3 Marks Question

Question 13 Marks

Find the equation of straight lines which passes through point of intersection of lines $y-3 x+5$ $=0$ and $y-2 x+2=0$ and is at a distance of $7 / \sqrt{2}$ unit from origin.

Answer

Equation of straight line passing through point of intersection of lines
\[
\begin{array}{ll}
& y-3 x+5=0 \\
\text { and } & y-2 x+2=0 \text { is given by } \\
& (y-3 x+5)+\lambda(y-2 x+2)=0 \\
\Rightarrow \quad & y(1+\lambda)-x(3+2 \lambda)+5+2 \lambda=0
\end{array}
\]
Length of perpendicular drawn from point $(0,0)$ on
\[
\begin{array}{l}
\text { line }(1)=\frac{7}{\sqrt{2}} \quad \text { (Given) } \\
\Rightarrow \quad \frac{0-0+5+2 \lambda}{\sqrt{(1+\lambda)^2+(3+2 \lambda)^2}}=\frac{7}{\sqrt{2}} \\
\Rightarrow \quad \frac{5+2 \lambda}{\sqrt{1+\lambda^2+2 \lambda+9+4 \lambda^2+12 \lambda}}=\frac{7}{\sqrt{2}}
\end{array}
\]
On squaring both sides
\[
\begin{array}{ll}
\Rightarrow & \frac{(5+2 \lambda)^2}{5 \lambda^2+14 \lambda+10}=\frac{49}{2} \\
\Rightarrow & 2\left(25+4 \lambda^2+20 \lambda\right)=49\left(5 \lambda^2+14 \lambda+10\right) \\
\Rightarrow & 50+8 \lambda^2+40 \lambda=245 \lambda^2+686 \lambda+490 \\
\Rightarrow & 245 \lambda^2-8 \lambda^2+686 \lambda-40 \lambda+490-50=0 \\
\Rightarrow & 237 \lambda^2+646 \lambda+440=0 \\
\Rightarrow & 237 \lambda^2+316 \lambda+330 \lambda+440=0 \\
\Rightarrow & 79 \lambda(3 \lambda+4)+110(3 \lambda+4)=0 \\
\Rightarrow & (79 \lambda+110)(3 \lambda+4)=0 \\
\Rightarrow & \lambda=\frac{-110}{79} \text { and }-\frac{4}{3}
\end{array}
\]
On putting value of $\lambda$ in equation (1) :
\[
\begin{aligned}
& y\left(1-\frac{110}{79}\right)-x\left(3-\frac{2 \times 110}{79}\right)+5-\frac{2 \times 110}{79}=0 \\
\Rightarrow & y\left(\frac{79-110}{79}\right)-x\left(\frac{237-220}{79}\right)+\frac{395-220}{79}=0 \\
\Rightarrow & \frac{-y \times 31}{79}-\frac{x \times 17}{79}+\frac{175}{79}=0 \\
\Rightarrow & -31 y-17 x+175=0 \\
\Rightarrow & 17 x+31 y=175 \text { Ans. }
\end{aligned}
\]
and putting $\lambda=-\frac{4}{3}$
\[
y\left(1-\frac{4}{3}\right)-x\left(3-2 \times \frac{4}{3}\right)+5-2 \times \frac{4}{3}=0
\]
\[
\begin{array}{l}
\Rightarrow \quad y\left(-\frac{1}{3}\right)-x\left(\frac{9-8}{3}\right)+\frac{15-8}{3}=0 \\
\Rightarrow \quad-\frac{y}{3}-\frac{x}{3}+\frac{7}{3}=0 \quad \Rightarrow-y-x+7=0 \\
\Rightarrow \quad x+y=7
\end{array}
\]
Hence, equations (2) and (3) are required equations of lines.

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Question 23 Marks

Find the distance of point $(1,2)$ from straight line $3 x+y+4=0$, when it is drawn parallel to line $3 x-4 y+8=0$.

