Introduction
Rene Descartes (1596-1650) ले 1600 मा बिकास गरेको Geometry लाई Analytic Geometry भनिन्छ। यसलाई Coordinate Geometry वा Cartesian Geometry पनि भनिन्छ ।
Geometrical Object हरुलाई Position र Value (स्थान र मान) दिनु नै यस Geometry को मुख्य बिशेषता थियो। यस Geometry को बिकासले गर्दा geometrical object हरुलाई जस्तै point, line, circle, …. आदिलाई algebraic expression बाट number को प्रयोग गरि ब्याख्या बिश्लेषण गर्न सकिने भयो ।
यसको लागी Descartes ले Geometry को अध्ययनमा Algebra को प्रयोग गरेका थिए। Descartes को यो योगदानलाई “the turning point of modern mathematics” पनि भनिन्छ। यस सताब्दी लाई गणितको बिकासमा सबैभन्दा धनी सताब्दी पनि मानिन्छ । Descartes ले बिकास गरको Analytic Geometry को मुख्य आधार “there is one-to-one correspondence between the points of a line and real numbers” भन्ने तथ्य हो ।
जस्तै, Two dimensional plane मा कुनै एउटा point A भएमा point A लाई X-axis बाट एउटा number x र Y-axis बाट एउटा number y correspond गर्न सकिन्छ। यसै number (x,y) लाई A को coordinate भनिन्छ।
In the figure above, there are two scales – One is the X-axis which is running across the plane and the other one is the Y-axis which is at the right angles to the X-axis. This is similar to the concept of the rows and columns.
Another mathematician Legendre is considered one of the fathers of modern analytic geometry, a geometry that incorporates all the inherent power of both algebra and calculus.
By the use of this coordinate geometry, number of things are possible, some of them are given below.
- Determine the distance between these points.
- Find the equation, midpoint, and slope of the line segment.
- Determine if the given lines are perpendicular or parallel.
- Find the perimeter and the area of the polygon formed by the points on the plane.
- Transform the shape by reflecting, moving and rotating it.
- Define the equations of ellipses, curves, and circles.
Gradient
gradient =\( \frac{\text{change in y}}{\text{change in x}} \)
Gradient
We know that the gradient of a straight line is a measure of how steep it is. To calculate the gradient of a straight line, we choose any two points on it and find the run and the rise from the first point to the second point. The run is the change in the x-coordinates, and the rise is the change in the y-coordinates, as illustrated in Figure below [Figure 1].
Then,
gradient =\( \frac{rise}{run} \) (1)
The run and the rise from one point to another on a straight line can be positive, negative or zero, depending on whether the relevant
coordinates increase, decrease or stay the same from the first point to the second point.
A straight-line graph that slopes down from left to right has a negative gradient, one that’s horizontal has a gradient of zero, and one that slopes
up from left to right has a positive gradient.
If the run is zero, which happens when the line is vertical, then the gradient of the line is undefined, since division by zero isn’t possible. Vertical lines are the only straight lines that don’t have gradients.
If the two points that
we choose to calculate the gradient of a line are
\( (x_1, y_1)\) and \( (x_2, y_2)\) as illustrated in Figure below (Figure 2).
Then
run = x2 − x1 and rise = y2 − y1.
Now,
gradient =\( \frac{y_2-y_1}{x_2-x_1} \) (2)
Remember that when we use this formula to calculate the gradient of a straight line, it doesn’t matter which point we take
to be the first point, \( (x_1, y_1)\), and which you take to be the second point, \( (x_2, y_2)\), as we get the same result either way.
For example, using the formula to calculate the gradient of the line through the points (1, 8) and (5,
2), which is shown in Figure below [Figure 3]
It gives either
gradient =\( \frac{y_2-y_1}{x_2-x_1}=\frac{2-8}{5-1}=-\frac{3}{2} \)
gradient =\( \frac{y_2-y_1}{x_2-x_1}=\frac{8-2}{1-5}=-\frac{3}{2} \)
The gradient of a straight line
The gradient of a line is a measure of its slope or steepness. It is defined as a ratio of its vertical displacement or Rise, to its horizontal displacement or Runs.
