5 Marks Questions

Question 15 Marks

Find equation of the line which is equidistant from parallel lines $9 x+6 y-7=0$ and $3 x+2 y+$ $6=0$.

Answer

The equations of given lines are
$9 x+6 y-7=0\ldots(i)$
$3 x+2 y+6=0\ldots(ii)$
Let $P(h, k)$ be the arbitrary point is equidistant from lines (1) and (2). The perpendicular distance of $P(h, k)$ from line (1) is given by
$d_1 =\left|\frac{9 h+6 k-7}{(9)^2+(6)^2}\right|$
$=\left|\frac{9 h+6 k-7}{\sqrt{117}}\right|$
$=\left|\frac{9 h+6 k-7}{3 \sqrt{13}}\right|$
The perpendicular distance of $P(h, k)$ from line (ii) is given by
$d_2=\left|\frac{3 h+2 k+6}{(3)^2+(2)^2}\right|=\left|\frac{3 h+2 k+6}{\sqrt{13}}\right|$
Since $P(h, k)$ is equidistant from lines (i) and (ii),
Therefore, $\quad d_1=d_2$
$\therefore$ $\left|\frac{9 h+6 k-7}{3 \sqrt{13}}\right|=\left|\frac{3 h+2 k+6}{\sqrt{13}}\right|$
⇒ $|9 h+6 k-7|=3|3 h+2 k+6|$
⇒ $|9 h+6 k-7|= \pm 3(3 h+2 k+6)$
$\text {Either } 9 h+6 k-7=3(3 h+2 k+6)$
$\text {or }\qquad 9 h+6 k-7=-3(3 h+2 k+6)$
The case $9 h+6 k-7=3(3 h+2 k+6)$ is not possible
$\text{as,}\quad 9 h+6 k-7=3(3 h+2 k+6)$
⇒ $-7=18 \quad$ (which is absurd)
$\therefore$ $9 h+6 k-7=-3(3 h+2 k+6)$
$\therefore$ $9 h+6 k-7=-9 h-6 k-18$
⇒ $18 h+12 k+11=0$
Thus, the required equation of the line is
$18 x+12 y+11=0$

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Question 25 Marks

If sum of the perpendicular distance of a variable point $P (x, y)$ from the lines $x+y-5=0$ and $3 x-2 y$ $+7=0$ is always 10 . Show that $P$ must move on a line.

Answer

Equations of the given lines are
$\begin{array}{r}
x+y-5=0\ldots(i) \\
3 x-2 y+7=0\ldots(ii)
\end{array}$
The perpendicular distance of $P(x, y)$ from lines (i) and (ii) are respectively
$\begin{aligned} d_1 & =\frac{|x+y-5|}{\sqrt{(1)^2+(1)^2}} \text { and } d_2=\frac{|3 x-2 y+7|}{\sqrt{(3)^2+(-2)^2}} \\ \text { i.e., } d_1 & =\frac{|x+y-5|}{\sqrt{2}} \text { and } d_2=\frac{|3 x-2 y+7|}{\sqrt{13}}\end{aligned}$
It is given that $d_1+d_2=10$
$\begin{array}{l}
\therefore \frac{|x+y-5|}{\sqrt{2}}+\frac{|3 x-2 y+7|}{\sqrt{13}}=10 \\
\Rightarrow \sqrt{13}|x+y-5|+\sqrt{2}|3 x-2 y+7|-10 \sqrt{26}=0 \\
\Rightarrow \sqrt{13}(x+y-5)+\sqrt{2}(3 x-2 y+7)-10 \sqrt{26}=0
\end{array}$
[Assuming $(x+y-5)$ and $(3 x-2 y+7)$ are positive]
$\Rightarrow \sqrt{13} x+\sqrt{13} y-5 \sqrt{13}+3 \sqrt{2} x-2 \sqrt{2} y+7 \sqrt{2}-10 \sqrt{26}=0$
$\Rightarrow x(\sqrt{13}+3 \sqrt{2})+y(\sqrt{13}-2 \sqrt{2})+(7 \sqrt{2}-5 \sqrt{13}-10 \sqrt{26})=0,$
which is equation of a line.
Similarly, we can obtain the equation of line for any signs of $(x+y-5)$ and $(3 x-2 y+7)$.
Thus, point $P$ must move on a line. Hence Proved.

