drawn from the points \(\left(\sqrt{a^2-b^2},0\right)\) and \(\left(-\sqrt{a^2-b^2},0\right)\) to the line \(\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta=1\) is \(b^2\)

Solution

Let the given line be

\[ \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1 \]

Rewrite it in standard linear form by multiplying throughout by \(ab\):

\[ bx\cos\theta + ay\sin\theta - ab = 0 \]

The perpendicular distance from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

For the first point \(P\left(\sqrt{a^2 - b^2},\,0\right)\), the perpendicular distance is

\[ \begin{aligned} d_1 &= \frac{\left| b\sqrt{a^2 - b^2}\cos\theta - ab \right|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} \end{aligned} \]

For the second point \(Q\left(-\sqrt{a^2 - b^2},\,0\right)\), the perpendicular distance is

\[ \begin{aligned} d_2 &= \frac{\left| -b\sqrt{a^2 - b^2}\cos\theta - ab \right|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} \end{aligned} \]

Now consider the product \(d_1 d_2\):

\[ \begin{aligned} d_1 d_2 &= \frac{\left| \left(b\sqrt{a^2 - b^2}\cos\theta - ab\right)\left(-b\sqrt{a^2 - b^2}\cos\theta - ab\right) \right|}{b^2\cos^2\theta + a^2\sin^2\theta} \\ &= \frac{\left| a^2b^2 - b^2(a^2 - b^2)\cos^2\theta \right|}{b^2\cos^2\theta + a^2\sin^2\theta} \end{aligned} \]

Factor out \(b^2\) in the numerator:

\[ \begin{aligned} d_1 d_2 &= \frac{b^2\left| a^2 - (a^2 - b^2)\cos^2\theta \right|}{b^2\cos^2\theta + a^2\sin^2\theta} \end{aligned} \]

Simplify the expression inside the modulus:

\[ \begin{aligned} a^2 - (a^2 - b^2)\cos^2\theta &= a^2(1 - \cos^2\theta) + b^2\cos^2\theta \\ &= a^2\sin^2\theta + b^2\cos^2\theta \end{aligned} \]

Substituting back:

\[ \begin{aligned} d_1 d_2 &= \frac{b^2\left(a^2\sin^2\theta + b^2\cos^2\theta\right)}{a^2\sin^2\theta + b^2\cos^2\theta} \\ &= b^2 \end{aligned} \]

Hence, the product of the lengths of the perpendiculars from the given points to the line is

\[ \boxed{b^2} \]

Q23. A person standing at the junction (crossing) of two straight paths represented by the equations \(2x – 3y + 4 = 0\) and \(3x + 4y – 5 = 0 \) wants to reach the path whose equation is \(6x – 7y + 8 = 0 \) in the least time. Find equation of the path that he should follow.

Solution

The person is standing at the junction of the two paths given by

\[ 2x - 3y + 4 = 0 \quad \text{and} \quad 3x + 4y - 5 = 0 \]

The junction point is the point of intersection of these two lines. Solving simultaneously:

\[ \begin{aligned} 2x - 3y + 4 &= 0 \\ 3x + 4y - 5 &= 0 \end{aligned} \]

\[ \begin{aligned} 2x - 3y &= -4 \\ 3x + 4y &= 5 \end{aligned} \]

\[ \begin{aligned} 8x - 12y &= -16 \\ 9x + 12y &= 15 \end{aligned} \]

\[ \begin{aligned} 17x &= -1 \\ x &= -\frac{1}{17} \end{aligned} \]

\[ \begin{aligned} 2\left(-\frac{1}{17}\right) - 3y &= -4 \\ - \frac{2}{17} - 3y &= -4 \\ -3y &= -\frac{66}{17} \\ y &= \frac{22}{17} \end{aligned} \]

Thus, the junction point is \(P\left(-\frac{1}{17},\, \frac{22}{17}\right)\).

To reach the path \(6x - 7y + 8 = 0\) in the least time, the person must move along the perpendicular from \(P\) to this line, since the shortest distance between a point and a line is along the perpendicular.

The slope of the line \(6x - 7y + 8 = 0\) is

\[ \begin{aligned} -7y &= -6x - 8 \\ y &= \frac{6}{7}x + \frac{8}{7} \end{aligned} \]

So the slope is \(\frac{6}{7}\), and the slope of the required perpendicular path is

\[ -\frac{7}{6} \]

Equation of the required path passing through \(P\left(-\frac{1}{17}, \frac{22}{17}\right)\) is

\[ \begin{aligned} y - \frac{22}{17} &= -\frac{7}{6}\left(x + \frac{1}{17}\right) \end{aligned} \]

Multiplying throughout by \(102\) to clear fractions:

\[ \begin{aligned} 102y - 132 &= -119x - 7 \\ 119x + 102y - 125 &= 0 \end{aligned} \]

Therefore, the equation of the path that the person should follow is

\[ \boxed{119x + 102y - 125 = 0} \]