Q1. Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts
(i) \(x + 7y = 0\)
(ii) \(6x + 3y – 5 = 0\)
(iii) \(y = 0\).

Solution

For the first equation, start with \(x+7y=0\). To express it in slope–intercept form \(y=mx+c\), isolate \(y\) by shifting the term involving \(x\) to the right side and then dividing by \(7\). \[ \begin{aligned} x+7y=0\\ 7y=-x\\ y=-\frac{1}{7}x+0\\ m=-\frac{1}{7}\\ c=0 \end{aligned} \] This shows that the slope is \(-\frac{1}{7}\) and the line cuts the \(y\)-axis at the origin.

For the second equation, consider \(6x+3y-5=0\). Rearranging helps in identifying the slope and intercept clearly. Move the constant term and the \(x\)-term to the right, then divide throughout by \(3\) to obtain the required form. \[ \begin{aligned} 6x+3y-5=0\\ 3y=-6x+5\\ y=-\frac{6}{3}x+\frac{5}{3}\\ y=-2x+\frac{5}{3}\\ m=-2\\ c=\frac{5}{3} \end{aligned} \] Hence, the slope of the line is \(-2\) and the \(y\)-intercept is \(\frac{5}{3}\).

For the third equation, \(y=0\) already represents a horizontal line. Writing it explicitly in slope–intercept form makes the parameters evident. \[ \begin{aligned} y=0\cdot x+0\\ m=0\\ c=0 \end{aligned} \] Thus, the slope is \(0\), indicating a horizontal line, and the \(y\)-intercept is also \(0\).

Q2. Reduce the following equations into intercept form and find their intercepts on the axes.
(i) \(3x + 2y – 12 = 0\),
(ii) \(4x – 3y = 6\),
(iii) \(3y + 2 = 0\).

Solution

For the first equation, start with \(3x+2y-12=0\). To write it in intercept form \(\frac{x}{a}+\frac{y}{b}=1\), move the constant term to the right side and then divide each term by the constant so that the right side becomes \(1\). \[ \begin{aligned} 3x+2y-12=0\\ 3x+2y=12\\ \frac{3x}{12}+\frac{2y}{12}=1\\ \frac{x}{4}+\frac{y}{6}=1\\ a=4\\ b=6 \end{aligned} \] Thus, the intercept on the \(x\)-axis is \(4\) and the intercept on the \(y\)-axis is \(6\).

For the second equation, consider \(4x-3y=6\). Dividing every term by \(6\) helps in directly comparing it with the standard intercept form. \[ \begin{aligned} 4x-3y=6\\ \frac{4x}{6}-\frac{3y}{6}=1\\ \frac{2}{3}x-\frac{1}{2}y=1\\ \frac{x}{3/2}+\frac{y}{-2}=1\\ a=\frac{3}{2}\\ b=-2 \end{aligned} \] Hence, the line cuts the \(x\)-axis at \(\frac{3}{2}\) and the \(y\)-axis at \(-2\).

For the third equation, take \(3y+2=0\). First isolate \(y\) to understand the nature of the line and its intercepts. \[ \begin{aligned} 3y+2=0\\ 3y=-2\\ y=-\frac{2}{3} \end{aligned} \] This represents a horizontal line parallel to the \(x\)-axis. Such a line has no finite \(x\)-intercept, while its \(y\)-intercept is obtained by putting \(x=0\), giving \(b=-\frac{2}{3}\).

Q3. Find the distance of the point (–1, 1) from the line \(12(x + 6) = 5(y – 2)\).

Solution

Begin by rewriting the given line in the standard form \(Ax+By+C=0\), since this form is required for applying the distance formula. Starting from \(12(x+6)=5(y-2)\), expand both sides and rearrange all terms to one side. \[ \begin{aligned} 12(x+6)=5(y-2)\\ 12x+72=5y-10\\ 12x-5y+82=0 \end{aligned} \] Here, the coefficients are clearly identified as \(A=12\), \(B=-5\), and \(C=82\).