To determine the slope (or gradient), we can use:
- direct measurement OR
- count the number of units on the vertical and
horizontal lines when the line is drawn on the Cartesian plane or a grid OR - use slope formula as
\( m=\frac{y_2-y_1}{x_2-x_1}\)
if two points \( (x_1,y_1)\) and \( (x_2,y_2)\) is given.
Some Examples
The slope of a line can positive, negative, zero, or undefined.- Positive slope
If a line admits, \( y\) increases as \( x\) increases, then the slope upwards to the right. This slope will be a positive number. The line in Figure below has a slope of about \( +0.5\) , it goes up about 0.5 for every step of 1 along the x-axis. - Negative slope
If a line admits, \( y\) decreases as \( x\) increases, then the slope downwards to the right. This slope will be a negative number. The line in Figure below has a slope of about \( -0.5\), it goes down about \( 0.5\) for every step of 1 along the x-axis. - Zero slope
If a line admits, \( y\) does not change as \( x\) increases /decreases, then the line is exactly horizontal. The slope of any horizontal line is always zero. The line in Figure below goes neither up nor down as \( x\) increases, so its slope is zero. - Undefined slope
If a line is exactly vertical, it does not have a defined slope. In such a line any two \( x\) coordinates are the same, so the difference is zero. The line in the Figure below is exactly vertical, so it has no defined slope. We say "the slope is undefined".
Example 1
Find the slope of a line through the points (3, 4) and (5, 1).
Solution
We know that,slope of a line through the points \( (x_1,y_1)\) and \( (x_2,y_2)\) is
\( slope=\frac{y_2-y_1}{x_2-x_1}\)
So, using \( (3, 4)\) as \( (x_1,y_1)\) , the slope is
\( slope=\frac{y_2-y_1}{x_2-x_1}=\frac{1-4}{5-3}=-\frac{3}{2}\)
NOTE,
using \( (5, 1)\) as \( (x_1,y_1)\) , the slope is
\( slope=\frac{y_2-y_1}{x_2-x_1}=\frac{4-1}{3-5}=-\frac{3}{2}\)
Straight line
A straight line, drawn on a Cartesian Plane can be described by an equation. Such equations have a general form and vary depending on where the line cuts the axes and its degree of slope.
For detail study of a striaght line, following key points are essential.
Horizental Lines
Some linear equations have only one variable. They may have just y .
Let’s consider the equation y=2. This equation has only one variable, y. The equation says that y is always equal to 2, so its value does not depend on x. No matter what is the value of x, the value of y is always 2.
So to make a graph, draw a horizental line from y=2.
Horizontal lines are all parallel to the x-axis.
We know that the equation of the \( x\) -axis is
\( y = 0\) .
This is because all points on this axis have a \( y\) -coordinate zero, regardless of their different x-coordinates. So, the horizontal line which cuts the y-axis at \( 2\) has equation
\( y = 2\) .
Also, the horizontal line that cuts the y-axis at \( -1\) has equation
\( y = -1\)
and so on.
In general, the equation of a horizontal line is
\( y=a\)
where \( a\) is a constant.
This line cut the vertical axis at \( a\) and all points on the line have a \( y\) -coordinate of \( a\)
Vertical Line
Some linear equations have only one variable. They may have just x .
Let’s consider the equation x=−3. This equation has only one variable, x. The equation says that x is always equal to−3, so its value does not depend on y. No matter what is the value of y, the value of x is always −3.
So to make a graph, draw a verticasl line from x=-3.
Vertical Lines
Vertical lines are all parallel to the \( y\) -axis. We know that the equation of the \( y\) -axis is
\( x = 0\) .
This is because all points on this axis have an \( x\) -coordinate of zero, regardless of their different y-coordinates.
So, the vertical line which cuts the x-axis at \( 2\) has equation
\( x = 2\) .
Also, the vertical line that cuts the \( x\) -axis at \( -1\) has equation
\( x =-1\)
and so on.
In general, the equation of a vertical line is
\( x=b\)
where \( b\) is a constant.
This line cuts the horizontal axis at \( b\) and all points on the line have an \( x\) -coordinate of \( b\) .
