MCQ 11 Mark
$A$ point equidistant from the lines $4x + 3y + 10 = 0, 5x – 12y + 26 = 0$ and $7x + 24y – 50 = 0$ is:
- A
$(1, -1)$
- B
$(1, 1)$
- ✓
$(0, 0)$
- D
$(0, 1)$
AnswerCorrect option: C. $(0, 0)$
Given equation are
$4x + 3y + 10 = 0 .....(i)$
$5x - 12y + 26 = 0 .....(ii)$
and $7x + 27y - 50 = 0 .....(iii)$
Let ($x_1, y_1)$ be any point equidistant from $eq. (i), eq. (ii)$ and $eq. (iii).$
Distance of $(x_1, y_1)$ from $eq. (i)$
$=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{\sqrt{16+9}}\Big|=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|$
Distance of $(x_1, y_1)$ from $eq. (iii)$
$=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{\sqrt{25+144}}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{13}\Big|$
Distance of $(x_1, y_1)$ from $eq. (iii)$
$=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{\sqrt{49+576}}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$
If the point $(x_1, y_1)$ is equidistant from the given lines$,$ then
$\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_!+26}{13}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$
We see that putting $x_1 = 0$ and $y_1 = 0,$ the above relation is satisfied i.e.,
$=\frac{10}{5}=\frac{26}{13}=\frac{50}{25}=2$
Hence$,$ the correct option is $(c)$.
View full question & answer→MCQ 21 Mark
Equation of the line passing through $(1, 2)$ and parallel to the line $y = 3x - 1$ is:
- A
$y + 2 = x + 1$
- B
$y + 2 = 3 (x + 1)$
- ✓
$y - 2 = 3 (x - 1)$
- D
$y - 2 = x - 1$
AnswerCorrect option: C. $y - 2 = 3 (x - 1)$
Given equation is $y = 3x - 1$
Slope $= 3$
Slope of the line passing through the given point $(1, 2)$ and parallel to the given line $= 3$
So, the equation of the required line is
$y - 2 = 3(x - 1)$
Hence, the correct option is $(c).$
View full question & answer→MCQ 31 Mark
The tangent of angle between the lines whose intercepts on the axes are $a, -b$ and $b, -a,$ respectively, is
- A
$\frac{\text{a}^2-\text{b}^2}{\text{ab}}$
- B
$\frac{\text{b}^2-\text{a}^2}{2}$
- ✓
$\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$
- D
AnswerCorrect option: C. $\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$
Intercepts of line are a and $-b;$
i.e., line passes through the points $(a, 0), (0, -b).$
$\therefore$ Slope of line, $\text{m}_1=\frac{-\text{b}-0}{0-\text{a}}=\frac{\text{b}}{\text{a}}$
Intercepts of line are $b, -a;$
i.e., line passes through the points $(b, 0), (0, -a).$
$\therefore$ Slope of line, $\text{m}_2=\frac{-\text{a}-0}{0-\text{b}}=\frac{\text{a}}{\text{b}}$
If $\theta$ is the angle between the lines, then
$\tan=\theta=\frac{\frac{\text{b}}{\text{a}}-\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{a}}}=\frac{\frac{\text{b}^2-\text{a}^2}{\text{ab}}}{2}=\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$
View full question & answer→MCQ 41 Mark
The equation of the line passing through the point $(1, 2)$ and perpendicular to the line $x + y + 1 = 0$ is:
- A
$y - x + 1 = 0$
- ✓
$y - x - 1 = 0$
- C
$y - x + 2 = 0$
- D
$y - x - 2 = 0$
AnswerCorrect option: B. $y - x - 1 = 0$
Slope of the given line $+1 = 0$ is $-1.$
So, slope of line perpendicular to above line is $1.$
Line passes through the point $(1, 2).$
Therefore, equation of the required linens:
$\Rightarrow y - 2 = 1(x - 1)$
$\Rightarrow y - x - 1 = 0.$
