\displaystyle \textbf{Question 1: }\text{Find the distance of the point }(4,5)\text{ from the straight line }3x-5y+7=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given line }3x-5y+7=0.
\displaystyle \text{Comparing with }ax+by+c=0,\text{ we get }a=3,\ b=-5,\ c=7.
\displaystyle \text{Given point }P(4,5).
\displaystyle \text{Distance of }P(x_1,y_1)\text{ from the line }ax+by+c=0\text{ is}
\displaystyle d=\left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|.
\displaystyle \therefore d=\left|\frac{3(4)+(-5)(5)+7}{\sqrt{3^2+(-5)^2}}\right|
\displaystyle =\left|\frac{12-25+7}{\sqrt{34}}\right|
\displaystyle =\frac{6}{\sqrt{34}}.
\displaystyle \therefore \text{The required distance is }\frac{6}{\sqrt{34}}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the perpendicular distance from the origin to the line joining the points}
\displaystyle (\cos\theta,\sin\theta)\text{ and }(\cos\phi,\sin\phi).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(\cos\theta,\sin\theta)\text{ and }Q(\cos\phi,\sin\phi)\text{ be two distinct points.}
\displaystyle \text{The equation of the line joining }P\text{ and }Q\text{ is}
\displaystyle y-\sin\theta=\frac{\sin\phi-\sin\theta}{\cos\phi-\cos\theta}(x-\cos\theta).
\displaystyle (\cos\phi-\cos\theta)(y-\sin\theta)=(\sin\phi-\sin\theta)(x-\cos\theta).
\displaystyle (\sin\phi-\sin\theta)x-(\cos\phi-\cos\theta)y
\displaystyle \qquad+\sin\theta\cos\phi-\cos\theta\sin\phi=0.
\displaystyle \text{Therefore, the perpendicular distance }d\text{ from the origin is}
\displaystyle d=\left|\frac{\sin\theta\cos\phi-\cos\theta\sin\phi}{\sqrt{(\sin\phi-\sin\theta)^2+(\cos\phi-\cos\theta)^2}}\right|.
\displaystyle =\left|\frac{\sin(\theta-\phi)}{\sqrt{2-2\cos(\theta-\phi)}}\right|.
\displaystyle =\left|\frac{\sin(\theta-\phi)}{\sqrt{4\sin^2\left(\frac{\theta-\phi}{2}\right)}}\right|.
\displaystyle =\left|\frac{2\sin\left(\frac{\theta-\phi}{2}\right)\cos\left(\frac{\theta-\phi}{2}\right)}{2\left|\sin\left(\frac{\theta-\phi}{2}\right)\right|}\right|.
\displaystyle =\left|\cos\left(\frac{\theta-\phi}{2}\right)\right|.
\displaystyle \therefore \text{The required perpendicular distance is }\left|\cos\left(\frac{\theta-\phi}{2}\right)\right|.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the length of the perpendicular from the origin to the straight line}
\displaystyle \text{joining the points }(a\cos\alpha,a\sin\alpha)\text{ and }(a\cos\beta,a\sin\beta).
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the line joining }(a\cos\alpha,a\sin\alpha)\text{ and }(a\cos\beta,a\sin\beta)\text{ is}
\displaystyle y-a\sin\alpha=\frac{a\sin\beta-a\sin\alpha}{a\cos\beta-a\cos\alpha}(x-a\cos\alpha).
\displaystyle \text{Using }\sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2}\text{ and}
\displaystyle \cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2},
\displaystyle y-a\sin\alpha=-\cot\left(\frac{\alpha+\beta}{2}\right)(x-a\cos\alpha).
\displaystyle \therefore x\cot\left(\frac{\alpha+\beta}{2}\right)+y-a\sin\alpha
\displaystyle \qquad-a\cos\alpha\cot\left(\frac{\alpha+\beta}{2}\right)=0.
\displaystyle \text{Comparing with }Ax+By+C=0,\text{ we get}
\displaystyle A=\cot\left(\frac{\alpha+\beta}{2}\right),\quad B=1,
\displaystyle C=-a\sin\alpha-a\cos\alpha\cot\left(\frac{\alpha+\beta}{2}\right).