The distance \(d\) of a point \((x_1,y_1)\) from the line \(Ax+By+C=0\) is given by the formula \[ d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^{2}+B^{2}}} \] Substituting the coordinates of the point \((-1,1)\) and the values of \(A\), \(B\), and \(C\), we obtain \[ \begin{aligned} d=\frac{\left|12(-1)+(-5)(1)+82\right|}{\sqrt{12^{2}+5^{2}}}\\ =\frac{\left|-12-5+82\right|}{\sqrt{144+25}}\\ =\frac{65}{\sqrt{169}}\\ =\frac{65}{13}\\ =5 \end{aligned} \]

Therefore, the distance of the point \((-1,1)\) from the given line is \(5\) units.

Q4. Find the points on the x-axis, whose distances from the line \(\frac{x}{3}+\frac{y}{4}=1\) are 4 units.

Solution

Let the required point on the \(x\)-axis be \((a,0)\), since any point on the \(x\)-axis has ordinate equal to zero. First convert the given line into standard form so that the distance formula can be applied. \[ \begin{aligned} \frac{x}{3}+\frac{y}{4}=1\\ 4x+3y-12=0 \end{aligned} \]

The distance of the point \((a,0)\) from the line \(4x+3y-12=0\) is given as \(4\) units. Using the distance formula \(d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^{2}+B^{2}}}\), substitute \(A=4\), \(B=3\), \(C=-12\), \(x_1=a\), and \(y_1=0\). \[ \begin{aligned} 4=\frac{\left|4a+3\cdot 0-12\right|}{\sqrt{4^{2}+3^{2}}}\\ 4=\frac{\left|4a-12\right|}{5} \end{aligned} \]

Removing the modulus gives two possible cases, since distance is always positive. \[ \begin{aligned} 4a-12=20\\ 4a=32\\ a=8 \end{aligned} \] and \[ \begin{aligned} 4a-12=-20\\ 4a=-8\\ a=-2 \end{aligned} \]

Hence, the required points on the \(x\)-axis whose distances from the given line are \(4\) units are \((8,0)\) and \((-2,0)\).

Q5. Find the distance between parallel lines
(i) \(15x + 8y – 34 = 0\) and \(15x + 8y + 31 = 0\)
(ii) \(l (x + y) + p = 0 and l (x + y) – r = 0\).

Solution

For two parallel lines written in the form \(ax+by+c_1=0\) and \(ax+by+c_2=0\), the distance between them is obtained using the formula \[ d=\frac{\left|c_2-c_1\right|}{\sqrt{a^{2}+b^{2}}} \] This formula applies because both lines have the same coefficients of \(x\) and \(y\), ensuring they are parallel.

In the first case, the given lines are \(15x+8y-34=0\) and \(15x+8y+31=0\). Here, \(a=15\), \(b=8\), \(c_1=-34\), and \(c_2=31\). Substituting these values into the formula gives \[ \begin{aligned} d=\frac{\left|31-(-34)\right|}{\sqrt{15^{2}+8^{2}}}\\ =\frac{\left|65\right|}{\sqrt{225+64}}\\ =\frac{65}{\sqrt{289}}\\ =\frac{65}{17} \end{aligned} \]

In the second case, the given parallel lines are \(l(x+y)+p=0\) and \(l(x+y)-r=0\). Writing them in standard form gives coefficients \(a=l\), \(b=l\), with constants \(c_1=p\) and \(c_2=-r\). Applying the same distance formula, \[ \begin{aligned} d=\frac{\left|-r-p\right|}{\sqrt{l^{2}+l^{2}}}\\ =\frac{\left|p+r\right|}{\sqrt{2l^{2}}}\\ =\frac{\left|p+r\right|}{l\sqrt{2}} \end{aligned} \]

Thus, the distance between the given pairs of parallel lines is \(\frac{65}{17}\) units in the first case and \(\frac{\left|p+r\right|}{l\sqrt{2}}\) units in the second case.