View full question & answer→MCQ 51 Mark
The equations of the lines passing through the point $(1, 0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin, are
- ✓
$\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
- B
$\sqrt{3}\text{x}+\text{y}+\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0$
- C
$\text{x}+\sqrt{3}\text{y}-\sqrt{3}=0,\text{x}-\sqrt{3}\text{y}-\sqrt{3}=0$
- D
AnswerCorrect option: A. $\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
Equation of any line passing through $(1, 0)$ is
$\Rightarrow y - 0 = m(x - 1)$
$\Rightarrow mx - y - m = 0$
Distance of the line from origin is $\frac{\sqrt{3}}{2}$
$\therefore \frac{\sqrt{3}}{2} =\Big|\frac{\text{m}\times0-0-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$
$\Rightarrow \frac{\sqrt{3}}{2}=\Big|\frac{-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$
Squaring both sides, we get
$\frac{3}{4}=\frac{\text{m}^2}{1+\text{m}^2}$
$\Rightarrow 4\text{m}^2=3+3\text{m}^2$
$\Rightarrow 4\text{m}^2-3\text{m}^2=3$
$\Rightarrow \text{m}^2=3$
$\therefore \text{m}=\pm\sqrt{3}$
$\therefore$ Required equations are
$\pm\sqrt{3}\text{x}-\text{y}\mp\sqrt{3}=0$
i.e., $\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0=0$ and $-\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0=0$
$\Rightarrow \sqrt{3}\text{x}+\text{y}-\sqrt{3}=0$
Hence, the correct option is $(a).$
View full question & answer→MCQ 61 Mark
Slope of a line which cuts off intercepts of equal lengths on the axes is:
- ✓
$-1$
- B
$-0$
- C
$2$
- D
$\sqrt{3}$
AnswerIntercept form of a line is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\Rightarrow \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}=1(\because \text{a}=\text{b})$
$\Rightarrow x + y = a$
$\Rightarrow y =- -x + a$
$\therefore$ Slope is $-1$
Hence, the correct option is $(a).$
View full question & answer→MCQ 71 Mark
A line cutting off intercept $-3$ from the $y-$axis and the tengent at angle to the xaxis is $\frac{3}{5},$ its equation is:
- ✓
$5y - 3x + 15 = 0$
- B
$3y - 5x + 15 = 0$
- C
$5y - 3x - 15 = 0$
- D
AnswerCorrect option: A. $5y - 3x + 15 = 0$
Since the lines cut off intercepts $-3$ on $y-$axis then the line is passing through the point $(0, -3).$
Given that: $\tan\theta=\frac{3}{5}$
$\Rightarrow$ Slope of the line $\text{m}=\frac{3}{5}$
So$,$ the equation of the line is
$y - y_1 = m(x - x_1)$
$\Rightarrow \text{y}+ 3 = \frac{3}{5}(\text{x} -0)$
$\Rightarrow 5y + 15 = 3$x
$\Rightarrow 3x - 5y - 15 = 0$
$\Rightarrow 5y - 3x + 15 = 0$
Hence$,$ the correct option is $(a).$
View full question & answer→MCQ 81 Mark
One vertex of the equilateral triangle with centroid at the origin and one side as $x + y - 2 = 0$ is:
$[$Hint: Let $\text{ABC}$ be the equilateral triangle with vertex $A (h, k)$ and let $\text{D}(\alpha,\beta)$ be the point on $\text{BC.}$ Then $\frac{2\alpha+\text{h}}{3}=0=\frac{2\beta+\text{k}}{3}.$ Also $\alpha+\beta-2=0$ and $\frac{\text{k}-0}{\text{h}-0}\times(-1)=-1\Big].$
- A
$(-1, -1)$
- B
$(2, 2)$
- ✓
$(-2, -2)$
- D
$(2, -2)$
AnswerCorrect option: C. $(-2, -2)$
Let $\text{ABC}$ be the equilateral triangle with vertex $A(h, k).$
Also, centroid is $G(0, 0).$
Now, $\text{AG}\bot\text{BC}$
Slope of line $BC$ or $x + y - 2 = 0$ is $-1.$
$\therefore$ Slope of $\text{AG},\frac{\text{k}}{\text{h}}=1$ or $h = k.$
