Exercise 11 C
Question 1
Ans: (i) $\begin{aligned}& 3 x+4 y=9 \\\Rightarrow & 4 y=-3 x+9 \\\Rightarrow & y=-\frac{3}{4} x+\frac{9}{4}\end{aligned}$
which is in the form of $y=m x+c$
So slope $(m)=-\frac{3}{4}$
(ii) $3 x+2 y=4$
$\Rightarrow 2 y=-3 x+4$
$\Rightarrow y=\frac{-3}{2} x+\frac{4}{2}$
$\Rightarrow y=-\frac{3}{2} x+2$
so slope(m) $=\frac{-3}{2}$
Question 2
Ans: (i)
$\begin{aligned} & 3 x+2 y+4=0 \\ \Rightarrow & 2 y=-3 x-4 \\ \Rightarrow & y=-\frac{3}{2} x-2 \end{aligned}$
which is in the form of $y=m x+c$
(ii) slope (m) = $\frac{-3}{2}$ and y- intersect (c) = -2
(iii) The line has been drawn on the graph paper as given here.
(IMAGE TO BE ADDED)
Question 3
Ans: we know that the equation of a line is $y=m x+c$ where $m$ in the slope and $c$ is the $y$. intercept
Therefore
(a) $y$-intercept $(c)=2$
slope $(m)=7$
So equation of the line will be
$y=7 x+2 \Rightarrow 7 x-y+2=0$
(b) slope $(m)=-4$
and $y$ - intercept $(c)=-3$
So Equation on the line will be
$y=-4 x-3 \text { or } 4 x+y+3=0$
(c) y- intercept (c) = -1
If the line is parallel to the line y = 5x-7
slope(m) = 5
So the required line will be y = 5x-1
(d) y - intercept (c) = 2
So The equation inclined at $45^{\circ}$ to the $x$-axis So slope $(m)=\tan \theta=\tan 45^{\circ}=1$
So Equation of the required line will be
$\begin{aligned}&y=1 \quad x+2 \Rightarrow y=x+2 \\&x-y+2=0\end{aligned}$
(e) $y$ - intercept (c)=-5$
if The line is equally inclined to the axis
So slope $(m)=\tan \theta=\pm 1$
So Equation of the line will be
$\begin{aligned}&y=1 x-5 \Rightarrow y=x-5 \\&x-y=5 \\&\text { or } y=-1 x-5 \Rightarrow y=-x-5 \\&\text { or } x+y+5=0\end{aligned}$
Question 4
Ans: The equation of line is
$y=m x+c \quad-i 1$
so it passes through the points $(3,-4)$ and(-1,2)
Substituting the value of x and y in (i) -4 = 3m+ c................(ii)
and $2=-1 m+c \Rightarrow 2=-m+c$........(iii)
Subtracting we get ,
$-6=4 m$
$\Rightarrow m=\frac{-6}{4}=\frac{-3}{2}$
From (ii)
$\begin{aligned}-4 &=3 \times\left(\frac{-3}{2}\right)+c \\ & \Rightarrow-4=-\frac{9}{2}+c \\ & \Rightarrow c=-4+\frac{9}{2} \\ &=\frac{-8+9}{2}=\frac{1}{2} \end{aligned}$
So $m=-\frac{3}{2} \text { and } c=\frac{1}{2}$
Question 5
Ans: IF The line y = mx+c passes through the points $(1,4)$ and $(-2,-5)$
So substituting the value of x and y , we get
$4=1 m+c$
$\Rightarrow 4=m+c $............(i)
$-5=-2 m+c$
$\Rightarrow-5=-2 m+c$.........(ii)
Subtracting , we get
9=3m = $m=\frac{9}{3}=3$
From (i)
$\begin{aligned}&4=3+c \\&\Rightarrow c=4-3=1\end{aligned}$
Hence $m=3$ and $c=1$
Question 6
Ans: (a) The equation of the line will be
$y-y_{1}=m\left(x-x_{1}\right)$
= Here it passes through p (-4, 7) and m = $-\sqrt{3}$
So $y-7=-\sqrt{3}[x-(-4)]$
$\begin{aligned}&\Rightarrow y-7=-\sqrt{3} x-4 \sqrt{3} \\&\Rightarrow \sqrt{3} x+y=7-4 \sqrt{3}\end{aligned}$
