# Coordinate Geometry : Exercise (1.3) - Solutions

1.         Sketch the following lines.

$\begin{array} \text{(a)}\ y=3\\\\ \text{(b)}\ x=-2\\\\ \text{(c)}\ y=5x\\\\ \text{(d)}\ y=-3x\\\\ \text{(e)}\ y=\displaystyle\frac{1}{2}x \end{array}$

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$\text{(a)}\ y=3$

$\text{(b)}\ x=-2$

$\begin{array}{l} \text{(c)}\ y=5x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=5x & 0 & 5\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(d)}\ y=-3x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x & 0 & -3\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(e)}\ y=\displaystyle\frac{1}{2}x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x & 0 & -3\\ \hline \end{array}\end{array}$

2.         Graph each line in the same coordinate plane.

$\begin{array}{lll} \text{(a)}\ y=x+2 \\\\ \text{(b)}\ y=x+3 \\\\ \text{(c)}\ y=x+5 \\\\ \text{(d)}\ y=x-1 \\\\ \text{(e)}\ y=x-2 \\\\ \text{(f)}\ y=x-4 \end{array}$

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$\bullet\quad y=x$ is a line with slope equal to 1 and passing through origin.

$\bullet\quad y=x + 2$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 2).

$\bullet\quad y=x + 3$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 3).

$\bullet\quad y=x + 5$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 5).

$\bullet\quad y=x -1$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -1).

$\bullet\quad y=x -2$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -2).

$\bullet\quad y=x -4$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -4).

3.         Graph each line.

$\begin{array} \text{(a)}\ y=2x+2\\\\ \text{(b)}\ y=\displaystyle\frac{1}{2}x+2\\\\ \text{(c)}\ y=-\displaystyle\frac{1}{2}x+2 \end{array}$

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$\begin{array}{l} \text{(a)}\ y=2x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & -1\\ \hline y=2x+2 & 2 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(b)}\ y=\displaystyle\frac{1}{2}x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & -4\\ \hline y=\displaystyle\frac{1}{2}x+2 & 2 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(b)}\ y=-\displaystyle\frac{1}{2}x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & 4\\ \hline -\displaystyle\frac{1}{2}x+2 & 2 & 0\\ \hline \end{array}\end{array}$

4.         Find the slope and $y$-intercept for the following equations and sketch their graphs.

$\begin{array}{l} \text{(a)}\ y=x-2 \\\\ \text{(b)}\ y=-3x-3 \\\\ \text{(c)}\ y=\displaystyle\frac{1}{2}x+1\\\\ \text{(d)}\ y=-\displaystyle\frac{1}{2}x+1 \\\\ \text{(e)}\ y+3x=3 \\\\ \text{(f)}\ x-y=3 \end{array}$

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$\begin{array}{l} \text{(a)}\ y=x-2\\\\ \text{slope}= 1\\\\ y\text{-intercept}=-2\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 2\\ \hline y=x-2 & -2 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(b)}\ y=-3x-3\\\\ \text{slope}= -3\\\\ y\text{-intercept}=-3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x-3 & -3 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(c)}\ y=\displaystyle\frac{1}{2}x+1\\\\ \text{slope}= \displaystyle\frac{1}{2}\\\\ y\text{-intercept}=1\\\\ \begin{array}{|l|l|l|} \hline x & 0 & -2\\ \hline y=\displaystyle\frac{1}{2}x+1 & 1 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(d)}\ y=-\displaystyle\frac{1}{2}x+1\\\\ \text{slope}= -\displaystyle\frac{1}{2}\\\\ y\text{-intercept}=1\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 2\\ \hline y=-\displaystyle\frac{1}{2}x+1 & 1 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(e)}\ y+3x=3\\\\ \quad y=-3x+3\\\\ \text{slope}=-3\\\\ y\text{-intercept}=3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x+3 & 3 & 0\\ \hline \end{array}\end{array}$

$\begin{array}{l} \text{(f)}\ x-y=3\\\\ \quad y=x-3\\\\ \text{slope}=1\\\\ y\text{-intercept}=-3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 3\\ \hline y=x-3 & -3 & 0\\ \hline \end{array}\end{array}$

5.         Find the equation of the straight line with the given slope and $y$-intercept.

(a) slope 3, $y$-intercept 4

(b) slope 2, $y$-intercept 0

(c) slope 0, $y$-intercept 2

(d) slope 0, $y$-intercept 0

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$\begin{array}{l} \text{(a)}\quad m=3,\ c=4\\\\ \quad \quad \ \ y=m x+c\\\\ \quad \quad \ \ \ \ =3 x+4 \end{array}$

$\therefore$ The equation of the line is $y=3 x+4$.

