Rewrite $\sqrt{3}x + y + 4 = 0$ into normal form.

Rewrite √(3)x + y + 4 = 0 into normal form.

Question 129 of 129

\sqrt{3}x + y + 4 = 0

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2026Subjective· 3mImportance★★★★★

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Concept understanding — Straight Lines — Forms of the Equation

The general (linear) equation of a straight line is $ax+by+c=0$ , where $a,b$ are not both zero; the set of solutions of any such equation is a straight line in the plane. Because dividing through by $b$ (or $a$ ) removes one constant, every line's equation genuinely contains only two independent arbitrary constants — so exactly two independent pieces of information (two points, or a point and a slope, or two intercepts, etc.) are enough to pin a line down uniquely.

Slope. The angle of inclination $\theta$ of a line is the angle it makes with the $x$ -axis, measured counter-clockwise; the slope $m=\tan\theta$ (undefined when $\theta=\pi/2$ , i.e. for a vertical line). Equivalently, through two points $(x_1,y_1),(x_2,y_2)$ with $x_1\ne x_2$ , $m=\dfrac{y_2-y_1}{x_2-x_1}$ ; from the general form, $m=-a/b$ ( $b\ne0$ ). Three points are collinear exactly when the slope of any one pair equals the slope of another pair sharing a point.

Intercepts. The $x$ -intercept is where a line meets the $x$ -axis ( $y=0$ ); the $y$ -intercept is where it meets the $y$ -axis ( $x=0$ ). ( $x=0$ is itself the equation of the $y$ -axis; $y=0$ is the equation of the $x$ -axis.)

The six forms (two conditions each, all interconvertible by algebra):

Data given	Equation
Slope $m$ , $y$ -intercept $b$	$y=mx+b$
Slope $m$ , point $(x_1,y_1)$	$y-y_1=m(x-x_1)$
Two points $(x_1,y_1),(x_2,y_2)$	$\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$ , equivalently $\begin{vmatrix}x-x_1 & y-y_1\\ x_2-x_1 & y_2-y_1\end{vmatrix}=0$
$x$ -intercept $a$ , $y$ -intercept $b$ (both $\ne0$ )	$\dfrac{x}{a}+\dfrac{y}{b}=1$
Normal length $p$ , angle $\alpha$ of the normal with the $x$ -axis	$x\cos\alpha+y\sin\alpha=p$
Parametric, through $(x_1,y_1)$ at inclination $\theta$ , parameter $r$ = signed distance from $(x_1,y_1)$	$\dfrac{x-x_1}{\cos\theta}=\dfrac{y-y_1}{\sin\theta}=r$

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Dividing $\sqrt3x+y+4=0$ by $\sqrt{(\sqrt3)^2+1^2}=2$ and flipping the sign (so the constant is positive) gives the normal form $x\cos210^\circ+y\sin210^\circ=2$ .

The normal form of a line is $x\cos\alpha+y\sin\alpha=p$ , with $p\ge0$ .

Given line: $\sqrt3x+y+4=0$ , i.e. $\sqrt3x+y=-4$ .

Here $A=\sqrt3$ , $B=1$ . Compute $\sqrt{A^2+B^2}=\sqrt{3+1}=2$ .

Divide by 2: $\dfrac{\sqrt3}{2}x+\dfrac{1}{2}y=-2$ .

Since the right side is negative, multiply both sides by $-1$ to make $p$ positive:

$-\dfrac{\sqrt3}{2}x-\dfrac{1}{2}y=2$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$(2, 3)$
(a) $(-2, -3)$
(b) $(-3, -2)$
(c) $(3, 2)$
(d) $(-3, 2)$
›Reveal solutionSolution

Reflecting $(x,y)$ in the line $y=-x$ gives $(-y,-x)$ , so $(2,3)\to(-3,-2)$ .

The line $y=-x$ passes through the origin at $135^\circ$ to the x-axis. Reflection in this line swaps the coordinates and negates both: the general rule is $(x,y)\mapsto(-y,-x)$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$(2, 3)$
(a) $\alpha + 3\beta = 11$
(b) $\alpha + 2\beta = 7$
(c) $3\alpha + \beta = 11$
(d) $3\alpha + \beta = 9$
›Reveal solutionSolution

Find the equation of the line through the two given points and read off the condition satisfied by any point on it.

