Category : 11th Class
Two Dimensional Geometry (Coordinate and Straight Line)
Key Points to Remember
Let P(x, y) be any point
x\[\to \] abscissa
y\[\to \] ordinate
(a) The distance between two points \[A({{x}_{1}},\,{{y}_{1}})\] & \[B({{x}_{2}},\,{{y}_{2}})\]
\[Ab=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
(b) Distance between the origin 0(0, 0) and the point P(x, y) is OP
\[op=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
e.g.\[A=(5,3)\,\,\,B=(-2,5)\]
\[\therefore \,\,\,AB\]
\[=\sqrt{{{(-2-5)}^{2}}{{(5-3)}^{2}}}=\sqrt{49+4}=\sqrt{53}\]
\[x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\] \[y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\]
When P divides AB in the ratio m:n then
\[x=\frac{m{{x}_{2}}-n{{x}_{1}}}{m-n},\] \[y=\frac{m{{y}_{2}}-n{{y}_{1}}}{m-n},\]
When P divides AB in the ratio 1:1 i.e. P is the mid point of AB
\[\therefore \,\,\,\text{P}\equiv \text{(x,y)=}\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{2}}+{{y}_{1}}}{2} \right)\]
Area of ABC
\[=\frac{1}{2}\{{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})\}\]
[Using determinant form]
Note: The rule for writting the area of a quadrilateral is the same as that of a triangle.
Similarly, we can find the area of a polygon of n sides with vertices \[{{A}_{1}}({{x}_{1}},{{y}_{1}}),{{A}_{2}}({{x}_{2}},{{y}_{2}})...{{A}_{n}}({{x}_{n}},{{y}_{n}})\]is
If \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\,\,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] and \[{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0\] are the equation of the triangle. Then the area of the triangle be
Where \[{{c}_{1}},\,{{c}_{2}},\,{{c}_{3}}\] be the co-factor of \[{{c}_{1}},\,{{c}_{2}},\,{{c}_{3}}\] in the determinants:
\[\therefore \,\,\,{{c}_{1}}={{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}},\]
\[{{c}_{2}}={{a}_{3}}.{{b}_{1}}-{{a}_{1}}{{b}_{3}}\]
& \[{{c}_{3}}={{a}_{1}}.{{b}_{2}}-{{a}_{2}}{{b}_{1}}.\]
[In determinant form]
Solved Problem
Sol. Let \[A\equiv (a,b+c)\]
\[B\equiv (b,c+a)\]
\[C\equiv (c,a+b),\] be three point.
To show the collinear of the points A, B and C.
Area of \[\Delta ABC\] should be zero.
Now,
[Two columns are identical]
\[=(a+b+c)\times 0=0\]
Hence the points. A, B & C be collinear.
Straight Lines
\[\theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}\]
Note: If the lines are parallel then
\[\tan \theta =0\]
i.e. \[\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}=0\]
\[\Rightarrow {{m}_{1}}={{m}_{2}}\]
i.e. if two lines are parallel then their slopes are equal.
\[\tan \theta =\infty \] i.e.
\[\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}=\infty =\frac{1}{0}\]
\[\Rightarrow 1+{{m}_{1}}.{{m}_{2}}=0\]
\[\Rightarrow {{m}_{1}}.{{m}_{2}}=-1\]
i.e. The product of their slopes is equal to -1.
Sol. Given \[\theta \frac{\pi }{4}\]
& \[{{m}_{1}}=\frac{1}{2}\]
\[\because \] Angle between two lines in tangent form is written as
\[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}.{{m}_{2}}} \right|\] (when q is in acute angle) \[\tan \frac{\pi }{4}=\left| \frac{\frac{1}{2}-{{m}_{2}}}{1+\frac{1}{2}{{m}_{2}}} \right|\]
\[\Rightarrow \pm 1=\frac{\frac{1}{2}-{{m}_{2}}}{1+\frac{1}{2}{{m}_{2}}}\]
\[1+\frac{1}{2}{{m}_{2}}=\frac{1}{2}-{{m}_{2}}\] (Taking positive sign)
\[\Rightarrow \frac{1}{2}{{m}_{2}}+{{m}_{2}}=\frac{1}{2}-1=-\frac{1}{2}\]
\[\frac{3}{2}{{m}_{2}}=\frac{-1}{2}\]
\[\therefore \,\,\,\,\,{{m}_{2}}=\frac{-1}{3}\]
Again taking negative sign
\[\therefore \,\,\,\,\frac{\frac{1}{2}-{{m}_{2}}}{1+\frac{1}{2}{{m}_{2}}}=-1\Rightarrow {{m}_{2}}=3\]
Hence the slope of the other line is \[\frac{-1}{3}\] or 3.