$\begin{array}{l} \text{(b)}\quad m=2,\ c=0\\\\ \quad \quad \ \ y=m x+c\\\\ \quad \quad \ \ \ \ =2 x+0\\\\ \quad \quad \ \ \ \ =2 x \end{array}$

$\therefore$ The equation of the line is $y=2x$.

$\begin{array}{l} \text{(c)}\quad m=0,\ c=2\\\\ \quad \quad \ \ y=(0) x+2\\\\ \quad \quad \ \ \ \ =2 \end{array}$

$\therefore$ The equation of the line is $y=2$.

$\begin{array}{l} \text{(d)}\quad m=0,\ c=0\\\\ \quad \quad \ \ y=(0) x+0\\\\ \quad \quad \ \ \ \ =0 \end{array}$

$\therefore$ The equation of the line is $y=0$.

6.         Find the equation of the line which has a slope $m$ of $\displaystyle -\frac{2}{3}$ and passes through the point $(9,4)$.

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Let $\left(x_{1}, y_{1}\right)=(9,4). m=-\displaystyle \frac{2}{3}$

$y-y_{1}=m\left(x-x_{1}\right)$

$y-4=-\displaystyle \frac{2}{3}(x-9)=-\displaystyle \frac{2}{3} x+6$

$y=-\displaystyle \frac{2}{3} x+10$

$\therefore 2x+3y=30$

$\therefore$ The equation of the line is $2x+3y=30$.

7.         A line has slope $-2$ and $y$-intercept $6$, find its $x$-intercept.

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$$\begin{array}{l} m=-2,\ c=6 \\\\ \text{The equation of the line having slope} \\\\ m\ \text{and}\ y\text{-intercept}\ c\ \text{is}\ y=m x+c.\\\\ \therefore\ \text{The equation of given line is}\ y=-2x + 6.\\\\ \text { When }\ y=0,-2 x+6=0\Rightarrow x=3 \\\\ \therefore x \text {-intercept }=3 \end{array}$$

8.         Find the equation of the line which:

(a) has a slope of $5$ and passes through the point $(2,9)$.

(b) has a slope of $1$ and passes through the point $(1,-2)$.

(c) has a slope of $-3$ and passes through the point $(-1,6)$.

(d) has a slope of $-2$ and passes through the point $(-1,4)$.

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$$\begin{array}{l} \text { (a) } \text{Let}\ \left(x_{1}, y_{1}\right)=(2,9).\ m=5\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-9 =5(x-2)\\\\ \quad \quad y-9 =5 x-10 \\\\ \quad \quad y=5 x-1 \end{array}$$

$$\begin{array}{l} \text { (b) } \text{Let}\ \left(x_{1}, y_{1}\right)=(1,-2). \ m=1\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-(-2)=1(x-1)\\\\ \quad \quad y+2 =x-1 \\\\ \quad \quad y=x-3 \end{array}$$

$$\begin{array}{l} \text { (c) } \text{Let}\ \left(x_{1}, y_{1}\right)=(-1,6).\ m=-3\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-6=-3\left(x-(-1)\right)\\\\ \quad \quad y-6 =-3 x-3 \\\\ \quad \quad y=-3 x+3 \end{array}$$

$$\begin{array}{l} \text { (d) } \text{Let}\ \left(x_{1}, y_{1}\right)=(-1,4).\ m=-2\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-4=-2\left(x-(-1)\right) \\\\ \quad \quad y-4 =-2 x-2 \\\\ \quad \quad y=-2 x+2 \end{array}$$

