Straight Lines

A straight line is a line which is not curved or bent. In this lesson, all the basic and advanced concepts related to straight lines are covered here. This lesson can also be downloaded as PDF which will students to refer to the concepts in offline mode.

Table of Contents for Straight Lines

What is a Straight Line?

A line is simply an object in geometry that is characterized under zero width object that extends on both sides. A straight line is just a line with no curves. So, a line that extends to both sides till infinity and has no curves is called a straight line.

Straight Lines

Download this lesson as PDF:-Straight Lines PDF

Equation of Straight Line

The relation between variables x, y satisfy all points on the curve.

The general equation of straight line is as given below:

ax + by + c = 0 { equation of straight line.

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

Slope:-

Equation of Straight Line angle with + ve x-axis

‘tan θ’ is called slope of straight line.

Slope of Straight Line

Note 1 – If line is Horizontal, then slope = 0

If line is Horizontal, then slope = 0

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

If line is Perpendicular to x-axis, i.e. Vertical

Slope = 10\frac{1}{0}

= tanπ2\tan \frac{\pi }{2}

Note 3 – If line is passing through any of two points, then slope is

tanθ=y2y2x1x2\,\,\,\,\tan \theta =\frac{{{y}_{2}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}

Intercept Form

Equation of line with x – intercept as ‘a’ and y – intercept as ‘b’ can be given as 

xa+yb=1\frac{x}{a}+\frac{y}{b}=1

x – co-ordinate of point of intersection of line with x-axis is called x-intercept.

y – intercept will be y-co-ordinate of point of intersection of line with y-axis.

For example,

Intercept Form of a Straight Line

Along x-axis: x – Intercept = 5 and y – Intercept = 0

Along y-axis: y – Intercept = 5 and x – Intercept = 0

Also,

Length of x – intercept = |x1|

Length of y – intercept = |y1 |

Note: Line passes through origin, intercept = 0

x – Intercept = 0

y – Intercept = 0

Again,

Intercept form

ON = P

AON = α

Let length of \bot r from origin to S.L is ‘P’ and let this \botr make an angle with + vex- axis ‘α’, then equation of a line can be

xcosα+ysinα=px\cos \alpha +y\sin \alpha =p xpsecα+ypcosecα=1\frac{x}{p\sec \alpha }+\frac{y}{p\cos ec\alpha }=1 xcosα+ysinα=Px\cos \alpha +y\sin \alpha =P

Here ‘P’ should be the α E [0, 2π)

Learn More: Different Forms Of The Equation Of Line

Point form

Equation of line with slope ‘m’ and which passes through (x1, y1) can be given as

y – y= m(x – x1 )

Slope Point form (Equation of a Line with 2 Points)

Equation of a line passing through two points (x1, y1) & (x2, y2) is given as

yy1=(y2y1x2x1)(xx1)y-{{y}_{1}} = \left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)

For example,

Example: Find the equation of the line that passes through the points (-2, 4) and (1, 2).

Solution:

We know general equation of a line passing through two points is:

y = mx + b, where m = y2y1x2x1\frac{y_2-y_1}{x_2 – x_1}

Here m = (2-4)/(1-(-2)) = -2/3

We can find the equation (by solving first for “b”) if we have a point and the slope. So we need to choose one of the points and use it to solve for b. Using the point (–2, 4), we get:

4 = (– 2/3)(–2) + b

4 = 4/3 + b

4 – 4/3 = b

b = 8/3

so, y = ( – 2/3 ) x + 8/3.

Which is the required equation of the line.

Straight Lines Video Lesson


Relation between two Lines

Let L1 and L2 be the two lines as

L1 : a1x + b1y + c1 = 0

L2  : a2x + b2y + c2 = 0

  • For Parallel lines

Two lines are said to be parallel if satify the below condition,

a1a2=b1b2c1c2\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}} \neq \frac{{{c}_{1}}}{{{c}_{2}}}
  • For Intersection of lines

Two lines intersect a point if

a1a2b1b2\frac{{{a}_{1}}}{{{a}_{2}}} \neq \frac{{{b}_{1}}}{{{b}_{2}}}
  • For Coincident Lines