Intercept of a line on the axes: Intercept of a line on axis: If a straight line cuts x-axis at A and the y-axis at B then OA & OB is said to be the intercept of the line of x-axis & y-axis respectively.
·
(i) Equation of x-axis is \[y=0\]
(ii) Equation of y-axis is \[x=0\]
(iii) Equation of a line parallel to x-axis be \[y=b\]
(iv) If the line is parallel to x-axis, at the distance, b from it and is an negative side of y axis, then its equation is \[y=-b\]
(v) Equation of a Ine parallel to y-axis is \[x=+a\]
(vi) If the line is parallel to y-axis, at distance a from it and is on the negative side of x-axis, then its equation is \[x=-a\]
Equation of a straight line in various forms
Slope-Intercept form
The equation of a straight line whose slope is m and which cuts an intercept c on the y-axis is given by \[y=mx+c\]
If the line passes through the origin, then \[c=0\] and hence the equation of the line be \[y=mx\]
Solved Problem
Sol. Here \[c=2\] unit
\[\theta =30{}^\circ \]
so, \[m=\tan \theta =\tan 30{}^\circ =\frac{1}{\sqrt{3}}\]
Hence the required equation of straight line be
\[y=mx+c\]
\[y=\frac{1}{\sqrt{3}}x\times 2\]
\[\Rightarrow \,\,\,x-\sqrt{3}y+2\sqrt{3}=0\]
Solved Problem
Sol. Here, given be
\[P\equiv (2,2\sqrt{3}),\theta \equiv 75{}^\circ \]
Slope of straight line, \[m=\tan \theta =\tan 75{}^\circ \]
\[\tan (45{}^\circ +30{}^\circ )=\frac{\tan 45{}^\circ +\tan 30{}^\circ }{1-\tan 45{}^\circ .\tan 30{}^\circ }=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}\times 1}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}-1}\]
\[=\frac{\sqrt{3}+1{{)}^{2}}}{(\sqrt{3{{)}^{2}}-{{(1)}^{2}}}}=\frac{{{(\sqrt{3}+1)}^{2}}+{{1}^{2}}+2\sqrt{3}.1}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\]
Hence, the required equation of straight line be
\[y-{{y}_{1}}=m(x-{{x}_{1}})\]
\[y-2\sqrt{3}=(2+\sqrt{3})(x-2)\]
\[\Rightarrow \,\,\,(2+\sqrt{3})x-y+4-2\sqrt{3}\]
\[\Rightarrow \,\,\,(2+\sqrt{3})x-y+4-0\]
Solved Problem
Sol: Slope of AB line \[=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{-4-1}{2-(-1)}=\frac{-5}{3}\]
Thus, the equation of straight line AB be
\[y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})\]
\[\Rightarrow \,\,y-1=\frac{-5}{3}(x+1)\]
\[\Rightarrow \,\,5x+3y-3+5=0\]
\[\Rightarrow \,\,5x+3y+2=0\]
Intercept form
The equation of a straight line which cuts off intercepts a & b. an x-axis & y-axis respectively is written as \[\frac{x}{a}+\frac{y}{b}=1\]
Sol. The equation of straight line in intercept form be
\[\frac{x}{a}+\frac{y}{b}=1\]
since, line cuts off equal intercept on the coordinate axes.
i.e. \[a=b\]
then equation of straight line be \[\frac{x}{a}+\frac{y}{b}=1\]
\[\Rightarrow \,\,\,x+y=a\]
According to the question
straight line passes through (2, 3)
then \[2+3=a\Rightarrow a-5\]
Thus, the required equation of the straight line be
x+y=5=>x+y-5=0
Normal Form (or perpendicular Form):
The equation of a straight line having the length of the perpendicular from the origin is p and the perpendicular makes an angle a with the positive direction of x-axis is written as \[x.\cos \theta =y.\sin \theta =p\]
Note: In normal form of equation of a straight line is always taken as the positive and \[\theta \] is measured from the positive direction of x-axis in anticlockwise direction between 0 and \[2\pi \]
Sol. Here \[p=4\]unit
\[\theta =15\]
Equation of the straight line be
\[x.\cos \theta +y\sin \theta =p\]
\[x.\cos 15{}^\circ +y.\sin 15{}^\circ =4\]
\[\because \] Now, \[\cos 15{}^\circ =\cos (45{}^\circ -30{}^\circ )\]
\[\cos 45{}^\circ .\cos 30{}^\circ +\sin 45{}^\circ .\sin 30{}^\circ \]
\[=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}.\frac{1}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}\]
Similarly, \[\sin 15{}^\circ =\frac{\sqrt{3}-1}{2\sqrt{2}}\]
Thus, the required equation of the straight line in normal form be
\[x.\frac{\sqrt{3}+1}{2\sqrt{2}}+y.\frac{\sqrt{3}-1}{2\sqrt{2}}=4\]
\[(\sqrt{3}+1)x+(\sqrt{3}-1)y=8\sqrt{2}\]
Parametric Form (or Symmetric form)
The equation of a straight line passing through the point \[({{x}_{1}},\,{{y}_{1}})\] & making an angle a with the positive direction of x-axis is written as \[\frac{x-{{x}_{1}}}{\cos \alpha }=\frac{y-{{y}_{1}}}{\sin \alpha }=r\]
Where r is the distance of the point (x, y) from the point \[({{x}_{1}},\,{{y}_{1}}).\]
Note: If \[Q({{x}_{1}},\,{{y}_{1}})\] be a point on a line AB which makes an angle a with the positive direction of the x-axis, then there will be two points on line AB at distance r from \[Q({{x}_{1}},\,{{y}_{1}})\] and their coordinate will be
\[({{x}_{1}}+r\,\,\cos \theta ,{{y}_{1}}+r\sin \theta )\] and \[({{x}_{1}}-r\,\,\cos \theta ,{{y}_{1}}-r\sin \theta )\]
Reduction (Derived) of the General Equation to the different standard form.