Answer

Equations of straight lines
$\quad$$\quad$$3 x+y+4=0$.........(1)
and $\quad$$3 x-4 y+8=0$

According to question, We need to find the distance of line (1) from point (1, 2) parallel to line (2).
Now, equation of line parallel to line (2) :
$3 x-4 y+k=0$........(3)
This line passes through piont $(1,2)$. So on the satisfying the coordinates:
$\Rightarrow \quad 3 \times 1-4 \times 2+k=0$
$\Rightarrow \quad k=5$
Now putting value of $k$ in equation (3):
$3 x-4 y+5=0$
It is a straight line which is parallel to line (2) and passes through point $(1,2)$.
Now on solving equations (1) and (3),
$\begin{array}{r}3 x+y+4=0 \\ 3 x-4 y+5=0\end{array}$
$\frac{x}{5+16}=\frac{-y}{15-12}=\frac{1}{-12-3}$
$\begin{array}{ll}\Rightarrow & \frac{x}{21}=\frac{y}{-3}=\frac{1}{-15} \\ \Rightarrow & \frac{x}{7}=\frac{y}{-1}=\frac{1}{-5}\end{array}$
$\therefore \quad x=\frac{-7}{5}$ and $y=\frac{1}{5}$
Hence, on solving line (1) and line (3) the point of intersection is $\left(\frac{-7}{5}, \frac{1}{5}\right)$.
So required distance $=A B$
$=$ distance between $(1,2)$ and mid-point $\left(\frac{-7}{5}, \frac{1}{5}\right)$
$=\sqrt{\left(1+\frac{7}{5}\right)^2+\left(2-\frac{1}{5}\right)^2}$
$=\sqrt{\frac{144}{25}+\frac{81}{25}}=\sqrt{\frac{225}{25}}=\frac{15}{5}=3$
Hence, required distance $=3$ unit

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Question 33 Marks

Find the equation of the straight line which meets the foot of perpendicular drawn from the origin to the lines $3 x-4 y=25$ and $3 x+5 y=17$.

Answer

Given equation of lines
$\quad$$\quad$$\quad$$3 x-4 y=25$.......(1)
and$\quad$ $3 x+5 y=17$......(2)
Equations of lines passing through origin $(0,0)$ and perpendicular to lines (1) and (2) are respectively
$\quad$$\quad$$4 x+3 y=0$........(3)
and$\quad$$5 x-3 y=0$....(4)
Point of intersection of lines $(1)$ and $(3)$ is $(3,-4)$ and
point of intersection of lines (2) and (4) is $\left(\frac{3}{2}, \frac{5}{2}\right)$
According to question we need to find the equation of
line passing through points $(3,-4)$ and $\left(\frac{3}{2}, \frac{5}{2}\right)$.
$\therefore$ Equation of the required line
$\begin{aligned} & (y+4)=\frac{\frac{5}{2}+4}{\frac{3}{2}-3}(x-3) \\ \Rightarrow \quad & (y+4)=\frac{\frac{13}{2}}{\frac{-3}{2}}(x-3)\end{aligned}$
$\begin{array}{lrl}\Rightarrow & (y+4) & =\frac{-13}{3}(x-3) \\ \Rightarrow & 3 y+12 & =-13 x+39 \\ \Rightarrow & 13 x+3 y-27 & =0 \\ \Rightarrow & 13 x+3 y & =27 \quad \text {}\end{array}$

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Question 43 Marks

Prove that perpendiculars drawn from any point taken on line $7 x+4 y=3$ to the lines $3 x-4 y=$ 2 and $5 x-12 y=4$ are equal in length.

