NCERT Solutions: Class 11 Maths Chapter 9 - Straight Lines

Take the vertices in the given order: $(-4,5),(0,7),(5,-5),(-4,-2)$. By the shoelace formula, $\sum x_iy_{i+1}=-28+0-10-20=-58$ and $\sum y_ix_{i+1}=0+35+20+8=63$. Hence area $=\dfrac12|-58-63|=\dfrac{121}{2}$ square units.

Since the base has length $2a$ and midpoint at the origin on the y-axis, its endpoints are $(0,a)$ and $(0,-a)$. The altitude of an equilateral triangle of side $2a$ is $\sqrt3a$, perpendicular to the y-axis. Hence the third vertex is on the x-axis at $(\sqrt3a,0)$ or $(-\sqrt3a,0)$.

If $PQ$ is parallel to the y-axis, then $x_1=x_2$, so the distance is the absolute difference of ordinates: $|y_2-y_1|$. If $PQ$ is parallel to the x-axis, then $y_1=y_2$, so the distance is $|x_2-x_1|$.

Let the point on the x-axis be $(x,0)$. Equating squared distances from $(7,6)$ and $(3,4)$ gives $(x-7)^2+6^2=(x-3)^2+4^2$. Thus $x^2-14x+85=x^2-6x+25$, so $x=\dfrac{15}{2}$. The point is $\left(\dfrac{15}{2},0\right)$.

The midpoint of $(0,-4)$ and $(8,0)$ is $\left(\dfrac{0+8}{2},\dfrac{-4+0}{2}\right)=(4,-2)$. The slope of the line through the origin and $(4,-2)$ is $\dfrac{-2-0}{4-0}=-\dfrac12$.

Let $A=(4,4)$, $B=(3,5)$ and $C=(-1,-1)$. The slope of $AB$ is $\dfrac{5-4}{3-4}=-1$, and the slope of $AC$ is $\dfrac{-1-4}{-1-4}=1$. Since the product of these slopes is $-1$, $AB\perp AC$. Therefore the triangle is right-angled at $A=(4,4)$.

The positive y-axis makes an angle of $90^\circ$ with the positive x-axis. Measuring $30^\circ$ anticlockwise from the positive y-axis gives an inclination of $120^\circ$ with the positive x-axis. Hence the slope is $\tan120^\circ=-\sqrt3$.

Let the points in order be $A=(-2,-1)$, $B=(4,0)$, $C=(3,3)$ and $D=(-3,2)$. The slope of $AB$ is $\dfrac{0+1}{4+2}=\dfrac16$, and the slope of $CD$ is $\dfrac{2-3}{-3-3}=\dfrac16$, so $AB\parallel CD$. The slope of $BC$ is $\dfrac{3-0}{3-4}=-3$, and the slope of $AD$ is $\dfrac{2+1}{-3+2}=-3$, so $BC\parallel AD$. Hence both pairs of opposite sides are parallel, and the points form a parallelogram.

The slope of the line is $\dfrac{-2-(-1)}{4-3}=-1$. If $\theta$ is the inclination with the positive x-axis, then $\tan\theta=-1$. Taking $0^\circ\le\theta<180^\circ$, $\theta=135^\circ$.

Let the slopes be $m$ and $2m$. Then $\tan\theta=\left|\dfrac{2m-m}{1+2m^2}\right|=\dfrac13$. Hence $\left|\dfrac{m}{1+2m^2}\right|=\dfrac13$. This gives $3m=1+2m^2$ or $-3m=1+2m^2$. Solving, $m=1,\dfrac12,-1,-\dfrac12$. Therefore the possible pairs $(m,2m)$ are $(1,2)$, $\left(\dfrac12,1\right)$, $(-1,-2)$ and $\left(-\dfrac12,-1\right)$.

The slope of the line through $(x_1,y_1)$ and $(h,k)$ is $m=\dfrac{k-y_1}{h-x_1}$, provided $h\ne x_1$. Multiplying by $h-x_1$ gives $k-y_1=m(h-x_1)$.

Every point on the x-axis has ordinate $0$, so its equation is $y=0$. Every point on the y-axis has abscissa $0$, so its equation is $x=0$.

Using point-slope form, $y-3=\dfrac12(x+4)$. Multiplying by 2 gives $2y-6=x+4$, so $x-2y+10=0$.

A line through the origin with slope $m$ has equation $y-0=m(x-0)$, hence $y=mx$.