Equation of Straight Lines
Point Slope Form
Let \( l\) is a straight line passing through a point \( A=(x_1,y_1)\) with slope \( m\) , then the equation of straight line \( l\) is
\( y-y_1=m(x-x_1)\)
Proof
Given that \( l \) is a straight line passing through a point \( A(x_1,y_1)\) with slope \(m\) . Suppose that, \( P(x,y)\) is arbitrary point on the straight line \( l \) , then
slope of line = slope of PA
or
\( m=\frac{y_2-y_1}{x_2-x_1}\)
We take the point \( P(x,y)\) as \( (x_2,y_2)\) , thus
\( m=\frac{y-y_1}{x-x_1}\)
or
\( y-y_1=m(x-x_1)\)
Two Point Form
Let \( l \) is a straight line passing through two points \( (x_1,y_1)\) and \( (x_2,y_2)\) , then the equation of straight line \( l \) is
\( y-y_1= \frac{y_2-y_1}{x_2-x_1} (x-x_1)\)
Proof
Given that \( l \) is a straight line passing through two points \( A(x_1,y_1)\) and \( B(x_2,y_2)\)
Suppose that, \( P(x,y)\) is arbitrary point on the straight line \( l \) , then
slope of PA = slope of AB
Here, slope of PA is
slope of PA\( =\frac{y_2-y_1}{x_2-x_1}\)
We take the point \( P(x,y)\) as \( (x_2,y_2)\) , and \( A(x_1,y_1)\) as \( (x_1,y_1)\) , thus
slope of PA\( =\frac{y-y_1}{x-x_1}\) (1)
Next,
slope of \( AB=\frac{y_2-y_1}{x_2-x_1}\) (2)
Now, Equating (1) and (2), we get
slope of PA = slope of AB
or
\( \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\)
or
\( y-y_1= \frac{y_2-y_1}{x_2-x_1} (x-x_1)\)
Example 1
Find the equation of a straight line passing through (-9, 5) and inclined at an angle of \( 120^0\) with the positive direction of x-axis.
Solution
Here slope of the line is
\( m = \tan 120^0 = \sqrt{3} \)
Now, the required equation of the straight lien is
\( y - y_1 = m (x - x_1)\)
or
\( y - 5 = \sqrt{ 3}(x - (-9))\)
or
\( \sqrt{ 3}x - y + 9\sqrt{ 3} + 5 = 0\)
Straight line: Example 2
Find the equation of a straight line passing through the points (2, 3) and (6, - 5).
Solution
The equation of the straight line passing through the points B(2, 3) and A(6, - 5) is
\( y-y_1 =\frac{}{}x-x_1 = x_1 -x_2\)
or
\( y-3=\frac{y_2-y_1}{x_2-x_1} (x-x_1) \)
or
\( y-3=\frac{-5 --3}{6-2}(x-2)\)
or
\( y - 3 = -2(x - 2)\)
or
\( 2x + y + 1 = 0\)
Intercepts
In coordinate geometry of two dimension, there are two types of intercepts. they are x-intercept and y-intercept. The x-intercepts are where the straight line crosses the x-axis [In the Figure below, a is x-intercept], and the y-intercepts are where the line crosses the y-axis [In the Figure below, b is y-intercept].