Answer

Let $\left(x_1, y_1\right)$ be any point taken on line
So,
\[
\begin{array}{r}
7 x+4 y=3 \\
7 x_1+4 y_1=3
\end{array}
\]
\[
\text { or } \quad 7 x_1+4 y_1-3=0
\]
Let the length of the perpendicular drawn from point $\left(x_1, y_1\right)$ be P .
\[
\therefore \quad P=\left|\frac{3 x_1-4 y_1-2}{\sqrt{(3)^2+(-4)^2}}\right|=\left|\frac{3 x_1-4 y_1-2}{5}\right|
\]
Similarly, if the length of perpendicular drawn from point $\left(x_1, y_1\right)$ on line $5 x-12 y=4$ be $P ^{\prime}$, then,
\[
P^{\prime}=\left|\frac{5 x_1-12 y_1-4}{\sqrt{(5)^2+(-12)^2}}\right|=\left|\frac{5 x_1-12 y_1-4}{13}\right|
\]
but $P = P ^{\prime}$
\[
\begin{array}{rlrl}
& \begin{aligned}
\therefore & \left|\frac{3 x_1-4 y_1-2}{5}\right|
\end{aligned} & =\left|\frac{5 x_1-12 y_1-4}{13}\right| \\
& \text { or } & \frac{3 x_1-4 y_1-2}{5} & = \pm \frac{5 x_1-12 y_1-4}{13} \\
{[\because \text { If }|x|} & =|y| \Rightarrow x= \pm y]
\end{array}
\]
Taking +ive sign,
\[
\begin{aligned}
& & 39 x_1-52 y_1-26 & =25 x_1-60 y_1-20 \\
& \Rightarrow & 14 x_1+8 y_1-6 & =0 \\
\Rightarrow & & 7 x_1+4 y_1-3 & =0
\end{aligned}
\]
It is the same relation as defined in equation (1). So it is true.
$\therefore$ Length of perpendiculars drawn from any point taken on line $7 x+4 y=3$ to the lines $3 x-4 y=2$ and $5 x-12 y=4$ are equal in length.
Hence proved.

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Question 53 Marks

Find the equation of lines passing through point $(0, a)$ on which the perpendicular drawn from the point $(2 a, 2 a)$ is of length $a$.

Answer

Let the slope of line passing through point $(0, a)$ is $m$.
The equation of line$\quad$$\quad$$y-a=m x$
$\Rightarrow \quad m x-y+a=0$.........(1)
According to question, the length of perpendicular drawn from point $(2 a, 2 a)=a$
$\begin{array}{ll}\Rightarrow & \frac{|2 a m-2 a+a|}{\sqrt{(m)^2+(-1)^2}}=a \\ \Rightarrow & \frac{|2 a m-a|}{\sqrt{m^2+1}}=a \\ \Rightarrow & \frac{|(2 m-1)|}{\sqrt{m^2+1}}=1 \\ \Rightarrow & |(2 m-1)|=\sqrt{m^2+1}\end{array}$
On squaring both sides,
$(2 m-1)^2=m^2+1$
$\begin{array}{ll}\Rightarrow & 4 m^2-4 m+1=m^2+1 \\ \Rightarrow & 3 m^2-4 m=0\end{array}$
$\Rightarrow \quad m(3 m-4)=0$
$\Rightarrow \quad m=0, m=\frac{4}{3}$
(i) When $m=0$, then equation of line is required
$\quad$$\quad$$y-a=0$[From equation (1)]
(ii) When $m=\frac{4}{3}$, then equation of the required line is
$\frac{4 x}{3}-y+a=0$
$\Rightarrow \quad 4 x-3 y+3 a=0$

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Question 63 Marks

Find the equation of the straight line which joints the point $(3,5)$ to the point of intersection of lines $4 x+y=1$ and $7 x-3 y-35=0$. Prove that this line is at equal distance from origin and point $(8,34)$.

