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The Straight Lines — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSThe Straight Lines4 Marks
Question
Show that the point (3, -5) lies between the parallel lines 2x + 3y - 7 = 0 and 2x + 3y + 12 = 0 and find the equation of lines through (3, -5) cutting the above lines at an angle of 45°.
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Answer

The distance fram (3, -5) to the line 2x + 3y - 7 = 0 is $\Big|\frac{\text{ax}_1+\text{by}_1+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$ $=\frac{|2(3)+3(-5)-7|}{\sqrt{(2)^2+(3)^2}}$ $=\frac{|6-15-7|}{\sqrt{13}}$ $=\frac{16}{\sqrt{13}} $ Also, distance of (3, -5) fram the second line 2x + 3y + 12 =0 $\Big|\frac{\text{ax}_1+\text{by}_1+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$ $=\frac{|2(3)+3(-5)+12|}{\sqrt{(2)^2+(3)^2}}$ $=\frac{|6-15+12|}{\sqrt{13}}$ $=\frac{21}{\sqrt{13}} \ ...(\text{i})$ Now, $\text{c}_1-\text{c}_2=12-7=15$ Also difference between (i) and (ii) is 5 $\therefore$ (3, -5) lies between the two lines equation of line through (3, -5) cutting the lines at 45º is $\tan\theta=\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}$ $\tan45^\circ=\frac{\text{m}-(\frac{-2}{3})}{1-\frac{2}{3}\text{m}}=\pm1$ $\text{m}+\frac{2}{3}=1-\frac{2}{3}\text{m}$ or, $\text{m}+\frac{2}{3}=-1+\frac{2}{3}\text{m}$ $\text{m}\Big(1+\frac{2}{3}\Big)=1-\frac{2}{3}$ or, $\text{m}\Big(1-\frac{2}{3}\Big)=-1-\frac{2}{3}$ $\text{m}\Big(\frac{5}{3}\Big)=\frac{1}{3}$ or, $\text{m}\Big(\frac{1}{3}\Big)=\frac{-5}{3}$ $\text{m=}\frac{1}{5}$ or, $\text{m}=-5$ $\therefore$ Equation is $\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$or, $\text{y}+5=-5(\text{x}-3)$ $\text{y}+5=\frac{1}{5}(\text{x}-3)$ or, $5\text{x}+\text{y}-10=0$ $5\text{y}-\text{x}+28=0$

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