Now distance of origin from $\text{BC}=\frac{|0+0-2}{\sqrt{1^2+1^2}|}=\sqrt{2}$
$\therefore$ Distance of $A$ form $\text{BC}=3\sqrt{2}=\frac{|\text{h}+\text{k}-2|}{\sqrt{1^2+1^2}}$
$\therefore |h + k - 2| = 6$
$\Rightarrow h + k - 8 = 0$ or $h + k + 4 = 0$
$\Rightarrow h + k - 8 = 0$ or $h + k + 4 = 0$
$\Rightarrow h = 4$ or $h = -2$
$\therefore$ Vertex is $(-2, -2).$ View full question & answer→MCQ 91 Mark
If the line $\frac{\text{x}}{\text{a}} + \frac{\text{y}}{\text{b}} =1$ passes through the points $(2, -3)$ and $(4, -5),$ then $(a, b)$ is:
- A
$(1, 1)$
- B
$(-1, 1)$
- C
$(1, -1)$
- ✓
$(-1, -1)$
AnswerCorrect option: D. $(-1, -1)$
Equation of line passing through the points $(2, -3)$ and $(4, -5)$ is
$\text{y}+3=\frac{-5+3}{4-2}(\text{x}-2)$
$\Rightarrow \text{y}+3=\frac{-2}{2}(\text{x}-2)$
$\Rightarrow \text{y}+3=-(\text{x}-2)$
$\Rightarrow \text{y}+3=-\text{x}+2$
$\Rightarrow \text{x}+\text{y}=-1$
$\Rightarrow \frac{\text{x}}{-1}+\frac{\text{y}}{-1}=1 ($intercept from$)$
$\therefore \text{a}=-1,\text{b}=-1$
Hence, the correct option is $(d).$
View full question & answer→MCQ 101 Mark
For specifying a straight line$,$ how many geometrical parameters should be known?
AnswerDifferent form of equation of straight line are slope intercept form$, y = mx + c,$ Paramerer $= 2$
Intercept form, $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ parameter $= 2$
One$-$point from$, y - y_1 = m(x - x_1),$ parameter $= 2$
Normal form$, \text{x}\cos \text{w}+\text{y}\sin\text{w}=\text{P},$ Parameter $= 2$
Hence$,$ the correct option is $(b).$
View full question & answer→MCQ 111 Mark
The distance between the lines $y = mx + c_1$ and $y = mx + c_2$ is:
- A
$\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{m}^2+1}}$
- B
$\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
- ✓
$\frac{\text{c}_2-\text{c}_1}{\sqrt{1+\text{m}^2}}$
- D
$0$
AnswerCorrect option: C. $\frac{\text{c}_2-\text{c}_1}{\sqrt{1+\text{m}^2}}$
Let any point on the line $y = mx + c_1$ be $P(x_1, y_1).$
The equation of the other line is: $y = mx + c_2$
$\Rightarrow mx - y + c_2 = 0$
Distance of point $P$ from this line$, \text{d}=\frac{|\text{mx}_1-\text{y}_1+\text{c}_2|}{\sqrt{\text{m}^2+1}}$
Since $P$ line on the first line$,$ we get
$\Rightarrow y_1 = mx_1 + c_1$
$\Rightarrow mx_1 - y_1 = -c_1$
$\therefore \text{d}=\frac{|\text{c}_1-\text{c}_2|}{\sqrt{\text{m}^2+1}}$
View full question & answer→MCQ 121 Mark
If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is $(3, 2),$ then the equation of the line will be:
- ✓
$2x + 3y = 12$
- B
$3x + 2y = 12$
- C
$4x - 3y = 6$
- D
$5x - 2y = 10$
AnswerCorrect option: A. $2x + 3y = 12$
Let the given the line meets the axes at $A(a, 0)$ and $B(0, b).$
Given that $C(3, 2)$ is the mid$-$point of $AB$
$\therefore 3=\frac{\text{a}+0}{2}$
$\Rightarrow \text{a}=6$
and $2=\frac{0+\text{b}}{2}$
$\Rightarrow \text{b}=4$
Intercept form of the line $AB$
$\Rightarrow \frac{\text{x}}{6}+\frac{\text{y}}{4}=1$
$\Rightarrow 2\text{x}+3\text{y}=12$
Hence, the correct option is $(a).$
View full question & answer→MCQ 131 Mark
Equations of diagonals of the square formed by the lines $x = 0, y = 0, x = 1$ and $y = 1$ are:
AnswerCorrect option: A. $y = x, y + x = 1$
Given lines are plotted on coordinate plane as shown in the adjacent figure.