Answer

Equation of the which passes through point of intersection of lines $\quad$$\quad$4 x+y-1=0
and $\quad$$\quad$7x-3y-35=0
$\quad$$(4 x+y-1)+\lambda(7 x-3 y-35)=0$
$\Rightarrow \quad 4 x+y-1+7 \lambda x-3 \lambda y-35 \lambda=0$
$\Rightarrow \quad x(4+7 \lambda)+y(1-3 \lambda)-1-35 \lambda=0$
$\Rightarrow \quad(4+7 \lambda) x+(1-3 \lambda) y-1-35 \lambda=0 \ldots.....(1)$
This line passes through point (3,5). So on satisfying point(3,5),
$(4+7 \lambda) \times 3+(1-3 \lambda) \times 5-1-35 \lambda=0$
$\Rightarrow \quad 12+21 \lambda+5-15 \lambda-1-35 \lambda=0$
$\Rightarrow \quad-29 \lambda+16=0$
$\Rightarrow \quad \lambda=\frac{16}{29}$
Putting this value of $\lambda$ in equation (1):
$\left(4+7 \times \frac{16}{29}\right) x+\left(1-3 \times \frac{16}{29}\right) y-1-35 \times \frac{16}{29}=0$
$\Rightarrow\left(\frac{116+112}{29}\right) x+\left(\frac{29-48}{29}\right) y-\left(\frac{29+560}{29}\right)=0$
$\Rightarrow \quad \frac{228}{29} x-\frac{19 y}{29}-\frac{589}{29}=0$
$\Rightarrow \quad 228 x-19 y-589=0$
$\Rightarrow \quad 19(12 x-y-31)=0$
$\Rightarrow \quad 12 x-y-31=0 \ldots(2)$
Equation (2) is the required equation of line.
Now distance of point $(0,0)$ from line (2),
$P_1=\left|\frac{0-0-31}{\sqrt{12^2+1^2}}\right|=\frac{31}{\sqrt{145}}$
$=\frac{31}{\sqrt{145}}$ unit
Now distance of point $(8,34)$ from line (2)
$\begin{aligned} P_2 & =\left|\frac{12 \times 8-34-31}{\sqrt{12^2+1^2}}\right| \\ & =\left|\frac{96-65}{\sqrt{145}}\right|\end{aligned}$
$=\frac{31}{\sqrt{145}}$
So,$P_1=P_2$
$\Rightarrow$ Distance of line (2) from origin $(0,0)$
$\quad$$=$ Distance of point $(8,34)$ from line (2).
Hence proved.

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Question 73 Marks

Find the image of point $(2,3)$ w.r.t. line $x-2 y+$ $1 = 0$.

Answer

Equation of given line AB (say)
$\quad$$\quad$$x-2 y+1=0$........(1)

If image of point $(2,3)$ w.r.t. to line $x-2 y+1=0$ be $Q (h, k)$, then line $PQ \perp AB$ and mid-point of line PQ will be at line AB .
Now equation of line perpendicular to line $A B$ is given by
$\quad$$\quad$$2 x+y+k=0$
$\because$ This line passes through point $P (2,3)$. So this point will satisfy the equation of line. Hence,
$4+3+k=0 \Rightarrow k=-7$
So, equation of line PQ
$\quad$$\quad$$2 x+y-7=0$.......(2)
Coordinates of point R of intersection of lines (1) and
$(2)=\left(\frac{13}{5}, \frac{9}{5}\right)$
$\because$ Point R is the mid-point of line PQ .
So,
$\frac{h+2}{2}=\frac{13}{5} \quad$ and $\quad \frac{k+3}{2}=\frac{9}{5}$
$\Rightarrow \quad 5 h+10=26 \quad \Rightarrow \quad 5 k+15=18$
$\Rightarrow \quad 5 h=16 \quad \Rightarrow \quad 5 k=3$
$\Rightarrow \quad h=\frac{16}{5} \quad \Rightarrow \quad k=\frac{3}{5}$
Hence, required point of image $\left(\frac{16}{5}, \frac{3}{5}\right)$

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