Q6. Find equation of the line parallel to the line \(3x - 4 + 2 = 0\) and passing through the point (–2, 3).

Solution

First rewrite the given line in a form that clearly shows its slope, since parallel lines must have the same slope. The given line is \(3x-4y+2=0\). Rearranging this equation helps in identifying the slope directly. \[ \begin{aligned} 3x-4y+2=0\\ 4y=3x+2\\ y=\frac{3}{4}x+\frac{1}{2}\\ m=\frac{3}{4} \end{aligned} \] Thus, the slope of the required line is \(\frac{3}{4}\).

The required line passes through the point \((-2,3)\). Using the point–slope form of the equation of a line, substitute the slope \(m=\frac{3}{4}\) and the given point \((x_0,y_0)=(-2,3)\). \[ \begin{aligned} y-y_0=m(x-x_0)\\ y-3=\frac{3}{4}(x+2) \end{aligned} \]

Now simplify the equation to bring it into standard form. \[ \begin{aligned} 4y-12=3x+6\\ 4y=3x+18\\ 3x-4y+18=0 \end{aligned} \]

Hence, the equation of the line parallel to the given line and passing through the point \((-2,3)\) is \(3x-4y+18=0\).

Q7. Find equation of the line perpendicular to the line \(x – 7y + 5 = 0\) and having \(x\) intercept 3.

Solution

First rewrite the given line in slope–intercept form in order to identify its slope. Starting with \(x-7y+5=0\), rearrange the terms so that \(y\) is isolated. \[ \begin{aligned} x-7y+5=0\\ -7y=-x-5\\ y=\frac{1}{7}x+\frac{5}{7}\\ m_1=\frac{1}{7} \end{aligned} \] Thus, the slope of the given line is \(\frac{1}{7}\).

If two lines are perpendicular, the product of their slopes is \(-1\). Let \(m_2\) be the slope of the required line. Then \[ \begin{aligned} m_1\cdot m_2=-1\\ \frac{1}{7}\cdot m_2=-1\\ m_2=-7 \end{aligned} \] Hence, the slope of the line perpendicular to the given line is \(-7\).

The required line has an \(x\)-intercept equal to \(3\), so it passes through the point \((3,0)\). Using the point–slope form of the equation of a line with slope \(-7\), \[ \begin{aligned} y-y_0=m(x-x_0)\\ y-0=-7(x-3)\\ y=-7x+21\\ y+7x-21=0 \end{aligned} \]

Therefore, the equation of the line perpendicular to the given line and having \(x\)-intercept \(3\) is \(y+7x-21=0\).

Q8. Find angles between the lines \(\sqrt{3}x+y=1\) and \(x + \sqrt{3} y = 1\).

Solution

To find the angle between the given lines, first express each equation in slope–intercept form so that their slopes can be identified clearly. \[ \begin{aligned} \sqrt{3}x+y=1\\ y=-\sqrt{3}x+1 \end{aligned} \] From this equation, the slope of the first line is \(m_1=-\sqrt{3}\).

Now rewrite the second equation in the same form. \[ \begin{aligned} x+\sqrt{3}y=1\\ \sqrt{3}y=1-x\\ y=-\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}} \end{aligned} \] Hence, the slope of the second line is \(m_2=-\frac{1}{\sqrt{3}}\).

The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| \] Substituting the values of \(m_1\) and \(m_2\), \[ \begin{aligned} \tan\theta&=\left|\frac{-\sqrt{3}-\left(-\frac{1}{\sqrt{3}}\right)}{1+\left(-\sqrt{3}\right)\left(-\frac{1}{\sqrt{3}}\right)}\right|\\ &=\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+1}\right|\\ &=\left|\frac{-\frac{2}{\sqrt{3}}}{2}\right|\\ &=\frac{1}{\sqrt{3}} \end{aligned} \]

Since \(\tan\theta=\frac{1}{\sqrt{3}}\), it follows that \(\theta=30^\circ\). Therefore, the angle between the given two lines is \(30^\circ\).

