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NCERT Solutions · Class 11 Mathematics Straight Lines

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EXERCISE 9.2 1–10 (part 2 of 7)

  1. In Exercises $\displaystyle 1$ to $\displaystyle 8$, find the equation of the line which satisfy the given conditions:

    Exercise 1

    Write the equations for the x\displaystyle x-and y\displaystyle y-axes.

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    NCERT’s answer
    $\displaystyle y=0$ and $\displaystyle x=0$
    Read off the defining property of each axis.The \(\displaystyle x\)-axis. Every point on the \(\displaystyle x\)-axis has ordinate \(\displaystyle 0\), and no other point does. So a point \(\displaystyle (x,y)\) lies on the \(\displaystyle x\)-axis if and only if \(\displaystyle y=0\). (Consistently: the \(\displaystyle x\)-axis passes through \(\displaystyle (0,0)\) with slope \(\displaystyle 0\), so \(\displaystyle y-0=0(x-0)\), i.e. \(\displaystyle y=0\).)The \(\displaystyle y\)-axis. Every point on the \(\displaystyle y\)-axis has abscissa \(\displaystyle 0\), and no other point does. So a point \(\displaystyle (x,y)\) lies on the \(\displaystyle y\)-axis if and only if \(\displaystyle x=0\). (This line is vertical, so it has no slope and cannot be written in slope form.)Equation of the \(\displaystyle x\)-axis: \(\displaystyle y=0\). Equation of the \(\displaystyle y\)-axis: \(\displaystyle x=0\).

  2. Exercise 2

    Passing through the point (−4,3)\displaystyle (-4,3) with slope 12\displaystyle \frac{1}{2}.

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    NCERT’s answer
    $\displaystyle x-2 y+10=0$
    Point-slope form.A line of slope \(\displaystyle m\) through \(\displaystyle (x_0,y_0)\) has equation\[y-y_0=m\left(x-x_0\right).\]Here \(\displaystyle (x_0,y_0)=(-4,3)\) and \(\displaystyle m=\dfrac12\):\[y-3=\frac{1}{2}\left(x-(-4)\right)=\frac{1}{2}(x+4).\]Multiply through by \(\displaystyle 2\):\[2y-6=x+4\ \Rightarrow\ x-2y+10=0.\]Check: at \(\displaystyle (-4,3)\), \(\displaystyle -4-6+10=0\). ✓\(\displaystyle x-2y+10=0\).

  3. Exercise 3

    Passing through (0,0)\displaystyle (0,0) with slope m\displaystyle m.

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    NCERT’s answer
    $\displaystyle y=m x$
    Point-slope form with the point at the origin.A line of slope \(\displaystyle m\) through \(\displaystyle (x_0,y_0)\) has equation \(\displaystyle y-y_0=m(x-x_0)\).Here \(\displaystyle (x_0,y_0)=(0,0)\), so\[y-0=m(x-0).\]\(\displaystyle y=mx\).

  4. Exercise 4

    Passing through (2,23)\displaystyle (2,2 \sqrt{3}) and inclined with the x\displaystyle x-axis at an angle of 75\displaystyle 75°.

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    NCERT’s answer
    $\displaystyle (\sqrt{3}+1) x-(\sqrt{3}-1) y=4(\sqrt{3}-1)$
    Point-slope form, with the slope obtained from the inclination.The slope is \(\displaystyle m=\tan 75^{\circ}\). Write \(\displaystyle 75^{\circ}=45^{\circ}+30^{\circ}\) and use \(\displaystyle \tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\):\[\tan 75^{\circ}=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ}\tan 30^{\circ}}=\frac{1+\dfrac{1}{\sqrt3}}{1-\dfrac{1}{\sqrt3}}=\frac{\sqrt3+1}{\sqrt3-1}=\frac{(\sqrt3+1)^{2}}{(\sqrt3-1)(\sqrt3+1)}=\frac{4+2\sqrt3}{2}=2+\sqrt3.\]Now apply \(\displaystyle y-y_0=m(x-x_0)\) with \(\displaystyle (x_0,y_0)=\left(2,\,2\sqrt3\right)\):\[y-2\sqrt3=\left(2+\sqrt3\right)(x-2)\] \[y-2\sqrt3=\left(2+\sqrt3\right)x-4-2\sqrt3\] \[y=\left(2+\sqrt3\right)x-4.\]\(\displaystyle \left(2+\sqrt{3}\right)x-y=4\).