The slope is $\tan75^\circ=2+\sqrt3$. Using point-slope form through $\left(\dfrac32,2\right)$ gives $y-2=(2+\sqrt3)\left(x-\dfrac32\right)$.

The line intersects the x-axis at $(-3,0)$ and has slope $-2$. Thus $y-0=-2(x+3)$, or $2x+y+6=0$.

The line passes through $(0,2)$ and has slope $\tan30^\circ=\dfrac1{\sqrt3}$. Therefore $y-2=\dfrac{x}{\sqrt3}$. Multiplying by $\sqrt3$ and rearranging gives $x-\sqrt3y+2\sqrt3=0$.

The slope through $(-1,1)$ and $(2,-4)$ is $\dfrac{-4-1}{2-(-1)}=-\dfrac53$. Hence $y-1=-\dfrac53(x+1)$. Multiplying by 3 gives $3y-3=-5x-5$, so $5x+3y+2=0$.

The median from $R$ passes through the midpoint of $PQ$. The midpoint of $P(2,1)$ and $Q(-2,3)$ is $(0,2)$. The slope through $R(4,5)$ and $(0,2)$ is $\dfrac{5-2}{4-0}=\dfrac34$. Thus $y-5=\dfrac34(x-4)$, which simplifies to $3x-4y+8=0$.

The slope of the line through $(2,5)$ and $(-3,6)$ is $\dfrac{6-5}{-3-2}=-\dfrac15$. A perpendicular line has slope $5$. Passing through $(-3,5)$, its equation is $y-5=5(x+3)$, or $5x-y+20=0$.

Let the segment endpoints be $A(1,0)$ and $B(2,3)$, and let the required line meet $AB$ at $P$ with $AP:PB=1:n$. Then $P=\left(\dfrac{n+2}{n+1},\dfrac{3}{n+1}\right)$. The slope of $AB$ is $3$, so the perpendicular slope is $-\dfrac13$. Hence $y-\dfrac{3}{n+1}=-\dfrac13\left(x-\dfrac{n+2}{n+1}\right)$. Simplifying gives $(n+1)x+3(n+1)y-(n+11)=0$.

If the intercepts on both axes are equal to $a$, then the line is $\dfrac{x}{a}+\dfrac{y}{a}=1$, or $x+y=a$. Since it passes through $(2,3)$, $2+3=a$, so $a=5$. Therefore the equation is $x+y=5$.

Let the intercepts be $a$ and $b$. Then $a+b=9$ and the line is $\dfrac{x}{a}+\dfrac{y}{b}=1$. Since it passes through $(2,2)$, $\dfrac2a+\dfrac2b=1$, so $\dfrac{2(a+b)}{ab}=1$. Thus $ab=18$. The numbers $a,b$ have sum 9 and product 18, so they are 3 and 6. The two possible lines are $\dfrac{x}{3}+\dfrac{y}{6}=1$ and $\dfrac{x}{6}+\dfrac{y}{3}=1$, i.e. $2x+y=6$ and $x+2y=6$.

The slope is $\tan\dfrac{2\pi}{3}=-\sqrt3$. Through $(0,2)$, the line is $y-2=-\sqrt3(x-0)$, so $y=-\sqrt3x+2$. A parallel line has the same slope and crosses the y-axis at $(0,-2)$, so its equation is $y+2=-\sqrt3x$, i.e. $y=-\sqrt3x-2$.

The line from the origin to $(-2,9)$ has slope $\dfrac{9}{-2}=-\dfrac92$. The required line is perpendicular to it, so its slope is $\dfrac29$. Passing through $(-2,9)$, the equation is $y-9=\dfrac29(x+2)$. This simplifies to $2x-9y+85=0$.

Treat $(C,L)$ as points on a line: $(20,124.942)$ and $(110,125.134)$. The slope is $\dfrac{125.134-124.942}{110-20}=\dfrac{0.192}{90}=0.002133\ldots$. Therefore $L-124.942=\dfrac{0.192}{90}(C-20)$.

Let demand $D$ be a linear function of price $p$. The two points are $(14,980)$ and $(16,1220)$. The slope is $\dfrac{1220-980}{16-14}=120$. At Rs 17 per litre, $D=1220+120(17-16)=1340$ litres.

Let the line cut the axes at $(\alpha,0)$ and $(0,\beta)$. Since $P(a,b)$ is the midpoint, $a=\dfrac{\alpha}{2}$ and $b=\dfrac{\beta}{2}$, so $\alpha=2a$ and $\beta=2b$. The intercept form is $\dfrac{x}{2a}+\dfrac{y}{2b}=1$. Multiplying by 2 gives $\dfrac{x}{a}+\dfrac{y}{b}=2$.