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Question 35 Marks

Find the equations of the lines through the point of intersection of the lines $x-y+1=0$ and $2 x-3 y$ $+5=0$ and whose distance from the point $(3,2)$ is $\frac{7}{5}.$

Answer

Given equation of lines $x-y+1=0\ldots(i)$
and $\quad 2 x-3 y+5=0\ldots(ii)$
From eq. (i), $\quad x=y-1$
Now, put the value of $x$ in eq. (ii), we get
$2(y-1)-3 y+5=0$
$\Rightarrow$ $2 y-2-3 y+5=0$
$\Rightarrow$ $3-y=0$
$\Rightarrow$ y = 3
Putting $y=3$ in eq. (i), we get
$x=2$
Since, the point of intersection is $(2,3)$.
Let slope of the required line be $m$.
$\therefore$ Equation of line is $y-3=m(x-2)$
$\Rightarrow$ $m x-y+3-2 m=0\ldots(iii)$
Since, the distance from $(3,2)$ to line (iii) is $\frac{7}{5}$.
$\therefore$ $\frac{7}{5}=\left|\frac{3 m-2+3-2 m}{\sqrt{1+m^2}}\right|$
$\Rightarrow$ $\frac{49}{25}=\frac{(m+1)^2}{1+m^2}$
$\Rightarrow$ $49+49 m^2=25\left(m^2+2 m+1\right)$
$\Rightarrow$ $49+49 m^2=25 m^2+50 m+25$
$\begin{array}{l}\Rightarrow \quad 24 m^2-50 m+24=0 \\ \Rightarrow \quad 12 m^2-25 m+12=0\end{array}$
$\therefore$ $m=\frac{25 \pm \sqrt{625-4 \cdot 12 \cdot 12}}{24}$
$\begin{array}{l}=\frac{25 \pm \sqrt{49}}{24} \\ =\frac{25 \pm 7}{24} \\ =\frac{32}{24} \text { or } \frac{18}{24} \\ =\frac{4}{3} \text { or } \frac{3}{4}\end{array}$
$\therefore$ First equation $y-3=\frac{4}{3}(x-2)$
$\Rightarrow$ $3 y-9=4 x-8$
$\Rightarrow$ $4 x-3 y+1=0$
and second equation $y-3=\frac{3}{4}(x-2)$
$\Rightarrow$ $4 y-12=3 x-6$
$\Rightarrow$ $3 x-4 y+6=0$

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Question 45 Marks

Prove that the product of the lengths of the perpendicular drawn from the points $\left(\sqrt{a^2-b^2}, 0\right)$ and $\left(-\sqrt{a^2-b^2}, 0\right)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^2$

Answer

Let $P_1$ and $P_2$ be the perpendicular distances of $\left(\sqrt{a^2-b^2}, 0\right)$ and $\left(-\sqrt{a^2-b^2}, 0\right)$ from $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta$$=1$, then we have
$P_1=\left|\frac{\frac{\sqrt{a^2-b^2}}{a} \cos \theta-1}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}\right|, P_2=\left|\frac{\frac{-\sqrt{a^2-b^2}}{a} \cos \theta-1}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}\right|$
$\begin{aligned} P_1 P_2 & =\left|\frac{\left(\frac{\sqrt{a^2-b^2}}{a} \cos \theta-1\right)\left(\frac{-\sqrt{a^2-b^2}}{a} \cos \theta-1\right.}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}} \sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}\right| \\ & =\left|\frac{-\left(\frac{\sqrt{a^2-b^2}}{a} \cos \theta-1\right)\left(\frac{\sqrt{a^2-b^2}}{a} \cos \theta+1\right.}{\left(\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}\right)}\right|\end{aligned}$
$\begin{array}{l}=\left|\frac{-\left(\frac{a^2-b^2}{a^2} \cos ^2 \theta-1\right)}{\left(\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}\right)}\right| \\=\left|\frac{-\left(\frac{a^2 \cos ^2 \theta-b^2 \cos ^2 \theta-a^2}{a^2}\right)}{\left(\frac{b^2 \cos ^2 \theta+a^2\sin ^2 \theta}{a^2 b^2}\right)}\right| \\=\left|\frac{b^2\left[\left(b^2-a^2\right) \cos ^2 \theta-a^2\right.}{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}\right| \\=\frac{b^2\left\{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right\}}{\left\{a^2 \cos ^2 \theta+a^2 \sin ^2 \theta\right\}} \\=b^2 \quad\quad\text{Hence Proved.}\end{array}$

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Question 55 Marks

Find the equation of a straight line which passes through the point $(a, 0)$ and whose perpendicular distance from $(2 a, 2 a)$ is $a$.