(b) So the line passes through the point p (-1,-5)
and has slope $=\frac{-6}{11}$
So equation of the line will be
$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y+5=\frac{-6}{11}(x+1) \\ \Rightarrow & 11 y+55=-6 x-6 \\ \Rightarrow & 6 x+11 y+55+6=0 \\ \Rightarrow & 6 x+11 y+61=0 \end{aligned}$
Question 7
Ans: (a) Slope of the line $x+2 y=4 \Rightarrow 2 y=-x+4$
$\Rightarrow y=-\frac{1}{2} x+2$
$m_{1}=\frac{-1}{2}$ and slope of the line perpendicular to it $\left(-\frac{1}{m}\right)=2$
if The required equation parses through the origin ( 0,0)
So equation of the line will be
$\begin{aligned}& y-y_{1}=m\left(x-x_{1}\right) \\\Rightarrow & y-0=2(x-0) \\\Rightarrow & y=2 x \\\Rightarrow & 2 x-y=0\end{aligned}$
(b) Slope of the given line $3 x+4 y=12$
$\begin{aligned}\Rightarrow 4 y &=-3 x+12 \\\Rightarrow y &=-\frac{3}{4} x+3 \\m &=-\frac{3}{4}\end{aligned}$
if it passes through the points $(4,3)$ So equation of the line will be
$\begin{aligned}y-y_{1} &=m\left(x-x_{1}\right) \\\Rightarrow y-3 &=\frac{-3}{4}(x-4)\end{aligned}$
$\Rightarrow 4 y-12=-3 x+12$
$\Rightarrow 3 x+4 y=12+12$
$\Rightarrow 3 x+4 y=24$
$\Rightarrow 3 x+4 y-24=0$
(c) Slope of the given line $3 x-2 y+5=0$
$\begin{gathered}\Rightarrow 2 y=3 x+5 \\\Rightarrow y=\frac{3}{2} x+\frac{5}{2} \\m=\frac{3}{2}\end{gathered}$
(i) Slope as the required line which as parallel to the given line $=m=\frac{3}{2}$
IF it passes through the point (4,5)
So equation as the line will be
$y-y_{1}=m\left(x-x_{1}\right)$
$\Rightarrow y-5=\frac{3}{2}(x-4) \Rightarrow 2 y-10=3 x-12$
$\Rightarrow 3 x-2 y+10-12=0$
$\Rightarrow 3 x-2 y-2=0$
(ii) slope as the line perpendicular to the given line = $\frac{-1}{m}$
$m_{1}=\frac{-2}{3}$
if it passes through the points (4,5)
so equation of the line will be
$y-y_{1}=m_{1}\left(x-x_{1}\right)$
$\Rightarrow y-5=\frac{-2}{3}(x-4)$
$\Rightarrow 3 y-15=-2 x+8 .$
$\Rightarrow 2 x+3 y-15-8=0$
$\Rightarrow 2 x+3 y-23=0$
$\Rightarrow 2 x+3 y=23$
Question 8
Ans: The equation of a line which pares through two pointe is $y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$
(a) If the required line passes through the points $A(1,1)$ and $B(2,3)$
So Equation of the lines will be
$\begin{aligned}&y-1=\frac{3-1}{2-1}(x-1) \Rightarrow y-1=\frac{2}{1}(x-1) \\&\Rightarrow y-1=2 x-2 \\&\Rightarrow 2 x-y-2+1=0 . \\&\Rightarrow 2 x-y-1=0\end{aligned}$
(b) If the required line passes through the points
$P(3,3)$ and $Q(7,6)$
So Equation of the line w/11 be
$\begin{aligned}y-y_{1} &=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right) \\\Rightarrow y-3 &=\frac{6-3}{7-3}(x-3)\end{aligned}$
$\begin{aligned} \Rightarrow y-3 &=\frac{3}{4}(x-3) \Rightarrow 4 y-12=3 x-9 \\ 3 x-4 y+12-9 &=0 \\ \Rightarrow 3 x-4 y+3 &=0 \end{aligned}$