Two lines intersect a point if

a1a2=b1b2=c1c2\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}

Angle between Straight lines

LetL1y=m1x+c1Let\,\,\,\,{{L}_{1}}\,\,\,\,\equiv \,\,\,y={{m}_{1}}x+{{c}_{1}} and

L2y=m2x+c2{{L}_{2}}\,\,\,\equiv \,\,\,y={{m}_{2}}x+{{c}_{2}}

Angle = θ=tan1(m2m11+m1+m2)\theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}+{{m}_{2}}} \right) \right|

Special Cases:

m2=m1linesareparallel\Rightarrow {{m}_{2}}={{m}_{1}}\,\,\,\,\,\,\,\to \,\,\,\,\,lines\,are\,parallel m1m2=1,linesL1&L2areperpendiculartoeachother\Rightarrow \,\,{{m}_{1}}{{m}_{2}}=-1,\,\,\,\,\,\,\,\,\,\,lines\,L1\,\And L2\,are\,perpendicular\,to\,each\,other

Length of Perpendicular from a Point on a Line

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

Length of Perpendicular from a Point on a Line

=ax1+by1+ca2+b2\ell =\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|

B (x, y) is foot of perpendicular is given by

xx1a=yy1b=(ax1+by1+c)(a2+b2)\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{-(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}

A’(h, k) is mirror image, given by

hx1a=ky1b=2(ax1+by1+c)(a2+b2)\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 distance from the two given lines.  To find the equation of the bisector of the angle between lines, we have

Angular Bisector of Straight lines

The equation of line L can be given

a1x+b1y+c1a12+b12=±a2x+b2y+c2a22+b22\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}}}

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 +λ L2 = 0

Where λ is a parameter.

Concurrency of Three Lines

Let the lines be

L1a1x+b1y+c1=0{{L}_{1}}\equiv {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0;

L2a2x+b2y+c2=0{{L}_{2}}\equiv {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 and

L3a3x+b3y+c3=0{{L}_{3}}\equiv {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0

So, condition for concurrency of linear is

a1b1c1a2b2c2a3b3c3=0\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0

Pair of Straight Lines

Pair of Straight Lines

Join equation of line L1 & L2 represents P. S. L (a1x + b1y+c1) (a2x+b2y+c2) = 0

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

Let defines a standard form of equation:-

ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represent conics curve equation

Condition for curve of being P.O.S.L Δ = abc + 2fgy – ag2 – bf2 – ch2 = 0

If Δ ≠ 0, (i) parabola h2 = ab

(ii) hyperbola h2 < ab

(iii) circle h2 = 0, a = b

(iv) ellipse h2 > ab

Now, lets see how did we get Δ = 0

General equation ax2 + 2gx + hxy + by2 + 2fy + c = 0

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

we can consider above equation as quadratic equation in x keeping y as constant.

x=(2g+hy)±(2g+hy)24a(by2+2fy+c)2ax=\frac{-(2g+hy)\pm \sqrt{{{(2g+hy)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}, so

x=(2g+hy)±Q(y)2ax=\frac{-(2g+hy)\pm \sqrt{Q(y)}}{2a}

Now, Q(y) has to be perfect square then only we can get two different line equation Q(y) in perfect square for that Δ value of Q(y) should be zero.

From there D = 0

abc + 2fgh – bg2 – af2 – ch2 = 0

Or

ahghbfgfc=0\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0

Note:
1. Point of intersection

To find point of intersection of two lines (P.O.S.L), solve the P.O.S.L, factorize it in (L1).(L2) = 0 or f(x, y) . g(u,y) = 0

2. Angle between the lines:

tanθ=(2h2aba+b)\tan \theta =\left( \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right| \right)

Special cases:

h2 = ab \to lines are either parallel or confident

h2 < ab \to imaginary line

h2 > ab \to Two distinct lines

a + b = 0 \Rightarrow perpendicular line

3. P.O.S.L passing through origin, then

\Rightarrow (y – m1x) (y – m2x) = 0

y2 – m2yx – m1xy – m1m2x2 = 0

y2 – (m1 + m2) xy – m1m2x2 = 0

\Rightarrow ax2 + 2hxy + by2 = 0

\Rightarrow y2+2hbxy+abbx2=0{{y}^{2}}+\frac{2h}{b}xy+\frac{ab}{b}{{x}^{2}}=0 m1+m2=2hb\Rightarrow m1 + m2 = \frac{2h}{b}

m1 m2 = ab\frac{a}{b} tanθ=m1m21+m1m2=(m1+m2)24m1m21+m1m2=2h2aba+b\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\sqrt{{{({{m}_{1}}+{{m}_{2}})}^{2}}4{{m}_{1}}{{m}_{2}}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|