Let the line cut the axes at $A(\alpha,0)$ and $B(0,\beta)$. If $R(h,k)$ divides $AB$ in the ratio $1:2$, then by the section formula $R=\left(\dfrac{2\alpha}{3},\dfrac{\beta}{3}\right)$. Thus $\alpha=\dfrac{3h}{2}$ and $\beta=3k$. The intercept form is $\dfrac{x}{3h/2}+\dfrac{y}{3k}=1$, or $\dfrac{2x}{3h}+\dfrac{y}{3k}=1$.

The line through $(3,0)$ and $(-2,-2)$ has slope $\dfrac{-2-0}{-2-3}=\dfrac25$. Its equation is $y-0=\dfrac25(x-3)$, or $2x-5y-6=0$. Substituting $(8,2)$ gives $16-10-6=0$. Hence $(8,2)$ lies on the same line, so the three points are collinear.

Write each equation in the form $y=mx+c$. (i) $x+7y=0$ gives $y=-\dfrac17x$. (ii) $6x+3y-5=0$ gives $3y=-6x+5$, so $y=-2x+\dfrac53$. (iii) $y=0$ is already in slope-intercept form with $m=0$ and $c=0$.

(i) $3x+2y=12$, so $\dfrac{x}{4}+\dfrac{y}{6}=1$. (ii) Dividing $4x-3y=6$ by 6 gives $\dfrac{x}{3/2}+\dfrac{y}{-2}=1$. (iii) $3y+2=0$ gives $y=-\dfrac23$, a horizontal line; it has y-intercept $-\dfrac23$ and no finite x-intercept.

Rewrite the line as $12x+72=5y-10$, or $12x-5y+82=0$. The distance from $(-1,1)$ is $\dfrac{|12(-1)-5(1)+82|}{\sqrt{12^2+(-5)^2}}=\dfrac{65}{13}=5$.

A point on the x-axis is $(a,0)$. The line is $4x+3y-12=0$. Its distance from $(a,0)$ is $\dfrac{|4a-12|}{\sqrt{4^2+3^2}}=\dfrac{|4a-12|}{5}$. Set this equal to 4: $|4a-12|=20$. Thus $4a-12=20$ or $4a-12=-20$, giving $a=8$ or $a=-2$.

For parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$, distance is $\dfrac{|C_2-C_1|}{\sqrt{A^2+B^2}}$. (i) $\dfrac{|31-(-34)|}{\sqrt{15^2+8^2}}=\dfrac{65}{17}$. (ii) The constants are $p$ and $-r$, with $A=B=l$, so distance is $\dfrac{|-r-p|}{\sqrt{l^2+l^2}}=\dfrac{|p+r|}{|l|\sqrt2}$.

A line parallel to $3x-4y+2=0$ has the form $3x-4y+c=0$. Substituting $(-2,3)$ gives $3(-2)-4(3)+c=0$, so $c=18$. Therefore the equation is $3x-4y+18=0$.

The line $x-7y+5=0$ has slope $\dfrac17$. A perpendicular line has slope $-7$. Having x-intercept 3 means it passes through $(3,0)$. Thus $y=-7(x-3)$, which gives $7x+y-21=0$.

The slopes are $m_1=-\sqrt3$ and $m_2=-\dfrac1{\sqrt3}$. Thus $\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|=\left|\dfrac{-1/\sqrt3+\sqrt3}{1+1}\right|=\dfrac1{\sqrt3}$. Hence the acute angle is $30^\circ$, and the other angle is $180^\circ-30^\circ=150^\circ$.

The slope of $7x-9y-19=0$ is $\dfrac79$, so a perpendicular line has slope $-\dfrac97$. The slope through $(h,3)$ and $(4,1)$ is $\dfrac{1-3}{4-h}=\dfrac{-2}{4-h}$. Therefore $\dfrac{-2}{4-h}=-\dfrac97$. Cross-multiplying gives $14=36-9h$, so $h=\dfrac{22}{9}$.

Every line parallel to $Ax+By+C=0$ has the same normal direction, so it is of the form $Ax+By+K=0$. Since it passes through $(x_1,y_1)$, $Ax_1+By_1+K=0$, giving $K=-(Ax_1+By_1)$. Substituting this value gives $Ax+By-Ax_1-By_1=0$, or $A(x-x_1)+B(y-y_1)=0$.