(c) If the required line passes through L(a,b) and M(b,a)
So equation of the line will be
$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$
$y-b=\frac{a-b}{b-a}(x-a) \Rightarrow y-b=\frac{(a-b)}{(a-b)}(x-a)$
$\Rightarrow y-b=-1(x-a)$
$\Rightarrow y-b=-x+a$
$\Rightarrow x+y=a+b$
Question 9
Ans: Slope of line $3 x+4 y=8$
$\begin{aligned}&\Rightarrow 4 y=-3 x+8 \\&\Rightarrow y=-\frac{3}{4} x+2 \\&\text { So }\left(m_{1}\right)=\frac{-3}{4}\end{aligned}$
and slope of line Px +2y = 7
$\Rightarrow 2 y=-p x+7$
$y=-\frac{p}{2} x+\frac{7}{2}$
$m_{2}=\frac{-p}{2}$
if the lines are parallel
So Their slope are equal
$\text { so } \begin{aligned}m_{1} &=m_{2} \Rightarrow \frac{-3}{4}=\frac{-p}{2} \\& \Rightarrow p=\frac{3 \times 2}{4}=\frac{3}{2}\end{aligned}$
Question 10
Ans: Co- ordinates of two points are E$(0,4), f(3,7)$
(ii) so slope(m) $=\frac{y_{2}-y_{1}}{n_{2}-x_{1}}=\frac{7-4}{3-0}=\frac{3}{3}=1$
So gradient of EF = 1
(ii) Now equation of EF will be
$y-y_{1}=m\left(x-x_{1}\right)$
$\Rightarrow y-7=1(x-3)$
$\Rightarrow y-7=x-3$
$\Rightarrow x-y+7-3=0$
$\Rightarrow x-y+4=0$
(iii) Let the line EF intersect the x-axis at p then y- co-ordinates of P will be O
If P lines on EF
So it will satisfy the equation of EF
$x-0+4=0$
$\Rightarrow x+4=0$
$\Rightarrow x=-4$
So co-ordinates of P will be (-4,0)
Question 11
Ans: Co-ordinates of points are $P(-2, L), Q(2,2)$ and R(6,-2)
(i) Gradient of QR(M) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-2}{6-2}$
$=\frac{-4}{4}=-1$
(ii) Slope of line perpendicular to QR = $\frac{-1}{m}$
$=-(-1)=1$
So if it is passes through P
So equation of the line will be
$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y-1=1(x+2) \\ \Rightarrow & y-1=x+2 \\ & x-y+2+1=0 \\ \Rightarrow & x-y+3=0 \end{aligned}$
Question 12
Ans: The line $3 x-4 y+12=0$ meets x-axis at P
So y-coordinate of P will be O
Let the co-ordinates of P be (x,0)
if it lies on the $3 x-4 y+12=0$
So it will satisfy the equation
So $3 x-4 \times 0+12=0$
$\begin{aligned}&\Rightarrow 3 x+12=0 \\&\Rightarrow 3 x=-12\end{aligned}$
$\Rightarrow x=\frac{-12}{3}=-4$
So Co-ordinates of P will be (-4,0)
Now slope of the line $3 x+5 y-15=0$
$\Rightarrow 3 y=-3 x+15$
$\Rightarrow y=-\frac{3}{5} x+3$
$(m)=\frac{-3}{5}$
So slope of the line perpendicular to this line
$=-\frac{1}{m}=-\left(\frac{-5}{3}\right)=\frac{5}{3}$
So Equation of the perpendicular line through P will be
$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-0=\frac{5}{3}(x+4)$
$\Rightarrow y=\frac{5}{3}(x+4)$
$\Rightarrow 3 y=5 x+20$
$\Rightarrow 5 x-3 y+20=0$
Question 13
Ans: Point A divides the line joining the point p(5,2) and Q(9,-6) in the ratio 3:1
So $m_{1}: m_{2}=3: 1$
Now co-ordinates of $A$ will be
$\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}{m_{1}+m_{2}}\right)$