Straight Lines 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)
Angle Between Straight lines  θ=tan1(m2m11+m1+m2)\theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}+{{m}_{2}}} \right) \right|

Problems on Straight Lines

Question 1:

Find the equation to the straight line which passes through the point (-5, 4) and is such that the portion of it between the axes is divided by the given point in the ratio 1 : 2.

Solution:

Let the required straight line be xa+yb=1.\frac{x}{a}+\frac{y}{b}=1.

Using the given conditions, P(2a+1.02+1,2.0+1.b2+1)P\left( \frac{2a+1.0}{2+1},\frac{2.0+1.b}{2+1} \right) is the point which divides (a, 0) and (0, b) internally in the ratio 1 : 2.

But P is (-5, 4)

Hence -5 = 2a/3, 4 = b/3 a = -15/2, b = 12.

Hence the required equation is x(15/2)+y12=1\frac{x}{\left( -15/2 \right)}+\frac{y}{12}=1

Question 2:

Find 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 cosθ=13.\cos \theta =-\frac{1}{3}.

Solution:

Here cosθ=13.\cos \theta =-\frac{1}{3}. (a negative number) so that π2<θ<π\frac{\pi }{2}<\theta <\pi tanθ=8=\Rightarrow \tan \theta =-\sqrt{8}= slope of line.

We know that the equation of the straight line passing through the point (x1, y1) having slope m is

y – y1 = m(x – x1)

Therefore the equation of the required line is y2=8(x1)y-2=-\sqrt{8}\left( x-1 \right) 8x+y82=0.\Rightarrow \sqrt{8}x+y-\sqrt{8}-2=0.

Question 3:

Find the equation of the line joining the points (-1, 3) and (4, -2).

Solution:

Equation of the line passing through the points (x1, y1) and (x2, y2) is yy1=y1y2x1x2(xx1)y-{{y}_{1}}=\frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}\left( x-{{x}_{1}} \right)

Hence equation of the required line will be y3=3+214(x+1)x+y2=0y-3=\frac{3+2}{-1-4}\left( x+1 \right)\Rightarrow x+y-2=0

Question 4:

Which line is having greatest inclination with positive direction of x-axis?

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

(ii) line 3x – 4y + 3 = 0

Solution:

(i) Slope of line joining points A(1, 3) and B(4, 7) is 7341=43=tanα\frac{7-3}{4-1}=\frac{4}{3}=\tan \alpha

(ii) Slope of line is 34=34=tanβ-\frac{3}{-4}=\frac{3}{4}=\tan \beta

Now tan α > tan β

Question 5:

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

Solution:

Originally slope of line is tan θ = m

Now slope of line after rotation is 3.

Angle between old position and new position of lines is 45°.

∴ we have tan45=3m1+3m\tan 45{}^\circ =\frac{3-m}{1+3m}

1 + 3m = 3 – m

4m = 2

m = 1/2 = tan θ

θ = tan-1(1/2)

Question 6:

If line 3xay1=03x-ay-1=0 is parallel to the line (a+2)xy+3=0\left( a+2 \right)x-y+3=0 then find the values of a.

Solution:

Slope of line 3xay1=03x-ay-1=0 is 3a\frac{3}{a}

Slope of line (a+2)xy+3=0\left( a+2 \right)x-y+3=0is (a + 2)

Since lines are parallel then we have a+2=3aa+2=\frac{3}{a}

or a2+2a3=0{{a}^{2}}+2a-3=0

or (a1)(a+3)=0\left( a-1 \right)\left( a+3 \right)=0

or a = 1 or a = 3.

Question 7:

Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear.