Q9. The line through the points (h, 3) and (4, 1) intersects the line \(7x-9y-19=0\) at right angle. Find the value of h.

Solution

First find the slope of the line passing through the points \((h,3)\) and \((4,1)\). The slope of a line through two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(\frac{y_2-y_1}{x_2-x_1}\). Substituting the given coordinates, \[ \begin{aligned} m_1=\frac{1-3}{4-h}\\ =\frac{-2}{4-h} \end{aligned} \]

Next, determine the slope of the line \(7x-9y-19=0\). Rewrite the equation in slope–intercept form by isolating \(y\). \[ \begin{aligned} 7x-9y-19=0\\ -9y=-7x+19\\ y=\frac{7}{9}x-\frac{19}{9}\\ m_2=\frac{7}{9} \end{aligned} \]

Since the two lines intersect at a right angle, their slopes satisfy the condition \(m_1m_2=-1\). Substituting the values of \(m_1\) and \(m_2\), \[ \begin{aligned} \frac{-2}{4-h}\cdot\frac{7}{9}=-1\\ \frac{-14}{9(4-h)}=-1\\ -14=-9(4-h)\\ -14=-36+9h\\ 9h=22\\ h=\frac{22}{9} \end{aligned} \]

Therefore, the value of \(h\) for which the given lines intersect at right angles is \(\frac{22}{9}\).

Q10. Prove that the line through the point \((x_1, y_1)\) and parallel to the line Ax + By + C = 0 is \(A (x –x_1) + B (y – y_1) = 0\).

Solution

Consider the given line \(Ax+By+C=0\). To identify its slope, rewrite it in slope–intercept form by isolating \(y\). This helps because any line parallel to it must have the same slope. \[ \begin{aligned} Ax+By+C=0\\ By=-Ax-C\\ y=-\frac{A}{B}x-\frac{C}{B}\\ m=-\frac{A}{B} \end{aligned} \] Thus, the slope of the given line is \(-\frac{A}{B}\).

Let the required line pass through the point \((x_1,y_1)\) and be parallel to the given line. Since parallel lines have equal slopes, the slope of the required line is also \(-\frac{A}{B}\). Using the point–slope form of the equation of a line, \[ \begin{aligned} y-y_1=m(x-x_1)\\ y-y_1=-\frac{A}{B}(x-x_1) \end{aligned} \]

To express the equation in a symmetric and standard form, multiply both sides by \(B\) and rearrange the terms. \[ \begin{aligned} B(y-y_1)=-A(x-x_1)\\ A(x-x_1)+B(y-y_1)=0 \end{aligned} \]

Hence, the equation of the line passing through \((x_1,y_1)\) and parallel to \(Ax+By+C=0\) is \(A(x-x_1)+B(y-y_1)=0\), which proves the required result.

Q11. Two lines passing through the point (2, 3) intersects each other at an angle of 60o. If slope of one line is 2, find equation of the other line.

Solution

The given point of intersection of the two lines is \((2,3)\). The slope of one line is given as \(m_1=2\). Let the slope of the other line be \(m\). Since the angle between the two lines is \(60^\circ\), the relation between their slopes is given by \[ \tan 60^\circ=\left|\frac{m_1-m}{1+m_1m}\right| \] Substituting \(m_1=2\), \[ \begin{aligned} \sqrt{3}=\frac{2-m}{1+2m} \end{aligned} \]