Answer

Let the straight line be $A x+B y+C=0\quad\ldots\text{(i)}$
Since, (i) passes through $(a, 0)$, so
$A \cdot a+B \cdot 0+C=0$
⇒ $A a+C=0$
⇒ $C=-A a\quad\ldots\text{(ii)}$
Also, the $\perp$ distance of (i) from $(2 a, 2 a)$ is $a$.
∴ $a=\left|\frac{A \cdot 2 a+B \cdot 2 a+C}{\sqrt{A^2+B^2}}\right|$
$\begin{array}{l}⇒a=\left|\frac{A a+2 a B}{\sqrt{A^2+B^2}}\right| \quad[C=-A a] \\ ⇒1=\left|\frac{A+2 B}{\sqrt{A^2+B^2}}\right|\end{array}$
$\Rightarrow A^2+B^2 =A^2+4 A B+4 B^2$
$\Rightarrow -3 B^2 =4 A B$
$\Rightarrow 4 A B+3 B^2 =0$
$\Rightarrow B(4 A+3 B) =0$
$\Rightarrow$ $B=0$ or $4 A+3 B=0$
If $ 4 A+3 B=0$
⇒ $B=-\frac{4 A}{3}$
Substituting the values of $B$ and $C$ in eq. (i), we get
$A x-\frac{4}{3} A y-A a=0$
⇒ $3 x-4 y-3 a=0$
$\therefore$ Equation of line is $3 x-4 y-3 a=0$
Similarly, if $B=0$, then substituting the values of $B$ and $C$ in eq. (i), we get
$A x=-0 . y+A a$
⇒ $A(x-a)=0$
⇒ $x-a=0$
$\therefore$ Equation of required line is $x+a=0$.

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Question 65 Marks

Find the equation of the line through the points $(3,2)$ which make an angle of $45^{\circ}$ with the line $x-2 y=3 .$

Answer

Let the slope of the required line be $m_1$.
The given line can be written as
$y=\frac{1}{2} x-\frac{3}{2},$
which is of the form $\quad y=m x+c$
$\therefore$ Slope of the given line, $m_2=\frac{1}{2}$
It is given that the angle between the required line $x-2 y=3$ is $45^{\circ}$.
We know that if $\theta$ is the acute angle between lines $l_1$ and $l_2$ with slopes $m_1$ and $m_2$ respectively, then
$\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$
$\therefore \tan 45^{\circ}=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$
$\Rightarrow 1=\left|\frac{\frac{1}{2}-m_1}{1+\frac{m_1}{2}}\right|$
$\begin{array}{ll}\Rightarrow 1=\left|\frac{\left(\frac{1-2 m_1}{2}\right)}{\frac{2+m_1}{2}}\right| \\ \Rightarrow 1=\left|\frac{1-2 m_1}{2+m_1}\right|\end{array}$
$\begin{array}{ll}\Rightarrow 1= \pm\left(\frac{1-2 m_1}{2+m_1}\right) \\ \Rightarrow 1=\frac{1-2 m_1}{2+m_1}\end{array}$
$\begin{aligned} \text { or } 1 =-\left(\frac{1-2 m_1}{2+m_1}\right) \\ \Rightarrow 2+m_1 =1-2 m_1\end{aligned}$
$\text { or }\quad 2+m_1 =-1+2 m_1$
Case I: $m_1=3$
The equation of the line passing through $(3,2)$ and having a slope of 3 is :
$\begin{aligned} y-2 & =3(x-3) \\ y-2 & =3 x-9 \\ 3 x-y & =7\end{aligned}$
Case II : $m_1=-\frac{1}{3}$
The equation of the line passing through $(3,2)$ and having a slope of $-\frac{1}{3}$ is :
$\begin{aligned} y-2 & =-\frac{1}{3}(x-3) \\ 3 y-6 & =-x+3 \\ x+3 y & =9\end{aligned}$
Thus, the equations of the line are $3 x-y=7$ and $x+3 y=9$.

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Question 75 Marks

If three lines whose equations are $y=m_1 x+c_1$ $y=m_2 x+c_2$ and $y=m_3 x+c_3$ are concurrent, then show that $m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$.