Solution:

If points A(x, -1) B(2, 1), and C(4, 5) are collinear, then

1(1)2x=5142\Rightarrow \frac{1-\left( -1 \right)}{2-x}=\frac{5-1}{4-2} 22x=2x=1\Rightarrow \frac{2}{2-x}=2\Rightarrow x=1

Question 8:

The slope of a line is double of the slope of another line. It tangent of the angle between them is 13.\frac{1}{3}.Find the slopes of the lines.

Solution:

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

We know that if θ is the angle between the lines l1 and l2 with slopes m1 and m2, then tanθ=m2m11+m1m2\tan \theta =\left| \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right|

It is given that the tangent of the angle between the two lines is 13\frac{1}{3}

13=m2m1+(2m).m13=m1+2m2\frac{1}{3}=\left| \frac{m-2m}{1+\left( 2m \right).m} \right|\Rightarrow \frac{1}{3}=\left| \frac{-m}{1+2{{m}^{2}}} \right| 2m23m+1=0\Rightarrow 2{{\left| m \right|}^{2}}-3\left| m \right|+1=0 (m1)(2m1)=0\Rightarrow \left( \left| m \right|-1 \right)\left( 2\left| m \right|-1 \right)=0 m=1orm=1/2\Rightarrow \left| m \right|=1\,\,or\,\,\left| m \right|=1/2 m±1orm=±1/2\Rightarrow \left| m \right|\pm 1\,\,or\,\,m=\pm 1/2

Question 9:

Find equation of the line parallel to the line 3x4y+2=03x-4y+2=0 and passing through the point (2,3).\left( -2,3 \right).

Solution:

Line parallel to the line 3x4y+2=03x-4y+2=0 is 3x4y+t=03x-4y+t=0

It passes through the point (-2, 3), so 3(-2) – 4(3) + t = 0 or t = 18.

So equation of line is 3x4y+18=03x-4y+18=0

Question 10:

Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x4y16=0.3x-4y-16=0.

Solution:

Let (a, b) be the coordinates of the foot of the perpendicular from the point (-1, 3) to the line 3x4y16=0.3x-4y-16=0.

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

m1=b3a+1{{m}_{1}}=\frac{b-3}{a+1}

Slope of the line 3x4y16=03x-4y-16=0 is 34\frac{3}{4}

Since these two lines are perpendicular, m1m2 = -1

(b3a+1)×(34)=1\,\left( \frac{b-3}{a+1} \right)\times \left( \frac{3}{4} \right)=-1 4a+3b=5\Rightarrow 4a+3b=5 … (1)

Point (a, b) lies on line 3x4y=16.3x-4y=16.

3a4b=16\,\,3a-4b=16 … (2)

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

a=6825a=\frac{68}{25} and b=4925b=-\frac{49}{25}

Thus, the required coordinates of the foot of the perpendicular are (6825,4925)\left( \frac{68}{25},\frac{49}{25} \right)

Question 11:

Straight Lines Question Three lines x+2y+3=0,x+2y7=0x+2y+3=0,x+2y-7=0 and 2xy4=02x-y-4=0 form 3 sides of two squares. Find the equations of remaining sides of these squares.

Solution:

Distance between the two parallel lines is 7+35=25.\frac{\left| 7+3 \right|}{\sqrt{5}}=2\sqrt{5}.The equations of the sides forming the square are of the form 2xy+k=0.2x-y+k=0.

Since the distance between sides A and B = distance between sides B and C, k(4)5=25k+45=±25k=6,14.\frac{\left| k-\left( -4 \right) \right|}{\sqrt{5}}=2\sqrt{5}\Rightarrow \frac{k+4}{\sqrt{5}}=\pm 2\sqrt{5}\Rightarrow k=6,-14.

Hence the fourth side of the two squares is

(i) 2xy+6=0,2x-y+6=0, or (ii) 2xy14=0.2x-y-14=0.

Question 12:

For the straight lines 4x+3y6=04x+3y-6=0 and 5x+12y+9=0,5x+12y+9=0, find the equation of the

(i) bisector of the obtuse angle between them,

(ii) bisector of the acute angle between them,

(iii) bisector of the angle which contains (1, 2).

