Straight Lines

This page covers all the basic and advanced concepts related to straight lines. It can also be downloaded as a PDF, allowing students to refer to the concepts in offline mode.

Table of Contents:

Definition

Equation of Straight Line

Intercept Form of a Straight Line

Point Form of a Straight Line

Slope-Point Form of a Straight Line

Relation Between Two Lines

Angle Between Two Straight Lines

Length of Perpendicular from a Point on a Line

Angle Bisector

Family of Lines

Concurrency of Three Lines

Pair of Straight Lines

Formulas Related to Straight Lines

Straight Lines Problems

Video Lessons

FAQs

A straight line is a line in which all points are in the same plane and are equidistant from one another.

A straight line is a geometry object characterized by zero width that extends to both sides infinitely with no curves.

Equation of a Straight Line

The equation of a straight line is: y = mx + b

The general equation of the straight line is given below:

ax + by + c = 0

Where x and y are variables, a, b, and c are constants.

Slope:

The equation of a straight line in slope-intercept form is:

$$y=mx+b$$

y = mx + c

The slope of the line is denoted by m, and the y-intercept is denoted by c.

The slope of a straight line is given by the angle with the positive x-axis, tan θ.

Note 1 – If the line is Horizontal, then the slope is equal to 0.

If a line is Horizontal, then its slope is 0

Note 2 – If the line is perpendicular to the x-axis, i.e. vertical, then the slope is undefined.

Slope = $\infty$ = tan $\frac{\pi}{2}$

Note 3 – If the line passes through two points, then the slope can be calculated.

\(\tan\ \theta = \frac{y_2 - y_1}{x_2 - x_1}\)

Intercept Form

The equation of the line with x-intercept a and y-intercept b can be written as: y = (x-a)/b

($$\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1\end{array}$$)

The x-coordinate of the point of intersection of the line with the x-axis is referred to as the x-intercept.

The y-coordinate of the point at which the line intersects the y-axis is the y-intercept.

For example:

Along the x-axis:

• x-Intercept = 5
• y-Intercept = 0

Along y-axis:

• y-Intercept = 5
• x-Intercept = 0

Furthermore,

Length of x-intercept = |x₁|

Length of y-intercept = |y1|

Note: Line passes through the origin, intercept = 0

x-Intercept = 0

y-intercept = 0 he said

He said, “Again.”

![Equation of a straight line in intercept form]()

ON = P

∠AON = α

Let the length of the perpendicular from the origin to a straight line be P and let this perpendicular make an angle α with the positive x-axis, then the equation of the line can be:

($$\begin{array}{l}x\cos (\alpha )+y\sin (\alpha )=p\end{array}$$)

$$\frac{x}{p\sec\alpha} + \frac{y}{p\cos\alpha} = 1$$

‘$\begin{array}{l}x\cos \alpha +y\sin \alpha =P\end{array}$’

Explore Further: Different Forms Of The Equation Of Line

Point Form

The equation of a line with slope m that passes through the point (x1, y1) can be written as: y - y1 = m(x - x1).

y - y1 = m(x - x1)

Slope-Point Form (Equation of a Line Using Two Points)

Equation of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given as:

$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$

($$\begin{array}{l}y-y_1=(x-x_1)\left(\frac{y_2-y_1}{x_2-x_1}\right)\end{array}$$ )

Answer: The equation of the line that passes through the points (-2, 4) and (1, 2) is y = -2x + 6.

Given:

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Solution:

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The general equation of a line passing through two points is:

($$\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)\end{array}$$) she said

She said, “Now,”

(-2)/3

Thus, the equation of the line is:

y - 4 = \frac{-2}{3}(x + 2)

3y - 12 = -2x - 4

3y - 12 = -2x - 4

2x + 3y - 8 = 0

What is the equation of the line?

Relation between Two Lines

Let $L_1$ and $L_2$ be the two lines.

L1: a1x + b1y + c1 = 0

L2: $$a^{2}x+b^{2}y+c^2=0$$

For Parallel Lines

Two lines are said to be parallel if the following condition is satisfied:

($$\begin{array}{l}\frac{a_1}{a_2}\ne \frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$$)

For Intersecting Lines:

Two lines intersect at a point if they have at least one common point.