  5. Exercise 5

    Intersecting the x\displaystyle x-axis at a distance of 3\displaystyle 3 units to the left of origin with slope -2.

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    NCERT’s answer
    $\displaystyle 2 x+y+6=0$
    Point-slope form, after naming the point of intersection with the \(\displaystyle x\)-axis."Intersecting the \(\displaystyle x\)-axis at a distance of $\displaystyle 3$ units to the left of the origin" means the line passes through \(\displaystyle (-3,0)\). The slope is \(\displaystyle m=-2\).Using \(\displaystyle y-y_0=m(x-x_0)\):\[y-0=-2\left(x-(-3)\right)=-2(x+3)\] \[y=-2x-6\] \[2x+y+6=0.\]Check: at \(\displaystyle (-3,0)\), \(\displaystyle -6+0+6=0\). ✓\(\displaystyle 2x+y+6=0\).

  6. Exercise 6

    Intersecting the y\displaystyle y-axis at a distance of 2\displaystyle 2 units above the origin and making an angle of 30\displaystyle 30° with positive direction of the x\displaystyle x-axis.

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    NCERT’s answer
    $\displaystyle x-\sqrt{3} y+2 \sqrt{3}=0$
    Slope-intercept form.A line of slope \(\displaystyle m\) with \(\displaystyle y\)-intercept \(\displaystyle c\) has equation \(\displaystyle y=mx+c\).The line cuts the \(\displaystyle y\)-axis $\displaystyle 2$ units above the origin, so \(\displaystyle c=2\). It makes \(\displaystyle 30^{\circ}\) with the positive \(\displaystyle x\)-axis, so\[m=\tan 30^{\circ}=\frac{1}{\sqrt3}.\]Hence\[y=\frac{1}{\sqrt3}x+2.\]Multiplying by \(\displaystyle \sqrt3\): \(\displaystyle \sqrt3\,y=x+2\sqrt3\).\(\displaystyle x-\sqrt{3}\,y+2\sqrt{3}=0\).

  7. Exercise 7

    Passing through the points (−1,1)\displaystyle (-1, 1) and (2,−4)\displaystyle (2, -4).

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    NCERT’s answer
    $\displaystyle 5 x+3 y+2=0$
    Two-point form.The line through \(\displaystyle (x_1,y_1)\) and \(\displaystyle (x_2,y_2)\) is\[y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right).\]With \(\displaystyle (x_1,y_1)=(-1,1)\) and \(\displaystyle (x_2,y_2)=(2,-4)\), the slope is\[m=\frac{-4-1}{2-(-1)}=\frac{-5}{3}.\]So\[y-1=-\frac{5}{3}\left(x+1\right)\] \[3y-3=-5x-5\] \[5x+3y+2=0.\]Check: at \(\displaystyle (-1,1)\), \(\displaystyle -5+3+2=0\); at \(\displaystyle (2,-4)\), \(\displaystyle 10-12+2=0\). ✓\(\displaystyle 5x+3y+2=0\).

  8. Exercise 8

    The vertices of ΔPQR\displaystyle \Delta \mathrm{PQR} are P (2,1)\displaystyle (2, 1), Q (−2,3)\displaystyle (-2, 3) and R (4,5)\displaystyle (4, 5). Find equation of the median through the vertex R.