To clarify it algebraically,
- an x-intercept is a point on the graph where y is zero, [x-intercept is a point in the equation where the y-value is zero]
- a y-intercept is a point on the graph where x is zero.[y-intercept is a point in the equation where the x-value is zero]
Slope Intercept Form
Let l is a straight line with slope m and yintercept c, then the equation of straight line l is
y=mx+c
Proof
Given that l is a straight line with slope m andy intercept c
Then the line intersects y-axis at a point A(0,c)
Now, equation of the lie l is
\( y-y_1=m(x-x_1)\)
Taking the point A(0,c) as \( (x_1,y_1)\) , we get
y-c=m(x-0)
or
y=mx+c
Method 2
Slope Intercept Form
Let l is a straight line with slope m and y intercept c , then the equation of straight line l is
y=mx+c
Proof
Given that ll is a straight line with slope m and y intercept c
Then the line intersects y-axis at a point A(0,c)
Now, take an arbitrary point P(x,y) on l, then slope of the lie l is
slope of l=slope of l
Taking the point A(0,c) and P(x,y), we get
\( m= \frac{y_2-y_1}{x_2-x_1}\)
or
\( m= \frac{y-x}{x-0}\)
or
mx=y-c
or
y=mx+c
Double Intercept Form
Let \( l \) is a straight line whose x intercept is a and y intercept is b , then the equation of straight line \( l \) is
\( \frac{x}{a}+\frac{y}{b}=1\)
Solution
Given that \( l \) is a straight line whose x intercept is a and y intercept is b
Thus, the line intersects \( x\) -axis at a point \( A(a,0)\) and \( y\) -axis at a point \( B(0,b)\) [See Figure below
Now, equation of the lie \( l \) is
\( (y-y_1)=\frac{y_2-y_1}{x_2-x_1} (x-x_1)\)
Taking the point \( A(a,0)\) as \( (x_1,y_1)\) , and taking the point \( B(0,b)\) as \( (x_2,y_2)\) , the equation of straight line is
\( (y-0)=\frac{b-0}{0-a} (x-a)\)
or
\( y=-\frac{b}{a} (x-a)\)
or
\( ay=-b(x-a)\)
or
\( ay=-bx+ab\)
or
\( bx+ay=ab\)
or
\( \frac{x}{a}+\frac{y}{b}=1\)
Normal Form
The equation of the straight with length of the perpendicular from the origin \( p\) and this perpendicular makes an angle \( \alpha \) with x-axis is
\( x \cos \alpha + y \sin \alpha = p\)
Proof
Given that \( l \) is a straight line with length of the perpendicular from the origin \( p\) and this perpendicular makes an angle \( \alpha \)
Suppose the line \( l \) intersects the x-axis at \( A(a,0)\) and the y-axis at \( B(0,b)\)
. Now from the origin \( O\) draw \( OD\) perpendicular to \( l \) , whose \( x\) intercept is \( a\) and \( y\) intercept is \( b\) [See Figure below]
Thus, the equation of striaght line is
\( \frac{x}{a}+\frac{y}{b}=1\) (A)
Here, from the right-angled \( \triangle ODA\) , we get,
\( \frac{OD}{OA} = \cos \alpha \)
or
\( \frac{p}{a} = \cos \alpha \)
or
\( a=\frac{p}{\cos \alpha}\) (1)
Again, from the right-angled \( \triangle ODB\) , we get,
\( \frac{OD}{OB} = \cos \left(\frac{\pi}{2}-\alpha \right ) \)
or\( \frac{p}{b} = \sin \alpha \)
or
\( b=\frac{p}{\sin \alpha}\) (2)
Now, using (1) and (2) in (A), the equation of the straight line is
\( \frac{x}{a}+\frac{y}{b}=1\)
or
\( \frac{x}{\frac{p}{\cos \alpha}}+\frac{y}{\frac{p}{\sin \alpha}}=1\)
or
\( x \cos \alpha + y \sin \alpha = p\)
Bisector of the angles between two lines
Bisector of the angles between two lines (or Angle bisector of two lines) are the lines which bisects the angle between the two given lines. These Angle bisector of two lines are the locus of a point which is equidistant from the two lines. In other words, an angle bisector has equal perpendicular distance from the two lines.Equation of Angle Bisector
Let us consider a pair of straight lines given by
\(l_1 : a_1x + b_1y + c_1= 0\) and \(l_2 : a_2x + b_2y + c_2= 0\)
Also let, P(x, y) be a point lies on the angle bisectors, then length of perpendicular from the point P to both the lines \(l_1,l_2\) are equal.
Thus
\( \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2+b_2^2}}\)(1)
Solving (1), we will get two bisectors \(b_1,b_2\) as required.
General equation of second degree
Let us consider the general equation of the second degree in two variables x and y given by
\(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\)
where a, h, b, g, f and c are constants.
The study of this general equation of the second degree in two variables used
to be a major chapter in a course on analytic geometry. The equation usually represents
a pair of straight lines or a conic.