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

For Coincident Lines:

Two lines coincide if they have the same direction and contain the same points.

$$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$

Angle Between Straight Lines

($$\begin{array}{l}Let,,,,L_2,,,,\equiv ,,,y=m_{2}x+c_2\end{array}$$ )

($$\begin{array}{l}y,,,\equiv ,,,m_2{L}_2+c_2\end{array}$$ )

(\theta = \tan^{-1}\left| \left( \frac{m_2 - m_1}{1 + m_1m_2} \right) \right|)

Special Cases:

(\begin{array}{l} \Rightarrow ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,\

(\Rightarrow ,L1,\And,L2,are,perpendicular,to,each,other,if,,{{m}_{1}}{{m}_{2}}=-1)

Perpendicular Length from a Point on a Line

The length of the perpendicular from P(x1, y1) to the line ax + by + c = 0 is

(\ell = \left| \frac{a{{x}_1} + b{{y}_1} + c}{\sqrt{a^2 + b^2}} \right|)

The foot of the perpendicular at point B (x, y) is given by

(\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-(ax_1 + by_1 + c)}{a^2 + b^2})

The mirror image of A(h,k) is given by

(\frac{h-{{x}_{1}}}{a} = \frac{k-{{y}_{1}}}{b} = \frac{-2(a{{x}_{1}} + b{{y}_{1}} + c)}{\left( {{a}^{2}} + {{b}^{2}} \right)})

Angular Bisector of Straight Lines

An angle bisector has equal perpendicular distances from the two given lines.

The equation of line L can be expressed as:

($$\begin{array}{l}\frac{a_{1}x+b_{1}y+c_1}{\sqrt{{a_1}^2+{b_1}^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{{a_2}^2+{b_2}^2}}\end{array}$$)

Family of Lines:

The general equation of the family of lines through the point of intersection of two given lines, L1 & L2, is given by L1 + λL2 = 0

Where λ is a parameter.

Concurrency of Three Lines

#Let the Lines Be

($$\begin{array}{l}{L}_1\equiv a_{1}x+b_{1}y+c_1=0\end{array}$$)

($$\begin{array}{l}{L}_2\equiv a_{2}x+b_{2}y+c_2=0\end{array}$$) \ and

($$\begin{array}{l}{a}_{3}x+b_{3}y+c_3=L_3\end{array}$$)

So, the condition for the concurrency of lines is

(\left|$$\begin{array}{ccccccc}{a}_1& b_1& c_1{a}_2& b_2& c_2{a}_3& b_3& c_3\end{array}$$\right| = 0)

Pair of Straight Lines

The equation of lines $L_1$ and $L_2$ representing point $P$ is given by: $$a_{1}x+b_{1}y+c_1=0$$ $$a_{2}x+b_{2}y+c_2=0$$

i.e. f(x,y) $\cdot$ g(x,y) = 0

Let’s define a standard form of the equation:

The equation ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 represents a conics curve equation.

Δ = abc + 2fgh - af² - bg² - ch² = 0

If $\mathrm{\Delta }\ne 0$, then (i) the parabola is given by $y=a{x}^2+b$.

(ii) Hyperbola $h^2<ab$

(iii) Circle: $$h^2=0,a=b$$

(iv) Ellipse

h2 > ab

Now, let’s see how we got Δ = 0

ax² + 2gx + 2hxy + by² + 2fy + c = 0

ax² + (2g + 2hy)x + (by² + 2fy + c) = 0

We can consider the above equation as a quadratic equation in x, keeping y constant.

Therefore, $$x=\frac{-(2g+2hy)\pm \sqrt{{(2g+2hy)}^2-4a(b{y}^2+2fy+c)}}{2a}$$

($$\begin{array}{l}x=\frac{-(2g+2hy)\pm \sqrt{4ghy-4ah{y}^2}}{2a}\end{array}$$ )

Now, Q(y) must be a perfect square in order for us to get two different line equations Q(y). For this, the Δ value of Q(y) must be equal to zero.

From there, D = 0

abc + 2fgh - bg2 - af2 - ch2 = 0

Either

(\left| $$\begin{array}{ccccccc}a& h& gh& b& fg& f& c\end{array}$$ \right|=0)

Important:

1. Point of Intersection

Solve the Point of Intersection of Two Lines (P.O.S.L) by factorizing it as (L1).(L2) = 0 or f(x, y) . g(u,y) = 0.