9.         Find the slope and equation of the line joining the following pairs of points.

(a) $(2,4)$ and $(6,8)$

(b) $(-3,5)$ and $(6,-1)$

(c) $(-2, 1)$ and $(-4, -2)$

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(a) Let $\left(x_{1}, y_{1}\right)=(2,4)$ and $\left(x_{2}, y_{2}\right)=(6,8)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{8-4}{6-2}=\frac{4}{4}=1$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-4=1(x-2)$

$y-4=x-2$

$y=x+2$

(b) Let $\left(x_{1}, y_{1}\right)=(-3,5)$ and $\left(x_{2}, y_{2}\right)=(6,-1)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{-1-5}{6-(-3)}=-\frac{6}{9}=-\frac{2}{3}$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-5=-\displaystyle\frac{2}{3}\left(x-(-3)\right)$

$2x+3y = 9$

(c) Let $\left(x_{1}, y_{1}\right)=(-2,1)$ and $\left(x_{2}, y_{2}\right)=(-4,-2)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{-2-1}{-4-(-2)}=\frac{3}{2}$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-1=\displaystyle\frac{3}{2}\left(x-(-2)\right)$

$3x-2y = -8$

10.        Determine which of the pairs of lines in each case with given equations are parallel or perpendicular or neither.

(a) $y=3x-2$ and $y=3x+9$

(b) $y=\displaystyle \frac{2}{3}x-5$ and $y=\displaystyle \frac{3}{2}x-5$

(c) $y=3x-2$ and $y=-\displaystyle \frac{1}{3}x+9$

(d) $y=\displaystyle \frac{2}{3}x-5$ and $y=-\displaystyle \frac{3}{2}x-5$

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$\begin{array}{l}(\text{a})\ \ \ \ {{l}_{1}}:y=3x-2\ \text{and}\ {{l}_{2}}:y=3x+9\\\ \ \ \therefore \ \ {{m}_{1}}=3\ \text{and}\ {{m}_{2}}=3\\\ \ \ \therefore \ \ {{m}_{1}}={{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\parallel \ {{l}_{2}}\\\\(\text{b})\ \ \ \ {{l}_{1}}:y=\displaystyle \frac{2}{3}x-5\ \text{and}\ {{l}_{2}}:y=\displaystyle \frac{3}{2}x-5\\\ \ \ \therefore \ \ {{m}_{1}}=\displaystyle \frac{2}{3}\ \text{and}\ {{m}_{2}}=\displaystyle \frac{3}{2}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \ \ \ \ \text{Again,}\ {{m}_{1}}\cdot {{m}_{2}}=\displaystyle \frac{2}{3}\times \displaystyle \frac{3}{2}=1\ne -1\\\ \ \ \therefore \ \ {{l}_{1}}\ \text{and}\ {{l}_{2}}\ \text{are neither parallel nor }\\\ \ \ \ \ \ \text{perpendicular to each other}\text{.}\\\\(\text{c})\ \ \ \ {{l}_{1}}:y=3x-2\ \text{and}\ {{l}_{2}}:y=-\displaystyle \frac{1}{3}x+9\\\ \ \ \therefore \ \ {{m}_{1}}=3\ \text{and}\ {{m}_{2}}=-\displaystyle \frac{1}{3}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\nparallel \ {{l}_{2}}\\\ \ \ \ \ \ \text{But,}\ {{m}_{1}}\cdot {{m}_{2}}=3\left( {-\displaystyle \frac{1}{3}} \right)=-1\\\ \ \ \therefore \ \ {{l}_{1}}\bot \ {{l}_{2}}\\\\(\text{d})\ \ \ \ {{l}_{1}}:y=\displaystyle \frac{2}{3}x-5\ \text{and}\ {{l}_{2}}:y=-\displaystyle \frac{3}{2}x-5\\\ \ \ \therefore \ \ {{m}_{1}}=\displaystyle \frac{2}{3}\ \text{and}\ {{m}_{2}}=-\displaystyle \frac{3}{2}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\nparallel \ {{l}_{2}}\\\ \ \ \ \ \ \text{But,}\ {{m}_{1}}\cdot {{m}_{2}}=\displaystyle \frac{2}{3}\left( {-\displaystyle \frac{3}{2}} \right)=-1\\\ \ \ \therefore \ \ {{l}_{1}}\bot \ {{l}_{2}}\end{array}$

11.         Find the equation of the line which is parallel to the line:

(a) with equation $y=4x+2$ and passes through $(0,8)$.