Now solve this equation to find the value of \(m\). \[ \begin{aligned} \sqrt{3}(1+2m)=2-m\\ \sqrt{3}+2\sqrt{3}m=2-m\\ 2\sqrt{3}m+m=2-\sqrt{3}\\ m(1+2\sqrt{3})=2-\sqrt{3}\\ m=\frac{2-\sqrt{3}}{1+2\sqrt{3}} \end{aligned} \]

The required line passes through the point \((2,3)\) and has slope \(m=\frac{2-\sqrt{3}}{1+2\sqrt{3}}\). Using the point–slope form of the equation of a line, \[ \begin{aligned} y-y_0=m(x-x_0)\\ y-3=\frac{2-\sqrt{3}}{1+2\sqrt{3}}(x-2) \end{aligned} \]

To express the equation in a simplified linear form, multiply both sides by \((1+2\sqrt{3})\) and rearrange the terms. \[ \begin{aligned} (1+2\sqrt{3})y-(3+6\sqrt{3})=(2-\sqrt{3})x-4+2\sqrt{3}\\ (1+2\sqrt{3})y-(2-\sqrt{3})x=8\sqrt{3}-1\\ (1+2\sqrt{3})y-(2-\sqrt{3})x-(8\sqrt{3}-1)=0 \end{aligned} \]

Hence, the equation of the other line passing through \((2,3)\) and making an angle of \(60^\circ\) with the given line is \((1+2\sqrt{3})y-(2-\sqrt{3})x-(8\sqrt{3}-1)=0\).

Q12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Solution

First determine the midpoint of the line segment joining the points \((3,4)\) and \((-1,2)\), since the right bisector must pass through the midpoint. Using the midpoint formula, \[ \begin{aligned} x_1=\frac{3+(-1)}{2}=1\\ y_1=\frac{4+2}{2}=3 \end{aligned} \] Thus, the midpoint of the given line segment is \((1,3)\).

Next, find the slope of the line joining the given points. The slope \(m_1\) is calculated using the formula \(\frac{y_2-y_1}{x_2-x_1}\). \[ \begin{aligned} m_1=\frac{2-4}{-1-3}\\ =\frac{-2}{-4}\\ =\frac{1}{2} \end{aligned} \]

The slope of the perpendicular bisector, say \(m_2\), is the negative reciprocal of \(m_1\) because the bisector is perpendicular to the given line. \[ \begin{aligned} m_1m_2=-1\\ m_2=-\frac{1}{\frac{1}{2}}\\ =-2 \end{aligned} \]

Now write the equation of the perpendicular bisector using the point–slope form, since it passes through the midpoint \((1,3)\) and has slope \(-2\). \[ \begin{aligned} y-y_1=m(x-x_1)\\ y-3=-2(x-1)\\ y-3=-2x+2\\ y+2x-5=0 \end{aligned} \]

Hence, the equation of the right bisector of the given line segment is \(y+2x-5=0\).

Q13. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line \(3x – 4y – 16 = 0\).

Solution

First determine the slope of the given line \(3x-4y-16=0\) by rewriting it in slope–intercept form. This is necessary to obtain the slope of the perpendicular line. \[ \begin{aligned} 3x-4y-16=0\\ -4y=-3x+16\\ y=\frac{3}{4}x-4\\ m_1=\frac{3}{4} \end{aligned} \] Hence, the slope of the given line is \(\frac{3}{4}\).

The slope of the perpendicular drawn from the point \((-1,3)\) is the negative reciprocal of \(\frac{3}{4}\), that is \(-\frac{4}{3}\). Using the point–slope form for the perpendicular line through \((-1,3)\), \[ \begin{aligned} m_2=-\frac{4}{3}\\ y-y_0=m_2(x-x_0)\\ y-3=-\frac{4}{3}(x+1) \end{aligned} \] Simplifying, \[ \begin{aligned} 3y-9=-4x-4\\ 4x+3y-5=0 \end{aligned} \]