2. Angle between the lines:

(\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|)

Special Cases:

h2 = ab → Lines are either Parallel or Coincident

<ab → Imaginary Line

h2 > ab

a + b = 0 ⇒ Lines are perpendicular

3. Passing through the origin, the Point of Symmetry of the Line (P.O.S.L)

(y - m1x)(y - m2x) = 0

y^2 - m^2yx - m^1xy - m^1m^2x^2 = 0

y² - (m₁ + m₂)xy - m₁m₂x² = 0

ax^2 + 2hxy + by^2 = 0

($$\begin{array}{l}\Rightarrow y^2+xy\frac{2h}{b}+x^2\frac{ab}{b}=0\end{array}$$)

($$\begin{array}{l}m1+m2=\frac{2h}{b}\end{array}$$)

($$\begin{array}{l}{m}_1\times m_2=\frac{a}{b}\end{array}$$)

(\tan \theta = \left| \frac{2\sqrt{{h}^{2}-ab}}{a+b} \right|)

Straight Line Formulas

All Formulas Related to Straight Lines

|—| | |

| Equation of a Straight Line | ax + by + c = 0 |

| General Form or Standard Form | y = mx + c |

| Equation of a Line with 2 Points (Slope Point Form) | (y - y1) = m(x - x1) |

m = \dfrac{y_2 - y_1}{x_2 - x_1}

| Angle Between Straight Lines | (\theta = \tan^{-1}\left| \left( \frac{m_2 - m_1}{1 + m_1m_2} \right) \right| ) |

Problems on Straight Lines

Answer 1:

Find the equation of the straight line which passes through the point (-5, 4) and is such that the portion of it between the x-axis and the given point is twice the portion between the y-axis and the given point.

Given:

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Solution:

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($$\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1\end{array}$$)

The point (P\left( \frac{2a+1.0}{2+1},\frac{2.0+1.b}{2+1} \right)) divides the line segment joining the points ((a,0)) and ((0,b)) internally in the ratio 1 : 2.

But P is (-5, 4)

Hence, $$-5=\frac{2a}{3}\text{ and }4=\frac{b}{3}$$

a = -7.5, b = 12.

Hence, the required equation is:

$$\frac{x}{(-15/2)}+\frac{y}{12}=1$$

Answer 2:

The equation of the straight line which passes through the point (1, 2) and makes an angle θ with the positive direction of the x-axis where cos θ = -1/3 is:

y - 2 = -(1/3)(x - 1)

Given: This is bold

Solution: This is bold

Here cos θ = -1/3, so that π < θ < 3π/2.

The slope of the line is equal to $\tan \theta =-\sqrt{8}$

The equation of a straight line passing through the point $(x_1,y_1)$ with slope $m$ is $$y-y_1=m(x-x_1)$$

y - y1 = m(x - x1)

Therefore, the equation of the required line is:

$$y-2=-\sqrt{8}(x-1)$$

($$\begin{array}{l}x+y-2=\sqrt{8}(\sqrt{8}-1)\end{array}$$)

Answer 3:

The equation of the line joining the points (-1, 3) and (4, -2) is y = -(5/3)x + 5/3.

Given:

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Solution:

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($$\begin{array}{rl}y-y_1& =\frac{y_2-y_1}{x_2-x_1}(x-x_1)\end{array}$$)

Hence equation of the required line will be

$$y-3=\frac{-2-3}{4+1}(x+1)\Rightarrow x+y-2=0$$

Answer 4:

Which line has the greatest slope in the positive direction of the x-axis?

(i) Line joining points (1, 3) and (4, 7)

(ii) 3x - 4y + 3 = 0

Given:

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Solution:

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The slope of the line joining points A(1, 3) and B(4, 7) is $\frac{4}{3}=\tan \alpha$

(ii) Slope of line is $\tan \beta =\frac{3}{4}$

Now $\tan \alpha >\tan \beta$. So line (i) has more inclination.

Answer 5:

The angle of the line in the positive direction of the x-axis, θ, is rotated about some point on it in an anticlockwise direction by an angle of 45°, and its slope becomes 3. Find the value of θ.

Given:

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Solution:

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Originally, the slope of the line is $\tan \theta =m$

The slope of the line after rotation is now 3.

The angle between the old position and the new position of the lines is 45°.

(\therefore \ \tan 45{}^\circ = \frac{3-m}{1+3m})

1 + 3m = 3 - m

4m = 2

m = $\frac{1}{2}=\tan \theta$

θ = tan<sup>-1</sup>(1/2)

Answer 6:

If the line 3x - ay - 1 = 0 is parallel to the line (a + 2)x - y + 3 = 0, then what are the values of a?