\displaystyle \text{Hence the perpendicular distance from the origin is}
\displaystyle d=\left|\frac{C}{\sqrt{A^2+B^2}}\right|.
\displaystyle =\left|\frac{-a\sin\alpha-a\cos\alpha\cot\left(\frac{\alpha+\beta}{2}\right)}{\sqrt{\cot^2\left(\frac{\alpha+\beta}{2}\right)+1}}\right|.
\displaystyle =a\left|\sin\left(\frac{\alpha+\beta}{2}\right)\sin\alpha+\cos\left(\frac{\alpha+\beta}{2}\right)\cos\alpha\right|.
\displaystyle =a\left|\cos\left(\frac{\alpha+\beta}{2}-\alpha\right)\right|.
\displaystyle =a\left|\cos\left(\frac{\beta-\alpha}{2}\right)\right|.
\displaystyle =a\left|\cos\left(\frac{\alpha-\beta}{2}\right)\right|.
\displaystyle \therefore \text{The required perpendicular distance is }a\left|\cos\left(\frac{\alpha-\beta}{2}\right)\right|.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Show that the perpendiculars let fall from any point on the straight line}
\displaystyle 2x+11y-5=0\text{ upon the lines }24x+7y=20\text{ and }4x-3y-2=0
\displaystyle \text{are equal to each other.}
\displaystyle \text{Answer:}
\displaystyle \text{Given lines:}
\displaystyle 24x+7y-20=0\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle 4x-3y-2=0\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Let }P(a,b)\text{ be any point on the line }2x+11y-5=0.
\displaystyle \text{The perpendicular distance }d_1\text{ of }P(a,b)\text{ from line (i) is}
\displaystyle d_1=\left|\frac{24a+7b-20}{\sqrt{24^2+7^2}}\right|
\displaystyle =\left|\frac{24a+7b-20}{25}\right|.\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \text{The perpendicular distance }d_2\text{ of }P(a,b)\text{ from line (ii) is}
\displaystyle d_2=\left|\frac{4a-3b-2}{\sqrt{4^2+(-3)^2}}\right|
\displaystyle =\left|\frac{4a-3b-2}{5}\right|.\qquad\ldots\ldots\ldots\text{(iv)}
\displaystyle \text{Since }P(a,b)\text{ lies on }2x+11y-5=0,
\displaystyle 2a+11b-5=0
\displaystyle \therefore b=\frac{5-2a}{11}.
\displaystyle \text{Substituting }b=\frac{5-2a}{11}\text{ in (iii), we get}
\displaystyle d_1=\left|\frac{24a+7\left(\frac{5-2a}{11}\right)-20}{25}\right|
\displaystyle =\left|\frac{250a-185}{275}\right|
\displaystyle =\left|\frac{50a-37}{55}\right|.
\displaystyle \text{Substituting }b=\frac{5-2a}{11}\text{ in (iv), we get}
\displaystyle d_2=\left|\frac{4a-3\left(\frac{5-2a}{11}\right)-2}{5}\right|
\displaystyle =\left|\frac{50a-37}{55}\right|.
\displaystyle \therefore d_1=d_2.
\displaystyle \text{Hence, the perpendiculars drawn from any point on }2x+11y-5=0
\displaystyle \text{to the lines }24x+7y=20\text{ and }4x-3y-2=0\text{ are equal.}
\displaystyle \\

Q\displaystyle \textbf{Question 5: }\text{Find the distance of the point of intersection of the lines }2x+3y=21
\displaystyle \text{and }3x-4y+11=0\text{ from the line }8x+6y+5=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given lines:}
\displaystyle 2x+3y=21\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle 3x-4y+11=0\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Solving (i) and (ii), we obtain the point of intersection }(3,5).
\displaystyle \text{Comparing }8x+6y+5=0\text{ with }ax+by+c=0,\text{ we get }a=8,\ b=6,\ c=5.