The foot of the perpendicular is the point of intersection of the lines \(3x-4y-16=0\) and \(4x+3y-5=0\). Solving these two equations simultaneously, \[ \begin{aligned} 3x-4y-16=0\\ 4x+3y-5=0 \end{aligned} \] Multiply the first equation by \(3\) and the second by \(4\) to eliminate \(y\), \[ \begin{aligned} 9x-12y-48=0\\ 16x+12y-20=0 \end{aligned} \] Adding, \[ \begin{aligned} 25x-68=0\\ x=\frac{68}{25} \end{aligned} \]

Substitute \(x=\frac{68}{25}\) into \(4x+3y-5=0\), \[ \begin{aligned} 4\left(\frac{68}{25}\right)+3y-5=0\\ \frac{272}{25}+3y=\frac{125}{25}\\ 3y=-\frac{147}{25}\\ y=-\frac{49}{25} \end{aligned} \]

Therefore, the coordinates of the foot of the perpendicular from the point \((-1,3)\) to the given line are \(\left(\frac{68}{25},-\frac{49}{25}\right)\).

Q14. The perpendicular from the origin to the line \(y = mx + c\) meets it at the point (–1, 2). Find the values of \(m\) and \(c\).

Solution

The perpendicular from the origin to the given line meets it at the point \((-1,2)\). Hence, the slope of the perpendicular line is obtained using the slope formula between the points \((0,0)\) and \((-1,2)\). \[ \begin{aligned} m_2=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{2-0}{-1-0}\\ =-2 \end{aligned} \] Thus, the slope of the perpendicular from the origin is \(-2\).

If two lines are perpendicular, the product of their slopes is \(-1\). Let \(m\) be the slope of the given line \(y=mx+c\). Then, \[ \begin{aligned} m\cdot m_2=-1\\ m(-2)=-1\\ m=\frac{1}{2} \end{aligned} \] So, the equation of the line becomes \(y=\frac{1}{2}x+c\).

Since the line passes through the point \((-1,2)\), substitute these coordinates to find the value of \(c\). \[ \begin{aligned} 2=\frac{1}{2}(-1)+c\\ 2+\frac{1}{2}=c\\ c=\frac{5}{2} \end{aligned} \]

Hence, the values of the constants are \(m=\frac{1}{2}\) and \(c=\frac{5}{2}\).

Q15. If \(p\) and \(q\) are the lengths of perpendiculars from the origin to the lines \(x \cos \theta -y \sin \theta=k \cos 2\theta\) and \(x sec θ + y \;\text{cosec} θ = k\), respectively, prove that \(p^2 + 4q^2 = k^2\).

Solution

We are given the lines \[ x \cos \theta - y \sin \theta = k \cos 2\theta \quad \text{and} \quad x \sec \theta + y \;\text{cosec} \theta = k \] and \(p, q\) denote the lengths of perpendiculars from the origin to these lines respectively. We will compute each distance using the standard perpendicular distance formula.

First, rewrite the first line in general form: \[ x \cos \theta - y \sin \theta - k \cos 2\theta = 0 \] The perpendicular distance from the origin to a line \(ax + by + c = 0\) is given by \[ \frac{|c|}{\sqrt{a^2 + b^2}} \] Thus, \[ \begin{aligned} p &= \frac{| -k \cos 2\theta |}{\sqrt{\cos^2 \theta + \sin^2 \theta}} \\ &= |k \cos 2\theta| \end{aligned} \] Therefore, \[ p^2 = k^2 \cos^2 2\theta \]

Now rewrite the second line in general form: \[ x \sec \theta + y \;\text{cosec} \theta - k = 0 \] Using the same distance formula, \[ \begin{aligned} q &= \frac{| -k |}{\sqrt{\sec^2 \theta + \text{cosec}^2 \theta}} \end{aligned} \] Compute the denominator: \[ \begin{aligned} \sec^2 \theta + \text{cosec}^2 \theta &= \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} \\ &= \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \\ &= \frac{1}{\sin^2 \theta \cos^2 \theta} \end{aligned} \] Hence, \[ \begin{aligned} q &= |k| \sin \theta \cos \theta \\ &= \frac{|k|}{2} \sin 2\theta, \end{aligned} \] so \[ q^2 = \frac{k^2}{4} \sin^2 2\theta. \]