Given:

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Solution:

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The slope of line $3x-ay-1=0$ is $\frac{3}{a}$.

The slope of line $(a+2)x-y+3=0$ is $(a+2)$.

Since lines are parallel, then we have:

a + 2 = 3

a = 3/2

($$\begin{array}{l}{a}^2+2a-3=0\end{array}$$)

$$(a-1)(a+3)=0$$

a = 1 \|\| a = -3.

Question 7:

Find the value of x such that the points $(x,-1)$, $(2,1)$ and $(4,5)$ are collinear.

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Solution: This is a statement

If points A(x, -1), B(2, 1), and C(4, 5) are collinear, then the slope of the line connecting A and B is equal to the slope of the line connecting B and C.

Slope of AB = Slope of BC

($$\begin{array}{l}\frac{1+1}{2-x}=\frac{4}{2}\end{array}$$)

(\Rightarrow \frac{2}{2-x}=2 \Rightarrow x=1)

Answer 8:

The slope of one line is 2x and the slope of the other line is x. Therefore, the tangent of the angle between them is 1/3, which implies that tan(θ) = 1/3. Therefore, x = 3/2 and the slope of the other line is 2x = 3.

Given:

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Solution:

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Let m₁ and m be the slopes of the two given lines such that m₁ = 2m

($$\begin{array}{l}\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|\end{array}$$)

The tangent of the angle between the two lines is 1/3.

$$\therefore \frac{1}{3}=\left|\frac{-m}{1+2m^2}\right|$$

‘($$\begin{array}{l}\Rightarrow \left|m\right|=1\text{ or }\left|m\right|=-\frac{1}{2}\end{array}$$ )’

($$\begin{array}{l}\Rightarrow \left|m\right|(\left|m\right|-1)(2\left|m\right|-1)=0\end{array}$$ )

|m| = 1 or |m| = 1/2

($$\begin{array}{l}\Rightarrow \left|m\right|=1,,or,,m=\pm 1/2\end{array}$$)

Another slope will be: -2, -1, 2, 1.

Answer 9:

The equation of the line parallel to the line 3x - 4y + 2 = 0 and passing through the point (-2, 3) is 4y - 3x + c = 0, where c = 11.

Given: This is *italicized* text.

Solution: This is _italicized_ text.

The line parallel to the line 3x - 4y + 2 = 0 is 3x - 4y + t = 0.

3(-2) - 4(3) + t = 18

The equation of the line is:

3x - 4y + 18 = 0

Answer 10:

Find the coordinates of the foot of the perpendicular from the point (x₁, y₁) to the line ax + by + c = 0.

Where x₁ = -1, y₁ = 3, a = 3, b = -4, c = -16.

Let $(a,b)$ be the coordinates of the foot of the perpendicular from the point $(-1,3)$ to the line $3x-4y-16=0$.

Slope of the line joining $(-1,3)$ and $(a,b)$

($$\begin{array}{l}{m}_1=\frac{b-3}{a+1}\end{array}$$)

The slope of the line 3x - 4y - 16 = 0 is 3/4.

m1m2 = -1

(\therefore \left( \frac{b-3}{a+1} \right)\times \left( \frac{4}{3} \right)=-1)

4a + 3b = 5 ….(1) ⇒ 4a + 3b = 5 ….(1)

The point $(a,b)$ lies on the line $3x-4y=16$.

⇒ 3a - 4b = 16 …(2)

⇒ 3a - 4b = 16 …(2)

On solving equations (1) and (2), we obtain

($$\begin{array}{l}a=\frac{68}{25}\\ b=-\frac{49}{25}\end{array}$$ )

The coordinates of the foot of the perpendicular are (68/25, -49/25).

Answer 11:

These three equations form the sides of two squares:

1. x + 2y + 3 = 0 2. x + 2y - 7 = 0 3. 2x - y - 4 = 0

Find the equations of the remaining sides of these squares.

Given:

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Solution:

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The distance between the two parallel lines is $$|7+3|=2\sqrt{5}.$$

The equations of the sides forming the square are of the form:

2x - y + k = 0

Since the distance between sides A and B equals the distance between sides B and C.

