Question

Find the equation of straight line passing through (-2, -7) and having an intercept of length 3 between the straight lines 4x + 3y = 12 and 4x + 3y = 3.

Answer 1

$\text{Here,}(x_1,y_1)=A(-2,-7)$

$\text{The equation of any line passing through}(-2,-7)is$

$\frac{x-x_1}{cos\theta }=\frac{y-y_1}{sin\theta }=r$

$\frac{x+2}{cos\theta }=\frac{y+7}{sin\theta }=r$

$\text{B and C are at distance r and (r + 3)}$

$\text{Thus, coordinates of B and C are}(-2+rcos,-7+rsin\theta )and(-2+(r+3)cos\theta ,-7+(r+3\left)sin\theta \right)$

$\text{B lies on}4x+3y=12$

$\Rightarrow 4(-2+4cos\theta )+3(-7+rsin\theta )=12...\left(1\right)$

$\text{C lies on}4x+3y=3$

$\Rightarrow 4(-2+(r+3\left)cos\theta \right)+3(-7+(r+3\left)sin\theta \right)=3...\left(2\right)$

$\text{Subtracting (i) from (2)}$

$12cos\theta +9sin\theta =-9$

$\Rightarrow 4cos\theta =-3(1-sin\theta )$

$\Rightarrow 16co{s}^2\theta =9(1+si{n}^2\theta -2sin\theta )$

$\Rightarrow 16(1-si{n}^2\theta )=9(1+si{n}^2\theta -2sin\theta )$

$\Rightarrow 16-16si{n}^2\theta =9+9si{n}^2\theta -18sin\theta$

$\Rightarrow 25si{n}^2\theta -18sin\theta -7=0$

$\Rightarrow 25si{n}^2\theta -25sin\theta +7sin\theta -7=0$

$\Rightarrow 25sin\theta (sin\theta -1)-7(sin\theta -1)=0$

$sin\theta =1,sin\theta =\frac{7}{25}$

$\text{Now,}sin\theta =1\Rightarrow cos\theta =0$

$\therefore x+2=0$

$\text{and if}sin\theta =\frac{7}{25}thencos\theta =\frac{24}{25}$

$\therefore \frac{x+2}{\frac{24}{25}}=\frac{y+7}{\frac{7}{25}}$

$\Rightarrow 7x+24y+182=0$

$\text{P.Q. The slope of a straight line through A (3, 2) is}\frac{3}{4}.\text{Find the coordinates of the points on the line that are 5 units away from A:}$

$\text{The slope of the line}=\frac{3}{4}$

$\Rightarrow tan\theta =\frac{3}{4}$

$\Rightarrow sin\theta =\frac{3}{5}andcos\theta =\frac{4}{5}$

$\text{The coordinates of point that are 5 units away from A are}$

$\frac{x-3}{cos\theta }=\frac{y-3}{sin\theta }=\pm 5$

$or\frac{x-3}{\frac{4}{5}}=\frac{y-3}{\frac{3}{5}}=\pm 5$

$\frac{5(x-3)}{4}=\frac{5(y-3)}{3}=\pm 5$

$i.e.,\frac{5(y-3)}{4}=\pm 5and\frac{5(y-3)}{3}=\pm 5$

$\therefore x=7andy=5$

$x=-1ory=-1$

$\therefore \text{coordinates of point on the line that are 5 units away from A are (-1, -1) and (7, 5)}$

$\text{P.Q. Find the direction in which a straight line must be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of}$

$\sqrt{\frac{2}{3}}\text{from this point.}$

$\text{The equation of line passing through}$

$A(1,2)andr=\sqrt{\frac{2}{3}}is$

$\text{Let}\theta \text{be the slope of the line, so, the equation of the line that has slope}\theta \text{and passes through}A(1,2)is$

$\frac{x-x_1}{cos\theta }=\frac{y-y_1}{sin\theta }$

$\text{Coordinates of p are given by}$

$\frac{x-1}{cos\theta }=\frac{y-y_1}{sin\theta }=r=\sqrt{\frac{2}{3}}$

$\frac{x-1}{sin\theta }=\frac{y-2}{cos\theta }=\sqrt{\frac{2}{3}}$

$\Rightarrow x=\sqrt{\frac{2}{3}}sin\theta +1andy=\sqrt{\frac{2}{3}}cos\theta +2\text{and p with coordinates}$

$(\sqrt{\frac{2}{3}}sin\theta +1,\sqrt{\frac{2}{3}}cos\theta +2)$

$\text{lie in}x-y=4$

$\therefore \sqrt{\frac{2}{3}}sin\theta +1+\sqrt{\frac{2}{3}}cos\theta +2=\pm 4$

$\sqrt{\frac{2}{3}}(sin\theta +cos\theta )=\pm 1$

$\sqrt{\frac{2}{3}}(\frac{sin\theta }{\sqrt{2}}+\frac{cos\theta }{\sqrt{2}})=\pm 1$

$sin(\theta +\frac{\pi }{4})=\pm \frac{\sqrt{3}}{2}$

$[\because sinAcos,B+cosAsinB=sin(A+B\left)\right]$

$or\theta +\frac{\pi }{4}=\pm \frac{\pi }{3}$

$\therefore \theta =\frac{\pi }{12}or\frac{5\pi }{12}$

${15}^{\circ }or{75}^{\circ }$

$\text{Hence, the direction of the lines with the positive direction of the x-axis is}{15}^{\circ }or{75}^{\circ }$

$\text{P.Q. In what direction whould a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is}$

$\text{at a distance}\frac{\sqrt{6}}{3}\text{from the given point.}$

$\text{Let}P=(1,2)$

$\text{and let AB be the given line}$

$x+y=4...\left(i\right)$

$\text{Let the line through P making an angle Q with x axis cuts the line AB at Q at}$

$\text{a distance}\sqrt{\frac{2}{3}}\text{from P}$

$\text{Then,}$

$Q=(1+\sqrt{\frac{2}{3}}cos\theta ,2-\sqrt{\frac{2}{3}}sin\theta )$

$\text{Since Q lies on line (i)}$

$\therefore 1+\sqrt{\frac{2}{3}}cos\theta +2+\sqrt{\frac{2}{3}}sin\theta =4$

$\Rightarrow cos\theta +sin\theta =\sqrt{\frac{3}{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}cosq+\frac{1}{\sqrt{2}}sin\theta =\frac{\sqrt{3}}{2}$

$\Rightarrow {45}^{\circ }cos\theta +sin{45}^{\circ }sin\theta =\frac{\sqrt{3}}{2}$

$\Rightarrow cos(\theta -{45}^{\circ })=cos{30}^{\circ }$

$\Rightarrow q-{45}^{\circ }=2n\pi \pm {30}^{\circ },n=\theta \pm 1,\pm 2$

$\Rightarrow \theta ={15}^{\circ },{75}^{\circ }$

$\left(\text{taking only values between}{0}^{\circ }and{180}^{\circ }\right)$