From the figure, equation of diagonal $OB$ is $y = x.$
Equation of the diagonal $AC$ is $x + y = 1 ($using intercept form$).$
View full question & answer→MCQ 141 Mark
The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between the lines $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:
- A
$1 : 2$
- ✓
$3 : 7$
- C
$2 : 3$
- D
$2 : 5$
AnswerCorrect option: B. $3 : 7$
Given lines are:
$3x + 4y + 5 = 0 .....(i)$
$3x + 4y - 5 = 0 .....(ii)$
The third line is: $3x + 4y + 2 = 0$
Distance between the line $(i)$ and $(iii) =\frac{|5-2|}{\sqrt{9+16}}=\frac{3}{5}$
Distance between the lines $(ii)$ and $(iii) =\frac{|-5-2|}{\sqrt{9+16}}=\frac{7}{5}$
Hence, the required ratio is $\frac{3}{5}:\frac{7}{5}$ or $3 : 7.$
View full question & answer→MCQ 151 Mark
The point $(4, 1)$ undergoes the following two successive transformations:
Then the final coordinates of the point are:
$i.$ Reflection about the line $y = x.$
$ii.$ Translation through a distance $2$ units along the positive $x-$axis.
AnswerCorrect option: B. $(3, 4)$
Reflection of $A(4, 1)$ in $y = x$ is $5(1, 4).$
Now translation of point $B$ through a distance $'2\ '$ units along the positive $x-$axis shifts $B$ to $C(1 + 2, 4)$ or $C(3, 4).$
View full question & answer→MCQ 161 Mark
The distance of the point of intersection of the lines $2x - 3y + 5 = 0$ and $3x + 4y = 0$ from the line $5x - 2y = 0$ is:
AnswerCorrect option: A. $\frac{130}{17\sqrt{29}}$
Given lines are:
$2x - 3y + 5 = 0 .....(i)$
and $3x + 4y = 0 .....(ii)$
Solving these lines, we get point of intersection as $\Big(\frac{-20}{17},\frac{15}{17}\Big).$
$\therefore$ Distance of this point from the line $'5x - 2y = 0'$
$=\frac{\Big|5\times\Big(-\frac{20}{17}\Big)-2\Big(\frac{15}{17}\Big)\Big|}{\sqrt{25+4}}$
$=\frac{\Big|\frac{-100}{17}-\frac{30}{17}\Big|}{\sqrt{25}}$
$=\frac{130}{17\sqrt{29}}$
View full question & answer→MCQ 171 Mark
The coordinates of the foot of perpendiculars from the point $(2, 3)$ on the line $y = 3x + 4$ is given by
- A
$\frac{37}{10},\frac{-1}{10}$
- ✓
$\frac{-1}{10},\frac{37}{10}$
- C
$\frac{10}{37},-10$
- D
$\frac{2}{3},-\frac{1}{3}$
AnswerCorrect option: B. $\frac{-1}{10},\frac{37}{10}$
Let the foot of perpendicular from the point $P(2, 3)$ on the line $3x - y + 4 = 0$ be $M(h, k).$
$M(h, k)$ lies on the given line,
$\therefore 3h - k + 4 = 0 .....(i)$
Also, slopw of the given line is $3.$
$\therefore$ Slope of $\text{PM}=-\frac{1}{3}=\frac{\text{k}-3}{\text{h}-2}$ or $h + 3k - 11 = 0 .....(ii)$