Answer

The equation of the given lines are
$y=m_1 x+c_1\ldots(i)$
$\begin{array}{l}
y=m_2 x+c_2\ldots(ii) \\
y=m_3 x+c_3\ldots(iii)
\end{array}$
On subtracting equation (i) from (ii), we obtain
0 = (m₂ - m₁ x + (c₂ - c₁)
⇒(m₁ - m₂)x = c₂ - c₁
⇒$x=\frac{c_2-c_1}{m_1-m_2}$
On substituting this value of x in (i), we obtain
$y=m_1\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_1$
$y=\frac{m_1 c_2-m_1 c_1}{m_1-m_2}+c_1$
$y=\frac{m_1 c_2-m_1 c_1+m_1 c_1-m_2 c_1}{m_1-m_2}$
$y=\frac{m_1 c_2-m_2 c_1}{m_1-m_2}$
$\therefore \quad\left(\frac{c_2-c_1}{m_1-m_2}, \frac{m_1 c_2-m_2 c_1}{m_1-m_2}\right)$ is the point of intersection of lines (i) and (ii).
It is given that lines (i), (ii) and (iii) are concurrent. Hence, the point of intersection of lines (i) and (ii) will also satisfy equation (iii),
$\frac{m_1 c_2-m_2 c_1}{m_1-m_2}=m_3\left(\frac{c_2-c_1}{m_1-m_2},\right)+c_3$
$\frac{m_1 c_2-m_2 c_1}{m_1-m_2}=\frac{m_3 c_2-m_3 c_1+c_3 m_1-c_3 m_2}{m_1-m_2}$
$\begin{aligned} m_1 c_2-m_2 c_1-m_3 c_2+m_3 c_1-c_3 m_1+c_3 m_2 & =0 \\ m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right) & =0\end{aligned}$
Hence, $m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$.

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Question 85 Marks

Find the value of $k$ for which the line
$(k-3) x-\left(4-k^2\right) y+k^2-7 k+6=0 \text { is }$
(i) Parallel to $X$-axis,
(ii) Parallel to $Y$-axis,
(iii) Passing through the origin.

Answer

The given question of line is
$(k-3) x-\left(4-k^2\right) y+k^2-7 k+6=0\ldots(i)$
(i) If the given line is parallel to the $X$-axis, then Slope of the given line $=$ Slope of the $X$-axis The given line can be written as
$\begin{aligned}\left(4-k^2\right) y & =(k-3) x+k^2-7 k+6 \\ y & =\frac{(k-3)}{\left(4-k^2\right)} x+\frac{k^2-7 k+6}{\left(4-k^2\right)},\end{aligned}$
which is of the form
$y=m x+c$
$\therefore$ Slope of the given line $=\frac{(k-3)}{\left(4-k^2\right)}$
Slope of the $X$-axis $=0$
∴$\frac{(k-3)}{\left(4-k^2\right)}=0$
⇒$k-3=0$
⇒k = 3
Thus, the given line is parallel to $X$-axis, then the value of $k$ is 3 .
(ii) If the given line is parallel to the $Y$-axis, it is vertical. Hence, its slope will be undefined
The slope of the given line is $\frac{(k-3)}{\left(4-k^2\right)}$
Now, $\frac{(k-3)}{\left(4-k^2\right)}$ is undefined at $k^2=4$
$k^2=4$
⇒$k= \pm 2$
Thus, if the given line is parallel to the $Y$-axis, then the value of $k$ is $\pm 2$.
(iii) If the given line is passing through the origin, then point $(0,0)$ satisfies the given equation of the line.
$(k-3)(0)-\left(4-k^2\right)(0) +k^2-7 k+6=0$
$k^2-7 k+6 =0$
$k^2-6 k-k+6 =0$
$(k-6)(k-1) =0$
$k =1 \text { or } 6$
Thus, if the given line is passing through the origin, then the value of $k$ is either 1 or 6.

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Question 95 Marks

Draw a quadrilateral in the cartesian plane, whose vertices are $(-4,5),(0,7),(5,-5)$ and $(-4,-2)$. Also find its area.

Answer

Let the given points be $A(-4,5), B(0,7), C(5,-5)$ and $D(-4,-2) . A, B, C, D$ are plotted on a graph paper. $A$, $B ; B, C ; C, D$ and $D, A$ are joined.
Now, quadrilateral $A B C D$ is divided in two triangles, $\triangle A B D$ and $\triangle B C D$.

Area of $\triangle A B D$
$\begin{array}{l}
=\frac{1}{2}|\{-4(7+2)+0(-2-5)-4(5-7)\}| \\
=\frac{1}{2}|(-36+8)| \\
=14 \text { sq. units }
\end{array}$
area of $\triangle B C D$
$\begin{array}{l}
=\frac{1}{2}|\{0(-5+2)+5(-2-7)-4(7+5)\}| \\
=\frac{1}{2}|-45-48| \\
=\frac{93}{2} \text { sq.units }
\end{array}$
Area of quadrilateral $A B C D=$ Area of $\triangle A B C+$ $\triangle B C D$
$\begin{array}{l}=14+\frac{93}{2} \\ =\frac{121}{2} \\ =60.5 \text { sq. units. }\end{array}$

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Question 105 Marks

The vertices of a triangle are $(6,0),(0,6)$ and $(6,6)$. Find the distance between its circumcentre and centroid.

Answer

Let the vertices of the triangle are A( x₁, y₁)= (6, 0) , B(x₂, y₂) = (0, 6) and C(x₃,y₃) = (6, 6) .