\displaystyle \text{Distance of }(3,5)\text{ from the line is}
\displaystyle d=\left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|
\displaystyle =\left|\frac{8(3)+6(5)+5}{\sqrt{8^2+6^2}}\right|
\displaystyle =\frac{24+30+5}{\sqrt{64+36}}
\displaystyle =\frac{59}{10}.
\displaystyle \therefore \text{The required distance is }\frac{59}{10}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the length of the perpendicular from the point }(4,-7)\text{ to the line joining}
\displaystyle \text{the origin and the point of intersection of the lines }2x-3y+14=0\text{ and }5x+4y-7=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given lines:}
\displaystyle 2x-3y+14=0\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle 5x+4y-7=0\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Solving (i) and (ii), we obtain the point of intersection}
\displaystyle \left(-\frac{35}{23},\frac{84}{23}\right).
\displaystyle \text{The line joining }(0,0)\text{ and }\left(-\frac{35}{23},\frac{84}{23}\right)\text{ is}
\displaystyle y=\frac{\frac{84}{23}}{-\frac{35}{23}}x=-\frac{12}{5}x.
\displaystyle \therefore 12x+5y=0.
\displaystyle \text{Comparing with }ax+by+c=0,\text{ we get }a=12,\ b=5,\ c=0.
\displaystyle \text{Distance of }(4,-7)\text{ from the line }12x+5y=0\text{ is}
\displaystyle d=\left|\frac{12(4)+5(-7)+0}{\sqrt{12^2+5^2}}\right|
\displaystyle =\left|\frac{48-35}{\sqrt{169}}\right|
\displaystyle =\frac{13}{13}=1.
\displaystyle \therefore \text{The required perpendicular distance is }1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{What are the points on the }x\text{-axis whose perpendicular distance from the straight}
\displaystyle \text{line }\frac{x}{a}+\frac{y}{b}=1\text{ is }a?
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(t,0)\text{ be a point on the }x\text{-axis.}
\displaystyle \text{Given line: }\frac{x}{a}+\frac{y}{b}=1.
\displaystyle \text{Multiplying by }ab,\text{ we get}
\displaystyle bx+ay-ab=0.
\displaystyle \text{The perpendicular distance of }P(t,0)\text{ from this line is }a.
\displaystyle \therefore \left|\frac{bt+a(0)-ab}{\sqrt{b^2+a^2}}\right|=a.
\displaystyle \left|\frac{b(t-a)}{\sqrt{a^2+b^2}}\right|=a.
\displaystyle |b(t-a)|=a\sqrt{a^2+b^2}.
\displaystyle \text{Assuming }a>0\text{ and }b>0,
\displaystyle |t-a|=\frac{a}{b}\sqrt{a^2+b^2}.
\displaystyle t-a=\pm\frac{a}{b}\sqrt{a^2+b^2}.
\displaystyle \therefore t=a\pm\frac{a}{b}\sqrt{a^2+b^2}
\displaystyle =\frac{a}{b}\left(b\pm\sqrt{a^2+b^2}\right).
\displaystyle \therefore \text{The required points are}
\displaystyle \left(\frac{a}{b}\left(b+\sqrt{a^2+b^2}\right),0\right)
\displaystyle \text{and }\left(\frac{a}{b}\left(b-\sqrt{a^2+b^2}\right),0\right).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Show that the product of the perpendiculars drawn to the line}
\displaystyle \frac{x}{a}\cos\theta+\frac{y}{b}\sin\theta=1\text{ from the points }\left(\pm\sqrt{a^2-b^2},0\right)\text{ is }b^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\left(\sqrt{a^2-b^2},0\right)\text{ and }Q\left(-\sqrt{a^2-b^2},0\right).
\displaystyle \text{The given line is}
\displaystyle \frac{\cos\theta}{a}x+\frac{\sin\theta}{b}y-1=0.
\displaystyle \text{Let }d_1\text{ be the perpendicular distance of }P\text{ from the given line.}
\displaystyle d_1=\left|\frac{\frac{\cos\theta}{a}\sqrt{a^2-b^2}-1}{\sqrt{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}}}\right|.