Let the slope of the other line be $m$. Since the angle between slopes $2$ and $m$ is $60^\circ$, $\left|\dfrac{m-2}{1+2m}\right|=\sqrt3$. Solving $\dfrac{m-2}{1+2m}=\sqrt3$ gives $m=-\dfrac{8+5\sqrt3}{11}$, and solving $\dfrac{m-2}{1+2m}=-\sqrt3$ gives $m=\dfrac{5\sqrt3-8}{11}$. Through $(2,3)$, the required equations are $y-3=m(x-2)$ for these two values of $m$.

The midpoint of $(3,4)$ and $(-1,2)$ is $(1,3)$. The slope of the segment is $\dfrac{2-4}{-1-3}=\dfrac12$, so the perpendicular bisector has slope $-2$. Through $(1,3)$, its equation is $y-3=-2(x-1)$, or $2x+y-5=0$.

For line $Ax+By+C=0$ and point $(x_0,y_0)$, the foot is $\left(x_0-\dfrac{A d}{A^2+B^2},y_0-\dfrac{B d}{A^2+B^2}\right)$ where $d=Ax_0+By_0+C$. Here $A=3$, $B=-4$, $C=-16$, $(x_0,y_0)=(-1,3)$ and $d=3(-1)-4(3)-16=-31$. Thus the foot is $\left(-1-\dfrac{3(-31)}{25},3-\dfrac{(-4)(-31)}{25}\right)=\left(\dfrac{68}{25},-\dfrac{49}{25}\right)$.

The slope of the perpendicular from the origin to $(-1,2)$ is $\dfrac{2}{-1}=-2$. Since this is perpendicular to the line $y=mx+c$, we have $m(-2)=-1$, so $m=\dfrac12$. The point $(-1,2)$ lies on the line, hence $2=\dfrac12(-1)+c$, giving $c=\dfrac52$.

For $x\cos\theta-y\sin\theta-k\cos2\theta=0$, the distance from the origin is $p=\dfrac{|k\cos2\theta|}{\sqrt{\cos^2\theta+\sin^2\theta}}=|k\cos2\theta|$. For $x\sec\theta+y\cosec\theta-k=0$, $q=\dfrac{|k|}{\sqrt{\sec^2\theta+\cosec^2\theta}}=|k\sin\theta\cos\theta|$. Hence $p^2+4q^2=k^2\cos^22\theta+4k^2\sin^2\theta\cos^2\theta=k^2(\cos^22\theta+\sin^22\theta)=k^2$.

The slope of $BC$ is $\dfrac{2-(-1)}{1-4}=-1$. Therefore the altitude from $A$ has slope $1$ and passes through $A(2,3)$, so $y-3=x-2$, i.e. $x-y+1=0$. The line $BC$ is $x+y-3=0$. The altitude length is the distance from $A(2,3)$ to $BC$: $\dfrac{|2+3-3|}{\sqrt{1^2+1^2}}=\sqrt2$.

The line with intercepts $a$ and $b$ is $\dfrac{x}{a}+\dfrac{y}{b}=1$, or $bx+ay-ab=0$. The distance from the origin to this line is $p=\dfrac{|ab|}{\sqrt{a^2+b^2}}$. Therefore $p^2=\dfrac{a^2b^2}{a^2+b^2}$, and hence $\dfrac1{p^2}=\dfrac{a^2+b^2}{a^2b^2}=\dfrac1{a^2}+\dfrac1{b^2}$.

For a line parallel to the x-axis, the coefficient of $x$ must be 0: $k-3=0$, so $k=3$. For a line parallel to the y-axis, the coefficient of $y$ must be 0: $4-k^2=0$, so $k=\pm2$. For the line to pass through the origin, the constant term must be 0: $k^2-7k+6=0$, so $(k-1)(k-6)=0$ and $k=1$ or $6$.

Let the intercepts be $a$ and $b$. Then $a+b=1$ and $ab=-6$, so $a,b$ are roots of $t^2-t-6=0$, giving $3$ and $-2$. The intercept form gives $\dfrac{x}{3}+\dfrac{y}{-2}=1$, i.e. $2x-3y=6$, or $\dfrac{x}{-2}+\dfrac{y}{3}=1$, i.e. $-3x+2y=6$.

Let the point on the y-axis be $(0,b)$. The line is $4x+3y-12=0$. The distance is $\dfrac{|3b-12|}{5}$. Set this equal to 4: $|3b-12|=20$. Hence $3b-12=20$ or $3b-12=-20$, giving $b=\dfrac{32}{3}$ or $b=-\dfrac83$.