Solving $(1)$ and $(ii),$ we get $(\text{h},\text{k})\equiv\Big(-\frac{1}{10},\frac{37}{10}\Big)$ View full question & answer→MCQ 181 Mark
The equations of the lines which pass through the point $(3, -2)$ and are inclined at $60^\circ$ to the line $\sqrt{3} \text{x} + \text{y} = 1$ is:
- ✓
$\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
- B
$\sqrt{3}\text{x}+\text{y}+\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0$
- C
$\text{x}+\sqrt{3}\text{y}-\sqrt{3}=0,\text{x}-\sqrt{3}\text{y}-\sqrt{3}=0$
- D
AnswerCorrect option: A. $\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
Slope of the given line $\sqrt{3}\text{x}+\text{y}=1$ is, $\text{m}=-\sqrt{3}.$
Let the slope of the required line which makes an angle of $60^\circ$ with the above line is $m.$
$\therefore \tan 60^\circ=\bigg|\frac{-\sqrt{3}-\text{m}}{1-\sqrt{3}\text{m}}\bigg|$
$\Rightarrow \bigg|\frac{-\sqrt{3}-\text{m}}{1-\sqrt{3}\text{m}}\bigg|=\sqrt{3}$
$\Rightarrow -\sqrt{3}-\text{m}=\sqrt{3}-3\text{m}$ or $-\sqrt{3}-\text{m}=-\sqrt{3}+3\text{m}$
$\Rightarrow \text{m}=\sqrt{3}$ or $m = 0$
Line is passing throught the point $(3, -2).$
Thus, the equation of the required line is : $\text{y}+2=\sqrt{3}(\text{x}-3)$ or $y + 2 = 0$
$\Rightarrow \sqrt{3}\text{x}-\text{y}-2-3\sqrt{3}=0$ and $y + 2 = 0$
View full question & answer→MCQ 191 Mark
A line passes through $(2, 2)$ and is perpendicular to the line $3x + y = 3.$ Its $y$ intercept is:
- A
$\frac{1}{3}$
- ✓
$\frac{2}{3}$
- C
$1$
- D
$\frac{4}{3}$
AnswerCorrect option: B. $\frac{2}{3}$
Any line perpendicular to $3x + y = 3$
$\text{x}-3\text{y}=\lambda \ (\lambda=\text{constant})$
If is passes through the point $(2, 2)$ then
$2-3(2)=\lambda$
$\Rightarrow \lambda=-4$
$\therefore$ Required equation is $x - 3y = -4$
$\Rightarrow 3\text{y}=-\text{x}-4$
$\Rightarrow \text{y}=\frac{1}{3}\text{x}+\frac{4}{3}\big[\because \text{y}=\text{mx}+\text{c}\big]$
So, the $y-$intercept is $\frac{4}{3}.$
Hence, the correct option is $(d).$
View full question & answer→MCQ 201 Mark
The equation of the straight line passing through the point $(3, 2)$ and perpendicular to the line $y = x$ is:
- A
$x - y = 5$
- ✓
$x + y = 5$
- C
$x + y = 1$
- D
$x - y = 1$
AnswerCorrect option: B. $x + y = 5$
Slope of the given line $y = x$ is $1.$
Thus, slope of line perpendicular to $y = x$ is $-1.$
Line passes through the point $(3, 2).$
So, equation of the required line is:
$y - 2 = -1(x - 3)$
$\Rightarrow x + y = 5$
View full question & answer→