\displaystyle =b\left|\frac{\sqrt{a^2-b^2}\cos\theta-a}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}\right|.
\displaystyle \text{Let }d_2\text{ be the perpendicular distance of }Q\text{ from the given line.}
\displaystyle d_2=\left|\frac{-\frac{\cos\theta}{a}\sqrt{a^2-b^2}-1}{\sqrt{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}}}\right|.
\displaystyle =b\left|\frac{\sqrt{a^2-b^2}\cos\theta+a}{\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}}\right|.
\displaystyle \therefore d_1d_2=b^2\left|\frac{\left(\sqrt{a^2-b^2}\cos\theta-a\right)\left(\sqrt{a^2-b^2}\cos\theta+a\right)}{b^2\cos^2\theta+a^2\sin^2\theta}\right|.
\displaystyle =b^2\left|\frac{(a^2-b^2)\cos^2\theta-a^2}{b^2\cos^2\theta+a^2\sin^2\theta}\right|.
\displaystyle =b^2\left|\frac{a^2(\cos^2\theta-1)-b^2\cos^2\theta}{b^2\cos^2\theta+a^2\sin^2\theta}\right|.
\displaystyle =b^2\left|\frac{-a^2\sin^2\theta-b^2\cos^2\theta}{b^2\cos^2\theta+a^2\sin^2\theta}\right|.
\displaystyle =b^2\left|-1\right|.
\displaystyle \therefore d_1d_2=b^2.
\displaystyle \text{Hence, the product of the perpendiculars is }b^2.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the perpendicular distance from the origin to the perpendicular drawn}
\displaystyle \text{from the point }(1,2)\text{ upon the straight line }x-\sqrt{3}y+4=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given line: }x-\sqrt{3}y+4=0.\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{A line perpendicular to (i) is } \sqrt{3}x+y+\lambda=0.
\displaystyle \text{Since it passes through }(1,2),
\displaystyle \sqrt{3}(1)+2+\lambda=0
\displaystyle \therefore \lambda=-\sqrt{3}-2.
\displaystyle \text{Hence the equation of the perpendicular is}
\displaystyle \sqrt{3}x+y-\sqrt{3}-2=0.\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{The perpendicular distance from the origin to (ii) is}
\displaystyle d=\left|\frac{\sqrt{3}(0)+1(0)-(\sqrt{3}+2)}{\sqrt{(\sqrt{3})^2+1^2}}\right|
\displaystyle =\left|\frac{-(\sqrt{3}+2)}{\sqrt{4}}\right|
\displaystyle =\frac{\sqrt{3}+2}{2}.
\displaystyle \therefore \text{The required perpendicular distance is }\frac{\sqrt{3}+2}{2}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the distance of the point }(1,2)\text{ from the straight line with slope }5
\displaystyle \text{passing through the point of intersection of }x+2y=5\text{ and }x-3y=7.
\displaystyle \text{Answer:}
\displaystyle \text{Given lines:}
\displaystyle x+2y=5\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle x-3y=7\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Solving (i) and (ii), we obtain the point of intersection}
\displaystyle \left(\frac{29}{5},-\frac{2}{5}\right).
\displaystyle \text{The required line passes through }\left(\frac{29}{5},-\frac{2}{5}\right)\text{ and has slope }5.
\displaystyle y+\frac{2}{5}=5\left(x-\frac{29}{5}\right).
\displaystyle \Rightarrow 5y+2=25x-145.
\displaystyle \Rightarrow 25x-5y-147=0.
\displaystyle \text{Comparing with }ax+by+c=0,\text{ we get }a=25,\ b=-5,\ c=-147.
\displaystyle \text{Distance of }(1,2)\text{ from the line is}
\displaystyle d=\left|\frac{25(1)-5(2)-147}{\sqrt{25^2+(-5)^2}}\right|.
\displaystyle =\frac{132}{\sqrt{650}}
\displaystyle =\frac{132}{5\sqrt{26}}.