$=\left(\frac{3 \times 9 + 1 \times 5}{3+1}, \frac{3 \times 6 \times +1 \times(-21)}{3 \times 1}\right)$
$=\left(\frac{27+5}{4}+\frac{18-2}{4}\right)$ or (8,4)
Slope of the line x =$3 y+4=0$
$\Rightarrow 3 y=x+4$
$\Rightarrow y=\frac{1}{3} x+\frac{4}{3}$ will be $(m)\frac{-1}{3}$
So slope of the line perpendicular to it = $\frac{-1}{m}$
$=-\left(\frac{9}{3}\right)=-3$
So equation of the line through A will be
$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y-4=-3(x-8) \\ \Rightarrow & y-4=-3 x+2+\\ \Rightarrow & 3 x+y=2++4 \\ \Rightarrow & 3 x+y=28 \\ \Rightarrow & y+3 x-28=0 \end{aligned}$
Question 14
Ans: The slope of line x- 2y+ 8=0
$\Rightarrow 2 y=x+8$
$\Rightarrow y=\frac{1}{2} x+4$
$(m)=\frac{1}{2}$
So slope of the line parallel to it = $\frac{1}{2}$
IF it passes through the point (1,2)
So equation of the line will be
$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-2=\frac{1}{2}(x-1)$
$\Rightarrow 2 y-4=x-1$
$\Rightarrow x-2 y+4-1=0$
$\Rightarrow x-2 y+3=0$
Question 15
Ans: The equation of the line is given
$\begin{aligned} & \frac{x}{3}+\frac{y}{4}=1 \\ \Rightarrow & 4 x+3 y=12 \\ \Rightarrow & 3 y=-4 x+12 \\ \Rightarrow & y=-\frac{4}{3} x+4 \end{aligned}$
So Gradient (m) $=\frac{-4}{3}$
and y- intercept (c) = 4
Question 16
Ans: (i) $\quad 2 x-b y+5=0$
$\Rightarrow b y=2 x+5$
$\Rightarrow y=\frac{2}{b} x+\frac{5}{b}$
So slope $\left(m_{1}\right)=\frac{2}{b}$
$a x+3 y=2$
$\Rightarrow 3 y=-a x+2$
$\Rightarrow y=\frac{-a}{3}+\frac{2}{3}$
So Slope $\left(m_{2}\right)=\frac{-a}{3}$
IF these lines are parallel
So
$\begin{aligned} & m_{1}=m_{2} \\ \Rightarrow & \frac{2}{b}=\frac{-a}{3} \\ \Rightarrow &-a b=6 \\ \Rightarrow & a b=-6 \end{aligned}$
(ii) We know that equation of a line us $y=m x+c$ where m is slope and C is y - intercept
If it passes through (1,3) and has y - intercept= 5
So it will also pass through (0,5)
So slope (m) $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-3}{0-1}=\frac{2}{-1}=-2$
So equation of the line
$\begin{aligned}y &=m x+c \\& \Rightarrow y=-2 x+5 \\& \Rightarrow 2 x+y=5\end{aligned}$
Question 17
Ans: (i) Equation of the line is given $\frac{y}{2}=x-p$
If it passes through the point (-4,4)
So $\frac{4}{2}=-4-p$
$\Rightarrow 2=-4-p$
$\Rightarrow p=-4-2$
$\Rightarrow p=-6$
So p= -6
(ii) In equation
$\frac{y}{2}=x-p$
$\Rightarrow y=2 x-2 p$
Slope $\left(m_{1}\right)=2$
and in equation $a x+5=3 y$
$\begin{aligned} \Rightarrow & y=\frac{a}{3} x+\frac{5}{3} \\ & \text { slope }\left(m_{2}\right)=\frac{a}{3} \end{aligned}$
If these two lines are parallel
So $m_{1}=m_{2}$
$\Rightarrow 2=\frac{a}{3}$
$\Rightarrow 6=a$
So $a=6$
Question 18
Ans: The line passes through a point (-2,0) and has its y-intercept = 3
So it will pass through $(0,3)$.