(b) with equation $y=-x+3$ and passes through ($0,5)$.

(c) with equation $y=-2x-3$ and passes through $(0,-7)$.

(d) with equation $y=-\displaystyle \frac{4}{5}x-3$ and passes through $\displaystyle \left(0,\frac{1}{2}\right)$.

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$\begin{array}{l}(\text{a})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=4x+2\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=4\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=4\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,8)}\ \text{is}\\\\\ \ \ \ y-8=4(x-0)\\\\\ \ \ \ y=4x+8\\\\(\text{b})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-x+3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=-1.\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,5)}\ \text{is}\\\\\ \ \ \ y-5=-(x-0)\\\\\ \ \ \ y=-x+5\\\\(\text{c})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-2x-3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-2\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}},\ {{m}_{2}}=-2.\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}-\text{7)}\ \text{is}\\\\\ \ \ \ y-(-7)=-2(x-0)\\\\\ \ \ \ y=-2x-7\\\\(\text{d})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-\displaystyle \frac{4}{5}x-3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-\displaystyle \frac{4}{5}\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{4}{5}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}\displaystyle \frac{1}{2}\text{)}\ \text{is}\\\\\ \ \ \ y-\displaystyle \frac{1}{2}=-\displaystyle \frac{4}{5}(x-0)\\\\\ \ \ \ y=-\displaystyle \frac{4}{5}x+\displaystyle \frac{1}{2}\end{array}$

12.         Find the equation of the line which is perpendicular to the line:

(a) with equation $y=5x-4$ and passes through $(0,7)$

(b) with equation $y=-x+7$ and passes through $(0,4)$

(c) with equation $y=-2x+3$ and passes through $(0,-4)$

(d) with equation $\displaystyle y=x-\frac{3}{2}$ and passes through $\displaystyle \left( {0,\frac{5}{4}} \right)$

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$\begin{array}{l}(\text{a})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=5x-4\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=5\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=-\displaystyle \frac{1}{5}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,7)}\ \text{is}\\\\\ \ \ \ y-7=-\displaystyle \frac{1}{5}(x-0)\\\\\ \ \ \ y=-\displaystyle \frac{1}{5}x+7\\\\(\text{b})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-x+7\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=1\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,4)}\ \text{is}\\\\\ \ \ \ y-4=1(x-0)\\\\\ \ \ \ y=x+4\\\\(\text{c})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-2x+3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-2\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=\displaystyle \frac{1}{2}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}-\text{4)}\ \text{is}\\\\\ \ \ \ y-(-4)=\displaystyle \frac{1}{2}(x-0)\\\\\ \ \ \ y=\displaystyle \frac{1}{2}x-4\\\\(\text{d})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=x-\displaystyle \frac{3}{2}\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=-1\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point }\left( {\text{0,}\displaystyle \frac{5}{4}} \right)\ \text{is}\\\\\ \ \ \ y-\displaystyle \frac{5}{4}=-1(x-0)\\\\\ \ \ \ y=-x+\displaystyle \frac{5}{4}\end{array}$

13.         Show that the line through $(3n,0)$ and $(0,7n)$ is parallel to the line through $(0,21n)$ and $(9n,0)$.

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$\begin{array}{l}\text{Let }{{l}_{1}}\ \text{contain }(3n,\ 0)\ \text{and}\ (0,\ 7n)\ \\\text{and}\ {{l}_{2}}\ \text{contain }(0,\ 21n)\ \text{and}\ (9n,\ 0).\\\\\therefore \ {{m}_{1}}=\displaystyle \frac{{7n-0}}{{0-3n}}=\displaystyle \frac{{7n}}{{-3n}}=-\displaystyle \frac{7}{3}\\\\\therefore \ {{m}_{2}}=\displaystyle \frac{{0-21n}}{{9n-0}}=\displaystyle \frac{{-21n}}{{9n}}=-\displaystyle \frac{7}{3}\\\\\therefore \ {{m}_{1}}={{m}_{2}}\\\\\therefore \ {{l}_{1}}\parallel {{l}_{2}}\end{array}$