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    NCERT’s answer
    $\displaystyle 3 x-4 y+8=0$
    Median through a vertex. A median joins a vertex to the mid-point of the opposite side, so the median through \(\displaystyle \mathrm{R}\) joins \(\displaystyle \mathrm{R}\) to the mid-point of \(\displaystyle \mathrm{PQ}\).Mid-point of \(\displaystyle \mathrm{PQ}\), with P $\displaystyle (2, 1)$ and Q $\displaystyle (-2, 3)$: \[\mathrm{M}=\left(\frac{2+(-2)}{2},\ \frac{1+3}{2}\right)=(0,\,2). \]Slope of the line through M $\displaystyle (0, 2)$ and R $\displaystyle (4, 5)$: \[m=\frac{5-2}{4-0}=\frac{3}{4}. \]Using the point-slope form \(\displaystyle y-y_1=m(x-x_1)\) with the point M $\displaystyle (0, 2)$: \[y-2=\frac{3}{4}(x-0)\quad\Longrightarrow\quad 4y-8=3x. \]Both P and Q are checked against the result only through M; substituting R $\displaystyle (4, 5)$ gives \(\displaystyle 3(4)-4(5)+8=0\), so R lies on it as required.The median through R is \(\displaystyle 3x-4y+8=0\).

  9. Exercise 9

    Find the equation of the line passing through (−3,5)\displaystyle (-3,5) and perpendicular to the line through the points (2,5)\displaystyle (2, 5) and (−3,6)\displaystyle (-3, 6).

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    NCERT’s answer
    $\displaystyle 5 x-y+20=0$
    Point-slope form with the perpendicularity condition. Two lines are perpendicular when the product of their slopes is \(\displaystyle -1\).Slope of the line through $\displaystyle (2, 5)$ and $\displaystyle (-3, 6)$: \[m_1=\frac{6-5}{-3-2}=\frac{1}{-5}=-\frac{1}{5}. \]The required line is perpendicular to it, so its slope is \[m=-\frac{1}{m_1}=5. \]It passes through \(\displaystyle (-3,5)\), so by the point-slope form: \[y-5=5\bigl(x-(-3)\bigr)=5(x+3)\quad\Longrightarrow\quad y-5=5x+15. \]The required line is \(\displaystyle 5x-y+20=0\).

  10. Exercise 10

    A line perpendicular to the line segment joining the points (1,0)\displaystyle (1,0) and (2,3)\displaystyle (2,3) divides it in the ratio 1:n\displaystyle 1: n. Find the equation of the line.

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    NCERT’s answer
    $\displaystyle (1+n) x+3(1+n) y=n+11$
    Section formula, then point-slope form. Let the line meet the segment joining A $\displaystyle (1, 0)$ and B $\displaystyle (2, 3)$ at the point C that divides AB in the ratio \(\displaystyle 1:n\).By the section formula, the point dividing the join of \(\displaystyle (x_1,y_1)\) and \(\displaystyle (x_2,y_2)\) internally in the ratio \(\displaystyle m_1:m_2\) is \(\displaystyle \left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)\). With \(\displaystyle m_1:m_2=1:n\): \[\mathrm{C}=\left(\frac{1\cdot 2+n\cdot 1}{1+n},\ \frac{1\cdot 3+n\cdot 0}{1+n}\right)=\left(\frac{n+2}{n+1},\ \frac{3}{n+1}\right). \]Slope of AB is \(\displaystyle \dfrac{3-0}{2-1}=3\), so the slope of the perpendicular line is \(\displaystyle -\dfrac{1}{3}\).Point-slope form through C: \[y-\frac{3}{n+1}=-\frac{1}{3}\left(x-\frac{n+2}{n+1}\right). \]Multiplying through by \(\displaystyle 3(n+1)\): \[3(n+1)y-9=-(n+1)x+(n+2), \] \[(1+n)x+3(1+n)y=n+11. \]The required line is \(\displaystyle (1+n)x+3(1+n)y=n+11\).