so slope of the line $(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$=\frac{3-0}{0-(-2)}=\frac{3}{2}$
So Equation of the line will be
$\begin{gathered}y=m x+c \\y=\frac{3}{2} x+3 \Rightarrow 2 y=3 x+6\end{gathered}$
Question 19
Ans: (a) In the figure
(i) Co - ordinates of A are (2, 3) of B are (-1,2) and of C are (3,0)
(ii) Equation of a line through A and parallel to BC
Slope of BC (m) $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-2}{3+1}=\frac{-2}{4}$
$=\frac{-1}{2}$
So slope of the required line $=\frac{-1}{2}$
So $y-y_{1}=m\left(x-x_{1}\right)$
$\begin{aligned}&\Rightarrow y-3=-\frac{1}{2}(x-2) \\&\Rightarrow 2 y-6=-x+2 \\&\Rightarrow x+2 y=2+6 \Rightarrow x+2 y=8 \\&\Rightarrow x+2 y-8=0\end{aligned}$
(b) Let ratio be $m_{1}: m_{2}$ Which divides the join of
$A(0,3)$ and $B(4,1)$ by the $x$-axis at $p$
if $p$ lies on $x$-axis
So co-ordinates of $p$ be $(x, 0)$
so $\begin{aligned} & 0=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \Rightarrow \frac{m_{1} \times 1+m_{2} \times 3}{m_{1}+m_{2}}=0 \\ & \frac{m_{1}+3 m_{2}}{m_{1}+m_{2}}=0 \Rightarrow m_{1}+3 m_{2}=0 \\ & \Rightarrow m_{1}=-3 m_{2} \end{aligned}$
$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{-3}{1}$
So ratio $-3: 1$
So and $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{-3 \times 4+1 \times 0}{-3+1}$
$=\frac{-12+0}{-2}=\frac{-12}{-2}=6$
So co- ordinates of P will be (6,0)
Question 20
Ans: In the given Line 2y = 3x+5
$\Rightarrow y=\frac{3}{2} x+\frac{5}{2}$
slope $\left(m_{1}\right)=\frac{3}{2}$
So slope of the line AB perpendicular to the given line $\left(m_{2}\right)=\frac{-1}{m_{1}}=\frac{-2}{3}$
If it passes through (3,2)
So equation of the line will be
$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y-2=-\frac{2}{3}(x-3) \\ \Rightarrow & 3 y-6=-2 x+6 \\ \Rightarrow & 2 x+3 y=6+6 \\ \Rightarrow & 2 x+3 y=12 \\ \Rightarrow & 2 x+3 y-12=0 \end{aligned}$
If this lines meets x-axis at A and y-axis at B
So substituting the value of y = 0 and x =0
If y= 0 ,then
$2 x+0 \times y=12$
$\Rightarrow 2 x=12$
$\Rightarrow x=\frac{12}{2}=6$
If x =0, then
$2 \times 0+3 y=12$
$\Rightarrow 0+3 y=12$
$\Rightarrow 3 y=12$
$\Rightarrow y=\frac{12}{3}=4$
So co-ordinates of A and B will be A (6,0) and B (0,4)
IF 0 is the origin
So area of $\triangle O A B=\frac{1}{2} \times O A \times O B$
$=\frac{1}{2} \times 6 \times 4=12$ Sq.units
Question 21
Ans: (i) Let co-ordinates of point p be (x,y)
If p divides the line joining A $(-4,1)$ and $B(17.10)$
in the ratio $1: 2$ i.e. $m_{1}: m_{2}=1: 2$
$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 17+2 \times(-4)}{1+2}$
$=\frac{17-8}{3}=\frac{9}{3}=3$
and $y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 10+2 \times 1}{1+2}$
$=\frac{10+2}{3}=\frac{12}{3}=4$
So co-ordinates of p arc (3,4)
(ii) If O is the origin (0,0)
$\begin{aligned} O P &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ &=\sqrt{(3-0)^{2}+(4-0)^{2}} \\ &=\sqrt{(3)^{2}+(412} \\ &=\sqrt{9+16}=\sqrt{25}=5 \end{aligned}$
(ii) Let the ratio be $m_{1}: m_{2}$ In which y-axis divides the line segment AB. Let R be the point on y- axis which divides AB in the ratio $m_{1}: m_{2}$
co-ordinales of $R$ be $(0, y)$
$\text { so } x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$
$\begin{aligned} o &=\frac{m_{1} \times 17+m_{2} \times(-4)}{m_{1}+m_{2}} \\ & \Rightarrow 17 m_{1}-4 m_{2}=0 \\ & \Rightarrow 17 m_{1}=4 m_{2} \\ & \Rightarrow \frac{m_{1}}{m_{2}}=\frac{4}{17} \end{aligned}$
So Ratio = 4:17
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