Slope $=\dfrac{4-3}{-1-2}=\dfrac{1}{-3}=-\dfrac13$ .

Equation: $y-3=-\dfrac13(x-2)\Rightarrow 3(y-3)=-(x-2)\Rightarrow 3y-9=-x+2\Rightarrow x+3y=11$ . …

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CBSE 2025Set ANNUAL1 markMCQ
$(1, -1)$
(a) $4x+3y+7=0$
(b) $4x-3y-7=0$
(c) $3x+4y+7=0$
(d) $3x+4y-7=0$
›Reveal solutionSolution

Perpendicular slopes are negative reciprocals; use the point-slope form with the given point.

Rewrite $3x+4y=6$ as $y=-\dfrac34x+\dfrac64$ , so its slope is $-\dfrac34$ .

A line perpendicular to it has slope $\dfrac43$ (negative reciprocal).

Using point-slope form through $(1,-1)$ : …

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CBSE 2020Set ANNUAL1 markMCQ
$x - y + 5 = 0$
(a) $x - y - 5 = 0$
(b) $x + y - 5 = 0$
(c) $x + y + 5 = 0$
(d) $x + y + 10 = 0$
›Reveal solutionSolution

Find where the given line meets the $y$ -axis, then draw the perpendicular there.

The given line $x-y+5=0$ , i.e. $y=x+5$ , has slope $1$ . It meets the $y$ -axis where $x=0$ : $y=0+5=5$ , so the point of intersection is $(0,5)$ .

…

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CBSE 2019Set ANNUAL1 markMCQ
$\dfrac{x}{a}-\dfrac{y}{b}=0$
(a) $a=b$
(b) only for $a=1, b=1$
(c) $a>b$
(d) $a<b$
›Reveal solutionSolution

Writing $\dfrac{x}{a}-\dfrac{y}{b}=0$ as $y=\dfrac{b}{a}x$ shows the slope is $b/a$ ; setting this equal to 1 gives $a=b$ .

$\dfrac{x}{a}-\dfrac{y}{b}=0 \Rightarrow \dfrac{y}{b}=\dfrac{x}{a} \Rightarrow y = \dfrac{b}{a}x$ .

…

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CBSE 2019Set ANNUAL1 markMCQ
$(2, 3)$
(a) $\alpha+3\beta=11$
(b) $3\alpha+\beta=11$
(c) $\alpha+2\beta=7$
(d) $3\alpha+\beta=9$
›Reveal solutionSolution

The slope of the line through $(2,3)$ and $(-1,4)$ is $-1/3$ ; writing its equation and substituting $(\alpha,\beta)$ gives $\alpha+3\beta=11$ .

Slope $m = \dfrac{4-3}{-1-2} = \dfrac{1}{-3} = -\dfrac{1}{3}$ .

Using point-slope form through $(2,3)$ : $y-3 = -\dfrac{1}{3}(x-2)$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Which of the following has the greatest y-intercept in magnitude?$
(a) $3x + 4y = 5$
(b) $2x + 3y = 4$
(c) $4x + 5y = 6$
(d) $x + 2y = 3$
›Reveal solutionSolution
Computing $y$ -intercept $=c/b$ for each line, $x+2y=3$ gives the largest magnitude, $1.5$ .

For $ax+by=c$ , setting $x=0$ gives the $y$ -intercept $y=c/b$ .

$3x+4y=5 \Rightarrow y\text{-intercept} = 5/4 = 1.25$

$2x+3y=4 \Rightarrow y\text{-intercept} = 4/3 \approx 1.33$

$4x+5y=6 \Rightarrow y\text{-intercept} = 6/5 = 1.2$ …
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Q.Rewrite 3x+y+4=0\sqrt{3}x + y + 4 = 03​x+y+4=0 into normal form.

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Q.Rewrite $\sqrt{3}x + y + 4 = 0$ into normal form.

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