\displaystyle \therefore \text{The required distance is }\frac{132}{5\sqrt{26}}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{What are the points on the }y\text{-axis whose distance from the line}
\displaystyle \frac{x}{3}+\frac{y}{4}=1\text{ is }4\text{ units?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(0,t)\text{ be a point on the }y\text{-axis.}
\displaystyle \text{Given line: }\frac{x}{3}+\frac{y}{4}=1.
\displaystyle \text{Multiplying by }12,\text{ we get}
\displaystyle 4x+3y-12=0.
\displaystyle \text{The perpendicular distance of }P(0,t)\text{ from the line is }4\text{ units.}
\displaystyle \therefore \left|\frac{4(0)+3t-12}{\sqrt{4^2+3^2}}\right|=4.
\displaystyle \left|\frac{3t-12}{5}\right|=4.
\displaystyle |3t-12|=20.
\displaystyle 3t-12=\pm20.
\displaystyle \therefore 3t=12\pm20.
\displaystyle t=\frac{12\pm20}{3}.
\displaystyle \therefore t=\frac{32}{3}\text{ or }t=-\frac{8}{3}.
\displaystyle \therefore \text{The required points are }\left(0,\frac{32}{3}\right)\text{ and }\left(0,-\frac{8}{3}\right).
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In }\triangle ABC\text{ with vertices }A(2,3),\ B(4,-1)\text{ and }C(1,2),\text{ find the}
\displaystyle \text{equation and the length of the altitude from vertex }A.
\displaystyle \text{Answer:}
\displaystyle \text{Given vertices }A(2,3),\ B(4,-1)\text{ and }C(1,2).
\displaystyle \text{The equation of }BC\text{ is}
\displaystyle y-(-1)=\frac{2-(-1)}{1-4}(x-4).
\displaystyle y+1=-x+4.
\displaystyle \therefore x+y-3=0.\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{A line perpendicular to }x+y-3=0\text{ is}
\displaystyle x-y+\lambda=0.\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Since the altitude passes through }A(2,3),
\displaystyle 2-3+\lambda=0.
\displaystyle \therefore \lambda=1.
\displaystyle \therefore \text{The equation of the altitude from }A\text{ is }x-y+1=0.
\displaystyle \text{The length of the altitude is the perpendicular distance of }A(2,3)\text{ from }BC.
\displaystyle d=\left|\frac{2+3-3}{\sqrt{1^2+1^2}}\right|
\displaystyle =\frac{2}{\sqrt{2}}=\sqrt{2}.
\displaystyle \therefore \text{The equation of the altitude is }x-y+1=0\text{ and its length is }\sqrt{2}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Show that the path of a moving point whose distances from the two lines}
\displaystyle 3x-2y=5\text{ and }3x+2y+7=0\text{ are equal consists of straight lines.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(h,k)\text{ be a moving point whose distances from the two given lines are equal.}
\displaystyle \text{The given lines are}
\displaystyle 3x-2y-5=0\qquad\text{and}\qquad 3x+2y+7=0.
\displaystyle \therefore \left|\frac{3h-2k-5}{\sqrt{3^2+(-2)^2}}\right|=\left|\frac{3h+2k+7}{\sqrt{3^2+2^2}}\right|.
\displaystyle \left|3h-2k-5\right|=\left|3h+2k+7\right|.
\displaystyle \therefore 3h-2k-5=\pm(3h+2k+7).
\displaystyle \text{Taking the positive sign,}
\displaystyle 3h-2k-5=3h+2k+7.
\displaystyle -4k=12.
\displaystyle \therefore k=-3.
\displaystyle \text{Taking the negative sign,}
\displaystyle 3h-2k-5=-3h-2k-7.
\displaystyle 6h=-2.
\displaystyle \therefore h=-\frac{1}{3}.
\displaystyle \text{Replacing }h\text{ and }k\text{ by }x\text{ and }y,\text{ respectively, the locus is}
\displaystyle y=-3\qquad\text{or}\qquad x=-\frac{1}{3}.