14.         Prove that the triangle whose vertices are $H(-12,1)$, $K(9,3)$ and $M(11,-18)$ is a right triangle.

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$\begin{array}{l}H=(-12,1),\ K=(9,3),\ M=(11,-18)\\\\{{m}_{{HK}}}=\displaystyle \frac{{3-1}}{{9-(-12)}}=\displaystyle \frac{2}{{21}}\\\\{{m}_{{KM}}}=\displaystyle \frac{{-18-3}}{{11-9}}=-\displaystyle \frac{{21}}{2}\\\\\therefore {{m}_{{HK}}}\cdot {{m}_{{KM}}}=\displaystyle \frac{2}{{21}}\left( {-\displaystyle \frac{{21}}{2}} \right)=-1\\\\\therefore \ HK\bot KM\\\\\therefore \ \triangle HKM\ \text{is a right triangle}\text{.}\end{array}$

15.        Given the points $P(1,2)$, $Q(5,-6)$ and $R(b,b)$, determine the value of $b$ so that angle $PQR$ is a right angle.

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$\begin{array}{l}P=(1,2),\ Q=(5,-6),\ R=(b,b)\\\\{{m}_{{PQ}}}=\displaystyle \frac{{-6-2}}{{5-1}}=-2\\\\{{m}_{{QR}}}=\displaystyle \frac{{b-(-6)}}{{b-5}}=\displaystyle \frac{{b+6}}{{b-5}}\\\\\text{Since}\ \angle PQR\text{ is a right angle,}\\\\\therefore {{m}_{{PQ}}}\cdot {{m}_{{QR}}}=-1\\\\\therefore \ -2\left( {\displaystyle \frac{{b+6}}{{b-5}}} \right)=-1\\\\\therefore \ 2b+12=b-5\\\\\therefore b=17\end{array}$

16.        A right-angled isosceles triangle has vertices at $(0,5)$, $(5,0)$ and $(-5,0)$. Find the equation of each of the three sides.

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$\begin{array}{l}\ \ \ \ \text{Let}\ A=(0,\ 5),\ B=(-5,\ 0)\ \text{and}\ C=(5,\ 0).\\\\\therefore \ \ {{m}_{{AB}}}=\displaystyle \frac{{0-5}}{{-5-0}}\ =1\\\\\therefore \ \ \text{The equation of the line through }A\ \text{and}\ B\ \text{is}\\\ \ \text{ }\ y-5=1(x-0)\Rightarrow y=x+5\\\\\,\ \ \ \ {{m}_{{AC}}}=\displaystyle \frac{{0-5}}{{5-0}}\ =-1\\\\\therefore \ \ \text{The equation of the line through }A\ \text{and}\ C\ \text{is}\\\ \ \text{ }\ y-5=-1(x-0)\Rightarrow y=-x+5\\\\\ \ \ \ \text{Since }B\ \text{and}\ C\ \text{lie on }x\text{-axis, the equation }\\\ \ \ \ \text{of the line through }B\ \text{and}\ C\ \text{is}\ y=0.\end{array}$

17.         Determine the slope of each side of the quadrilateral whose vertices are $A(5,6)$, $B(13,6)$, $C(11,2)$ and $D(1,2)$. Can you tell what kind of a quadrilateral it is?

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$\begin{array}{l}A=(5,6),\ B=(13,6),\ C=(11,2),\ D=(1,2)\\\\\text{The slope of the line passing through }\\({{x}_{1}},\ {{y}_{1}})\ \text{and}\ ({{x}_{2}},\ {{y}_{2}})\ \text{is}\ m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}.\\\\\therefore \ \ {{m}_{{AB}}}=\displaystyle \frac{{6-6}}{{13-5}}=0.\\\\\,\ \ \ \ {{m}_{{BC}}}=\displaystyle \frac{{2-6}}{{11-13}}=2.\\\\\,\ \ \ \ {{m}_{{CD}}}=\displaystyle \frac{{2-2}}{{1-13}}=0.\\\\\,\ \ \ \ {{m}_{{AD}}}=\displaystyle \frac{{2-6}}{{1-5}}=1.\\\\\therefore \ \ {{m}_{{AB}}}={{m}_{{CD}}}\\\\\therefore \ \ AB\parallel CD\\\\\therefore \ \ ABCD\ \text{is a trapezium}\text{.}\end{array}$

18.         Given the points $D(-4,6)$, $E(1,1)$, and $F(4,6)$, find the slopes of $DE$ and $EF$. Are the points $D$, $E$ and $F$ collinear, explain why?