Solution:

Equations of bisectors of the angles between the given lines are

4x+3y642+32=±5x+12y+952+122\frac{4x+3y-6}{\sqrt{{{4}^{2}}+{{3}^{2}}}}=\pm \frac{5x+12y+9}{\sqrt{{{5}^{2}}+{{12}^{2}}}} 9x7y41=0\Rightarrow 9x-7y-41=0 and 7x+9y3=0.7x+9y-3=0.

If θ is the angle between the line 4x+3y6=04x+3y-6=0 and the bisector 9x7y41=0,9x-7y-41=0, then tanθ=43971+(43)97=113>1.\tan \theta =\left| \frac{-\frac{4}{3}-\frac{9}{7}}{1+\left( \frac{-4}{3} \right)\frac{9}{7}} \right|=\frac{11}{3}>1.

Hence

(i) The bisector of the obtuse angle is 9x7y41=0.9x-7y-41=0.

(ii) The bisector of the acute angle is 7x+9y3=0.7x+9y-3=0.

(iii) For the point (1, 2)

4x+3y6=4×1+3×26>0,4x+3y-6=4\times 1+3\times 2-6>0, 5x+12y+9=5×1+12×2+9>0.5x+12y+9=5\times 1+12\times 2+9>0.

Hence equation of the bisector of the angle containing the point (1, 2) is 4x+3y65=5x+12y+9139x7y41=0.\frac{4x+3y-6}{5}=\frac{5x+12y+9}{13}\Rightarrow 9x-7y-41=0.

Question 13:

Find the value of λ if 2x2+7xy+3y2+8x+14y+λ=02{{x}^{2}}+7xy+3{{y}^{2}}+8x+14y+\lambda =0 will represent a pair of straight lines

Solution:

The given equation ax2+2hxy+by2+2gx+2fy+c=0a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 represents a pair of lines

If abc + 2fgh – af2 – bg2 – bc2 = 0 i.e., if

6λ+2(7)(4)(72)2(7)23(4)2λ(72)2=06\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 6λ+196984849λ4=0\Rightarrow 6\lambda +196-98-48-\frac{49\lambda }{4}=0 49λ46λ=196146=50\Rightarrow \frac{49\lambda }{4}-6\lambda =196-146=50 25λ4=50λ=20025=8\Rightarrow \frac{25\lambda }{4}=50\,\,\lambda =\frac{200}{25}=8

Question 14:

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

Solution:

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 ax2+2hxy+by2=0,a{{x}^{2}}+2hxy+b{{y}^{2}}=0,

We have x2(a+2h+b)=0{{x}^{2}}\left( a+2h+b \right)=0 or a+b=2h.a+b=-2h.

Question 15:

If the angle between the two lines represented by 2x2+5xy+3y2+6x+7y+4=02{{x}^{2}}+5xy+3{{y}^{2}}+6x+7y+4=0 is tan-1(m), then find the value of m.

Solution:

The angle between the lines

2x2+5xy+3y2+6x+7y+4=02{{x}^{2}}+5xy+3{{y}^{2}}+6x+7y+4=0 is given by

tanθ=±225462+3θ=tan1(±15).\tan \theta =\frac{\pm 2\sqrt{\frac{25}{4}-6}}{2+3}\theta ={{\tan }^{-1}}\left( \pm \frac{1}{5} \right).

Question 16:

The pair of lines 3x24xy+3y2=0\sqrt{3}{{x}^{2}}-4xy+\sqrt{3}{{y}^{2}}=0 are rotated about the origin by pi/6 in the anticlockwise sense. Find the equation of the pair in the new position.

Solution:

The given equation of pair of straight lines can be rewritten as (3xy)(x3y)=0.\left( \sqrt{3}x-y \right)\left( x-\sqrt{3}y \right)=0. Their separate equations are y=3xy=\sqrt{3}x and y=1/3xy=1/\sqrt{3}xor y = tan 60° x and y = tan 30° x

After rotation, the separate equations are

y = tan 90° x and y = tan 60° x

or x = 0 and y=3xy=\sqrt{3}x

the combined equation in the new position is x(3xy)=0x\left( \sqrt{3}x-y \right)=0 or 3x2xy=0\sqrt{3}{{x}^{2}}-xy=0