\displaystyle \therefore \text{The locus consists of the two straight lines }y=-3\text{ and }x=-\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If the sum of the perpendicular distances of a variable point }P(x,y)\text{ from}
\displaystyle \text{the lines }x+y-5=0\text{ and }3x-2y+7=0\text{ is always }10,\text{ show that its locus}
\displaystyle \text{consists of straight-line segments.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(h,k)\text{ be a variable point satisfying the given condition.}
\displaystyle \left|\frac{h+k-5}{\sqrt{1^2+1^2}}\right|+\left|\frac{3h-2k+7}{\sqrt{3^2+(-2)^2}}\right|=10.
\displaystyle \therefore \left|\frac{h+k-5}{\sqrt2}\right|+\left|\frac{3h-2k+7}{\sqrt{13}}\right|=10.
\displaystyle \text{There are four possible sign combinations.}
\displaystyle \text{Case I: }h+k-5\geq0\text{ and }3h-2k+7\geq0.
\displaystyle \frac{h+k-5}{\sqrt2}+\frac{3h-2k+7}{\sqrt{13}}=10.
\displaystyle (\sqrt{13}+3\sqrt2)h+(\sqrt{13}-2\sqrt2)k
\displaystyle \qquad+7\sqrt2-5\sqrt{13}-10\sqrt{26}=0.
\displaystyle \text{Case II: }h+k-5\geq0\text{ and }3h-2k+7\leq0.
\displaystyle \frac{h+k-5}{\sqrt2}-\frac{3h-2k+7}{\sqrt{13}}=10.
\displaystyle (\sqrt{13}-3\sqrt2)h+(\sqrt{13}+2\sqrt2)k
\displaystyle \qquad-5\sqrt{13}-7\sqrt2-10\sqrt{26}=0.
\displaystyle \text{Case III: }h+k-5\leq0\text{ and }3h-2k+7\geq0.
\displaystyle -\frac{h+k-5}{\sqrt2}+\frac{3h-2k+7}{\sqrt{13}}=10.
\displaystyle (-\sqrt{13}+3\sqrt2)h+(-\sqrt{13}-2\sqrt2)k
\displaystyle \qquad+5\sqrt{13}+7\sqrt2-10\sqrt{26}=0.
\displaystyle \text{Case IV: }h+k-5\leq0\text{ and }3h-2k+7\leq0.
\displaystyle -\frac{h+k-5}{\sqrt2}-\frac{3h-2k+7}{\sqrt{13}}=10.
\displaystyle (\sqrt{13}+3\sqrt2)h+(\sqrt{13}-2\sqrt2)k
\displaystyle \qquad+7\sqrt2-5\sqrt{13}+10\sqrt{26}=0.
\displaystyle \text{Each of the four equations is a linear equation in }h\text{ and }k.
\displaystyle \text{Replacing }h\text{ and }k\text{ by }x\text{ and }y,\text{ respectively, each part of the locus is a straight line.}
\displaystyle \therefore \text{The locus of }P\text{ consists of portions of four straight lines forming a parallelogram.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If the length of the perpendicular from the point }(1,1)\text{ to the line}
\displaystyle ax-by+c=0\text{ is unity, show that }\frac{1}{c}+\frac{1}{a}-\frac{1}{b}=\frac{c}{2ab}.
\displaystyle \text{Answer:}
\displaystyle \text{The perpendicular distance from }(1,1)\text{ to the line }ax-by+c=0\text{ is }1.
\displaystyle \therefore \left|\frac{a-b+c}{\sqrt{a^2+b^2}}\right|=1.
\displaystyle \text{Squaring both sides,}
\displaystyle (a-b+c)^2=a^2+b^2.
\displaystyle a^2+b^2+c^2-2ab+2ac-2bc=a^2+b^2.
\displaystyle \therefore c^2-2ab+2ac-2bc=0.
\displaystyle \therefore 2ab+2bc-2ac=c^2.
\displaystyle \therefore ab+bc-ac=\frac{c^2}{2}.
\displaystyle \text{Dividing both sides by }abc,
\displaystyle \frac{1}{c}+\frac{1}{a}-\frac{1}{b}=\frac{c}{2ab}.
\displaystyle \therefore \frac{1}{c}+\frac{1}{a}-\frac{1}{b}=\frac{c}{2ab}.
\displaystyle \\

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