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$\begin{array}{l}D=(-4,6),\ E=(1,1),\ F=(4,6)\\\\\text{The slope of the line passing through }\\({{x}_{1}},\ {{y}_{1}})\ \text{and}\ ({{x}_{2}},\ {{y}_{2}})\ \text{is}\ m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}.\\\\\therefore \ \ {{m}_{{DE}}}=\displaystyle \frac{{1-6}}{{1-(-4)}}=-\displaystyle \frac{5}{5}=-1\\\\\,\ \ \ \ {{m}_{{EF}}}=\displaystyle \frac{{6-1}}{{4-1}}=\displaystyle \frac{5}{3}\\\\\therefore \ \ {{m}_{{DE}}}\ne {{m}_{{EF}}}\\\\\therefore \ \ D,\ E\ \text{and}\ F\ \text{are not collinear}\text{.}\end{array}$

19.        Prove that the quadrilateral with vertices $A(-2,2)$, $B(2,-2)$, $C(4,2)$ and $D(2,4)$ is a trapezoid with perpendicular diagonals.

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$D=(-4,6),\ E=(1,1),\ F=(4,6)$

The slope of the line passing through $({{x}_{1}},\ {{y}_{1}})$ and $({{x}_{2}},\ {{y}_{2}})$ is $m=\frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$.

\begin{aligned} \therefore m_{AB}&=\displaystyle \frac{-2-2}{2+2} = -1\\\\ m_{BC}&=\displaystyle \frac{2+2}{4-2} =2\\\\ m_{CD}&=\displaystyle \frac{4-2}{2-4} = -1\\\\ m_{AD}&=\displaystyle \frac{4-2}{2+2} = \displaystyle \frac{1}{2}\\\\ m_{AC}&=\displaystyle \frac{2-2}{4+2} = 0\\\\ m_{BD}&=\displaystyle \frac{4+2}{2-2} =\text{undefined} \end{aligned}

Since $m_{AB}=m_{CD},\ AB\parallel CD.$

$m_{AC} = 0\Rightarrow AC$ is a horizontal line.

$m_{BD} =\text{undefined}\Rightarrow BD$ is a vertical line.

$\therefore AC\bot BD$.

Hence, the quadrilateral $ABCD$ is a trapezoid with perpendicular diagonals.

20.         Find the slopes of the six lines determined by the points $A(-5,4)$, $B(3,5)$, $C(7,-2)$, $D(-1,-3)$. Prove that $ABCD$ is a rhombus.

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$A=(-5,4),\ B=(3,5),\ C=(7,-2),\ D=(-1,-3)$

The slope of the line passing through $({{x}_{1}},\ {{y}_{1}})$ and $({{x}_{2}},\ {{y}_{2}})$ is $m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$.

\begin{aligned} m_{AB}&=\displaystyle \frac{5-4}{3+5}=\displaystyle \frac{1}{8}\\\\ m_{BC}&=\displaystyle \frac{-2-5}{7-3} =-\displaystyle \frac{7}{4} \\\\ m_{CD}&=\displaystyle \frac{-3+2}{-1-7} = \displaystyle \frac{1}{8}\\\\ m_{AD}&=\displaystyle \frac{-3-4}{-1+5} = -\displaystyle \frac{7}{4}\\\\ m_{AC}&=\displaystyle \frac{-2-4}{7+5} = -\displaystyle \frac{1}{2}\\\\\ m_{BD}&=\displaystyle \frac{-3-5}{-1-3}=2 \end{aligned}

$\therefore\ m_{AB}=m_{CD}\Rightarrow AB\parallel CD$

$\quad\ m_{BC}=m_{AD}\Rightarrow BC\parallel AD$

$\quad\ m_{AC}\cdot m_{BD}=-\displaystyle \frac{1}{2}\times 2 = -1\Rightarrow AC\bot BD$

Hence $ABCD$ is a rhombus.