($$\begin{array}{l}|k-(-4)|=2\sqrt{5}\Rightarrow \frac{k+4}{\sqrt{5}}=\pm 2\sqrt{5}\Rightarrow k=6,-14.\end{array}$$ )

Therefore, the fourth side of the two squares is

(i) 2x - y = -6

either

or

neither

2x - y - 14 = 0

Answer 12:

The equation of the straight lines for 4x + 3y - 6 = 0 and 5x + 12y + 9 = 0 is:

y = (4x + 6)/3 - (5x + 9)/12

(i) Bisector of the obtuse angle between them.

(ii) Bisector of the acute angle between them,

The bisector of the angle containing the coordinates (1, 2).

Given:

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Solution:

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Equations of the bisectors of the angles between the given lines are

($$\begin{array}{l}\frac{\sqrt{4^2+3^2}}{4x+3y-6}=\pm \frac{\sqrt{5^2+{12}^2}}{5x+12y+9}\end{array}$$ )

9x - 7y - 41 = 0
7x + 9y - 3 = 0

If (\theta) is the angle between the line (4x + 3y - 6 = 0) and the bisector (9x - 7y - 41 = 0), then (\tan \theta = \left| \frac{-\frac{4}{3} - \frac{9}{7}}{1 + \left(\frac{-4}{3}\right)\frac{9}{7}} \right| = \frac{11}{3} > 1).

Therefore

(i) The equation of the bisector of the obtuse angle is 9x - 7y - 41 = 0.

(ii) The equation of the bisector of the acute angle is 7x + 9y - 3 = 0.

(iii) For the point $(1,2)$

($$\begin{array}{l}4x+3y-6>4\times 1+3\times 2-6,\end{array}$$ )

($$\begin{array}{l}5x+12y+9>5\times 1+12\times 2+9\end{array}$$ )

Hence equation of the bisector of the angle containing the point (1, 2) is \ ($$\begin{array}{l}\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}\Rightarrow 9x-7y-41=0.\end{array}$$ )

Answer 13:

Find the value of λ if the equation 2x² + 7xy + 3y² + 8x + 14y + λ = 0 represents a pair of straight lines.

Given:

The given equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines if its discriminant = 0.

abc + 2fgh - af2 - bg2 - ch2 = 0

($$\begin{array}{l}6\lambda +2\left(7\right)\left(4\right)\left(\frac{7}{2}\right)-2{\left(7\right)}^2-3{\left(4\right)}^2-\lambda {\left(\frac{7}{2}\right)}^2=0\end{array}$$)

‘($$\begin{array}{l}6\lambda +196-98-48-\frac{49\lambda }{4}=0\\ Rightarrow\lambda =\frac{196-98-48}{6-\frac{49}{4}}\end{array}$$)’

(\Rightarrow \frac{49\lambda}{4} - 6\lambda = 50)

($$\begin{array}{l}\Rightarrow \frac{25\lambda }{4}=50\Rightarrow \lambda =\frac{200}{25}=8\end{array}$$)

Answer 14:

If one of the lines of the pair ax² + 2hxy + by² = 0 bisects the angle between the positive direction of the axes, then find the relation between a, b and h.

Given:

The line of the bisector of the angle between the positive directions of the axes is y = x.

Since it is one of the lines of the given pair of lines, it can be expressed as ax² + 2hxy + by² = 0.

y = x

We have $$\begin{array}{l}{x}^2(a+2h+b)=0\end{array}$$ or $$\begin{array}{l}a+b=-2h.\end{array}$$

Answer 15:

If the angle between the two lines represented by $$2x^2+5xy+3y^2+6x+7y+4=0$$ is $${\tan }^{-1}(m)$$, then find the value of $$m$$.

Given:

The angle between the lines 2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0 is given by

$$\tan \theta =\frac{\pm \sqrt{\frac{25}{4}-6}}{2+3}\theta ={\tan }^{-1}\left|\frac{1}{5}\right|$$

Answer 16:

The pair of lines $$\sqrt{3}{x}^2-4xy+\sqrt{3}{y}^2=0$$ are rotated about the origin by $\frac{\pi }{6}$ in the anticlockwise sense. Find the equation of the pair in the new position.

Given:

The separate equations of the pair of straight lines can be rewritten as ($$\begin{array}{l}y=\sqrt{3}x\text{ and }y=\frac{x}{\sqrt{3}}.\end{array}$$)

y = tan 30° x or y = tan 60° x

After rotation, the equations are now separate.

y = tan 90° x and y = tan 60° x

or x = 0 and y = √3 x

The combined equation in the new position is $$x(\sqrt{3}x-y)=0$$ or $$\sqrt{3}{x}^2-xy=0$$