The chord of the unit circle through $(\cos\theta,\sin\theta)$ and $(\cos\phi,\sin\phi)$ has equation $x\cos\dfrac{\theta+\phi}{2}+y\sin\dfrac{\theta+\phi}{2}=\cos\dfrac{\theta-\phi}{2}$. Its distance from the origin is therefore $\dfrac{\left|\cos\dfrac{\theta-\phi}{2}\right|}{\sqrt{\cos^2\dfrac{\theta+\phi}{2}+\sin^2\dfrac{\theta+\phi}{2}}}=\left|\cos\dfrac{\theta-\phi}{2}\right|$.

From $3x+y=0$, $y=-3x$. Substitute in $x-7y+5=0$: $x+21x+5=0$, so $x=-\dfrac5{22}$. Then $y=\dfrac{15}{22}$. A line parallel to the y-axis through this point has equation $x=-\dfrac5{22}$.

The line $\dfrac{x}{4}+\dfrac{y}{6}=1$ meets the y-axis at $(0,6)$ and has slope $-\dfrac{6}{4}=-\dfrac32$. A perpendicular line has slope $\dfrac23$. Through $(0,6)$, its equation is $y-6=\dfrac23x$, or $2x-3y+18=0$.

The lines $y=x$ and $x+y=0$ meet at $(0,0)$. The line $x=k$ meets them at $(k,k)$ and $(k,-k)$. The vertical base has length $2|k|$, and the perpendicular distance from $(0,0)$ to $x=k$ is $|k|$. Therefore the area is $\dfrac12(2|k|)(|k|)=k^2$.

The intersection of $3x+y-2=0$ and $2x-y-3=0$ is found from $y=2-3x$ and $y=2x-3$. Hence $x=1$ and $y=-1$. For the third line to pass through this point, $p(1)+2(-1)-3=0$, so $p=5$.

Let the common point be $(x,y)$. Then $y=m_1x+c_1=m_2x+c_2=m_3x+c_3$. Hence $c_2-c_3=(m_3-m_2)x$, $c_3-c_1=(m_1-m_3)x$ and $c_1-c_2=(m_2-m_1)x$. Substituting in the required expression gives $x[m_1(m_3-m_2)+m_2(m_1-m_3)+m_3(m_2-m_1)]=0$. The bracket simplifies to 0, proving the result.

The line $x-2y=3$ has slope $\dfrac12$. Let the required slope be $m$. Since the angle is $45^\circ$, $\left|\dfrac{m-1/2}{1+m/2}\right|=1$. Solving gives $m=3$ or $m=-\dfrac13$. Through $(3,2)$, the equations are $y-2=3(x-3)$ and $y-2=-\dfrac13(x-3)$, i.e. $3x-y-7=0$ and $x+3y-9=0$.

Solving $4x+7y-3=0$ and $2x-3y+1=0$ gives the intersection point $\left(\dfrac1{13},\dfrac5{13}\right)$. A line with equal intercepts has equation $x+y=a$. Substituting the point gives $a=\dfrac1{13}+\dfrac5{13}=\dfrac6{13}$. Therefore $x+y=\dfrac6{13}$, or $13x+13y-6=0$.

The slope of $y=mx+c$ is $m$. If the required line through the origin has slope $M$ and makes an angle $\theta$ with it, then $M=\tan(\tan^{-1}m\pm\theta)$. Using the tangent addition formula, $M=\dfrac{m\pm\tan\theta}{1\mp m\tan\theta}$. Since the line passes through the origin, $y=Mx$, so $\dfrac{y}{x}=\dfrac{m\pm\tan\theta}{1\mp m\tan\theta}$.

A point on the segment from $A(-1,1)$ to $B(5,7)$ can be written as $(-1+6t,1+6t)$. It lies on $x+y=4$ when $(-1+6t)+(1+6t)=4$, so $12t=4$ and $t=\dfrac13$. Therefore $AP:PB=t:(1-t)=\dfrac13:\dfrac23=1:2$.

The point $(1,2)$ lies on $2x-y=0$. Intersect $2x-y=0$ with $4x+7y+5=0$. Since $y=2x$, $4x+14x+5=0$, so $x=-\dfrac5{18}$ and $y=-\dfrac59$. The required distance is the distance from $(1,2)$ to $\left(-\dfrac5{18},-\dfrac59\right)$: $\sqrt{\left(\dfrac{23}{18}\right)^2+\left(\dfrac{23}{9}\right)^2}=\dfrac{23\sqrt5}{18}$.