Finally, compute \(p^2 + 4q^2\): \[ \begin{aligned} p^2 + 4q^2 &= k^2 \cos^2 2\theta + 4 \cdot \frac{k^2}{4} \sin^2 2\theta \\ &= k^2 \left( \cos^2 2\theta + \sin^2 2\theta \right) \\ &= k^2 \end{aligned} \] Thus, we have proved that \[ \boxed{p^2 + 4q^2 = k^2} \]

Q16. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

Solution

Given the vertices of triangle \(A(2,3)\), \(B(4,-1)\), and \(C(1,2)\), we are required to find the equation and length of the altitude from vertex \(A\) to side \(BC\).

First, determine the slope of line \(BC\): \[ \begin{aligned} m_{BC} &= \frac{2 - (-1)}{1 - 4} \\ &= \frac{3}{-3} \\ &= -1 \end{aligned} \]

Since the altitude from \(A\) is perpendicular to \(BC\), its slope will be the negative reciprocal of \(-1\), which is \(1\). Thus, the equation of the altitude passing through \(A(2,3)\) is: \[ \begin{aligned} y - 3 &= 1(x - 2) \\ y &= x + 1 \end{aligned} \] So, the equation of the altitude from \(A\) is: \[ \boxed{y = x + 1} \]

Now, find the equation of line \(BC\). Using point \(B(4,-1)\) and slope \(-1\): \[ \begin{aligned} y + 1 &= -1(x - 4) \\ y &= -x + 3 \end{aligned} \] Rewriting in standard form: \[ x + y - 3 = 0 \]

The length of the altitude is the perpendicular distance from point \(A(2,3)\) to line \(BC\). Using the distance formula from a point to a line: \[ \begin{aligned} \text{Distance} &= \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \\ &= \frac{|2 + 3 - 3|}{\sqrt{1^2 + 1^2}} \\ &= \frac{2}{\sqrt{2}} \\ &= \sqrt{2} \end{aligned} \]

Therefore, the equation of the altitude from vertex \(A\) is \[ \boxed{y = x + 1} \] and its length is \[ \boxed{\sqrt{2}} \]

Q17. If \(p\) is the length of perpendicular from the origin to the line whose intercepts on the axes are \(a\) and \(b\), then show that \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\)

Solution

Let a line cut the \(x\)-axis at \(a\) and the \(y\)-axis at \(b\). The equation of the line in intercept form is \[ \frac{x}{a} + \frac{y}{b} = 1 \] Rewriting it in general form, we obtain \[ \begin{aligned} \frac{x}{a} + \frac{y}{b} - 1 &= 0 \\ bx + ay - ab &= 0 \end{aligned} \]

The perpendicular distance from the origin \((0,0)\) to a line of the form \(Ax + By + C = 0\) is given by \[ p = \frac{|C|}{\sqrt{A^2 + B^2}} \] Here, \(A = b\), \(B = a\), and \(C = -ab\). Substituting these values, we get \[ \begin{aligned} p &= \frac{| -ab |}{\sqrt{a^2 + b^2}} \\ &= \frac{ab}{\sqrt{a^2 + b^2}} \end{aligned} \]

Now square both sides: \[ \begin{aligned} p^2 &= \frac{a^2 b^2}{a^2 + b^2}. \end{aligned} \] Taking the reciprocal, we obtain \[ \begin{aligned} \frac{1}{p^2} &= \frac{a^2 + b^2}{a^2 b^2} \\ &= \frac{1}{a^2} + \frac{1}{b^2} \end{aligned} \]

Hence, it is proved that \[ \boxed{\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}} \]