Slope-Intercept form
The general equation of straight line,
\[Ax+By+c=0,\] can be reduced to the form \[y=mx+c\] by expressing y as:
\[y=\frac{-Ax-C}{B}=\left( \frac{-A}{B} \right)x+\left( \frac{-C}{B} \right)\]
i.e. \[m=\frac{-A}{B}\] & \[{{c}_{1}}=\frac{-C}{B}\]
i.e. the slope of the straight line
\[Ax+By+C=0\] be
\[m=\frac{-A}{B}=\frac{-coe\text{ff}icient\,of\,x}{coe\text{ff}icient\,of\,y}\]
& \[y-\operatorname{intercept},{{c}_{1}}=\frac{C}{B}=\frac{-cons\operatorname{tant}\,term}{coe\text{ff}icient\,of\,y}\]
If \[C\ne 0\]
\[\because \] equation of straight line be
\[Ax+By+C=0\]
\[Ax+By=-C\]
Dividing both sides by -C, we have
\[\frac{Ax}{-C}+\frac{By}{-C}=1\]
\[\Rightarrow \frac{x}{\left( \frac{-C}{A} \right)}+\frac{y}{\left( \frac{-C}{B} \right)}=1\]
which is of the form \[\frac{x}{a}+\frac{y}{b}=1\]
\[\therefore \,\,a=\frac{-C}{A}\] & \[b=\frac{-C}{B}\]
lf \[C=0\]
Then the general equation of the straight line \[ax+By+c=0,\] be
i.e. \[Ax+By=0\,\,(\because c=0)\] will be passing through the origin.
1st express it as:
\[Ax+By=-C\] ...... (i)
Case- I: If \[C<0\] or \[-C>0.\]
Dividing both sides of equation (i) by \[\sqrt{{{A}^{2}}+{{B}^{2}}},\] we have \[\frac{Ax}{\sqrt{{{A}^{2}}+{{B}^{2}}}}+\frac{By}{\sqrt{{{A}^{2}}+{{B}^{2}}}}=\frac{-C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]
which is of the form \[x.\cos \theta +y.\sin \theta =p\]
where \[\cos \theta =\frac{A}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]
\[\sin \theta =\frac{B}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\] & \[p=\frac{-C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]
Case- II: If \[C>0\]or \[-C<0\]
Then dividing the equation (i) by \[-\sqrt{{{A}^{2}}+{{B}^{2}}},\] we have \[\frac{-Ax}{\sqrt{{{A}^{2}}+{{B}^{2}}}}-\frac{By}{\sqrt{{{A}^{2}}+{{B}^{2}}}}.y=\frac{C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]
Which is of the form \[x\cos \theta +y\sin \theta =p\]
Carollary: The length of the perpendicular from the origin on the straight line
\[Ax+By+C=0\] is \[\frac{\left| C \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]
Tips: 1^{st} of all,
(i) Transpose the constant terms in the R.H.S.
(ii) Then make the constant term positive if it is not already so (this may be done by multiplying throughout by -1 if necessary).
(iii) Divide both sides by
\[\sqrt{{{(coe\text{ff}icient\,\,of\,\,x)}^{2}}+{{(coe\text{ff}icient\,\,of\,\,y)}^{2}}}\]
(iv) Now coefficient of \[x=\cos \theta \]
coefficient of \[y=\sin \theta \]
& R.H.S = p
Condition for two straight line to be coincident, parallel, perpendicular or intersecting.
Two lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]
i.e. They are neither coincident nor parallel.
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