Let the intersection point on $x+y=4$ be $(x,4-x)$. Its distance from $(-1,2)$ is 3, so $(x+1)^2+(2-x)^2=9$. This gives $2x^2-2x-4=0$, or $x^2-x-2=0$. Thus $x=2$ or $x=-1$, giving intersection points $(2,2)$ and $(-1,5)$. The directions from $(-1,2)$ to these points are horizontal and vertical, respectively.

If the legs are parallel to the coordinate axes, the right-angle vertex is formed by taking the x-coordinate of one endpoint and the y-coordinate of the other. Thus the right-angle vertex is either $(1,1)$ or $(-4,3)$. For $(1,1)$, the legs lie on $x=1$ and $y=1$. For $(-4,3)$, the legs lie on $x=-4$ and $y=3$.

For reflection of $(x_0,y_0)$ in $Ax+By+C=0$, use $(x',y')=\left(x_0-\dfrac{2A d}{A^2+B^2},y_0-\dfrac{2B d}{A^2+B^2}\right)$ where $d=Ax_0+By_0+C$. Here the line is $x+3y-7=0$, so $A=1$, $B=3$, $C=-7$, and for $(3,8)$, $d=3+24-7=20$. Thus $x'=3-\dfrac{40}{10}=-1$ and $y'=8-\dfrac{120}{10}=-4$.

The slopes of the two given lines are $3$ and $\dfrac12$. Equal inclination to $y=mx+4$ means $\left|\dfrac{m-3}{1+3m}\right|=\left|\dfrac{m-1/2}{1+m/2}\right|$. Squaring and simplifying gives $7m^2-2m-7=0$. Therefore $m=\dfrac{2\pm\sqrt{200}}{14}=\dfrac{1\pm5\sqrt2}{7}$.

The distances from $P(x,y)$ to the two lines are $\dfrac{|x+y-5|}{\sqrt2}$ and $\dfrac{|3x-2y+7|}{\sqrt{13}}$. Their sum is 10. In any region where the signs of $x+y-5$ and $3x-2y+7$ are fixed, this becomes $\pm\dfrac{x+y-5}{\sqrt2}\pm\dfrac{3x-2y+7}{\sqrt{13}}=10$, which is a linear equation in $x$ and $y$. Hence the variable point moves on a line.

Write the second line with the same $x,y$ coefficients as the first: $3(3x+2y+6=0)$ gives $9x+6y+18=0$. Points equidistant from the two parallel lines satisfy $|9x+6y-7|=|9x+6y+18|$. The non-trivial solution is $9x+6y-7=-(9x+6y+18)$, which gives $18x+12y+11=0$.

Reflect $(5,3)$ in the x-axis to get $(5,-3)$. The incident path from $(1,2)$ to the reflection point and the reflected path to $(5,3)$ correspond to the straight line from $(1,2)$ to $(5,-3)$. The slope is $-\dfrac54$, so the line is $y-2=-\dfrac54(x-1)$. Setting $y=0$ gives $x-1=\dfrac85$, hence $x=\dfrac{13}{5}$. Therefore $A=\left(\dfrac{13}{5},0\right)$.

Let $s=\sqrt{a^2-b^2}$. The line is $\dfrac{\cos\theta}{a}x+\dfrac{\sin\theta}{b}y-1=0$. The product of the distances from $(s,0)$ and $(-s,0)$ is $\dfrac{\left|\left(\dfrac{s\cos\theta}{a}-1\right)\left(-\dfrac{s\cos\theta}{a}-1\right)\right|}{\dfrac{\cos^2\theta}{a^2}+\dfrac{\sin^2\theta}{b^2}}$. The numerator simplifies to $1-\dfrac{(a^2-b^2)\cos^2\theta}{a^2}=\dfrac{a^2\sin^2\theta+b^2\cos^2\theta}{a^2}$. The denominator is $\dfrac{b^2\cos^2\theta+a^2\sin^2\theta}{a^2b^2}$. Their quotient is $b^2$.

First find the junction by solving $2x-3y+4=0$ and $3x+4y-5=0$. This gives $x=-\dfrac1{17}$ and $y=\dfrac{22}{17}$. The shortest path to the line $6x-7y+8=0$ is perpendicular to it. Since that line has slope $\dfrac67$, the required path has slope $-\dfrac76$ and passes through $\left(-\dfrac1{17},\dfrac{22}{17}\right)$. Thus its equation is $7x+6y-\dfrac{125}{17}=0$, or $119x+102y-125=0$.