Consider
\( \Delta =abc+2fgh-af^2-bg^2-ch^2=0\)
The detail classification are given below
- a circle if \( \Delta \ne 0, a=b,h=0\)
- a parabola if \( \Delta \ne 0, h^2=ab\)
- an ellipse if \( \Delta \ne 0, h^2 < ab\)
- a hyperbola if \( \Delta \ne 0, h^2 > ab\)
- a rectangular hyperbola if \( \Delta \ne 0, h^2 > ab, a+b=0\)
- a pair of st. lines if \( \Delta =0\)
- a pair of parallel line if \( \Delta =0,h^2=ab\)
- a pair of perpendicular line if \( \Delta =0,a+b=0\)
- a point if \( \Delta =0,h^2 < ab\)
NOTES
-
If the general equation of second degree \( ax^2 +2hxy + by^2 +2gx + 2fy + c = 0\) represents a pair of striaght lines then the discriminat must be perfect square, thus
the general equation is
\( ax^2 +2hxy + by^2 +2gx + 2fy + c = 0\)
or\( ax^2 +(2hy+2g)x + (by^2 + 2fy + c) = 0\)
Now, the discriminat must be perfect square, thus
\((2hy+2g)^2-4(a)(by^2+2fy+c)\) is perfect square
or \((hy+g)^2-a(by^2+2fy+c)\) is perfect square
or \((h^2-ab)y^2+2(hg-af)y +(g^2-ac)\) is perfect square
Again, \((h^2-ab)y^2+2(hg-af)y +(g^2-ac)\) is perfect square if its discriminant is zero
Thus,
\( 4(hg-af)^2-4(h^2-ab)(g^2-ac)=0\)
or \( (hg-af)^2-(h^2-ab)(g^2-ac)=0\)
or \( abc+2fgh-af^2-bg^2-ch^2=0\) -
If the straight lines \( ax^2 +2hxy + by^2 +2gx + 2fy + c = 0\) intersect on the X-axis, then
\( \Delta =abc+2fgh-af^2-bg^2-ch^2=0\) and \(y=0\)
Thus we have
\( abc+2fgh-af^2-bg^2-ch^2=0\) and \( ax^2 +2gx + c = 0\)
In the equation \( ax^2 +2gx + c = 0\), the discriminant must be zero, thus
\( abc+2fgh-af^2-bg^2-ch^2=0\) and \( g^2=ac\)
or\( abc+2fgh-af^2-bac-ch^2=0\) and \( g^2=ac\)
or\( 2fgh-af^2-ch^2=0\) and \( g^2=ac\)
or\( 2fgh=af^2+ch^2\) and \( g^2=ac\) -
If the straight lines \( ax^2 +2hxy + by^2 +2gx + 2fy + c = 0\) intersect on the Y-axis, then
\( \Delta =abc+2fgh-af^2-bg^2-ch^2=0\) and \(x=0\)
Thus we have
\( abc+2fgh-af^2-bg^2-ch^2=0\) and \(by^2 + 2fy + c = 0\)
In the equation \(by^2 + 2fy + c = 0\), the discriminant must be zero, thus
\( abc+2fgh-af^2-bg^2-ch^2=0\) and \( f^2=bc\)
or\( abc+2fgh-abc-bg^2-ch^2=0\) and \( f^2=bc\)
or\( 2fgh-bg^2-ch^2=0\) and \( f^2=bc\)
or\( 2fgh=bg^2+ch^2\) and \( f^2=bc\)
Homogeneous Equation
A polynomial is homogeneous if all its terms have the same degree. For example,
\( 7x^5y^2-3xy^6\) is homogeneous of degree 7.
Therefore, an equation is called homogeneous if all its terms have the same degree.
For example,
\( ax^2+2hxy+by^2=0\) is a homogeneous equation of degree 2.
NOTE
The general form of second degree equation in x and y is
\( ax^2+2hxy+by^2+2gx+2fy+c=0 \)
and it is non-homogeneous because its terms have the NOT same degree.
Pair of Straight Lines
Statement
Prove that \( ax^2+2hxy+by^2=0\) represents a pair of straight lines passing through the origin.
Solution
Given the equation is
\( ax^2+2hxy+by^2=0\)
or \( b \left ( \frac{y}{x}\right )^2 +2h\left ( \frac{y}{x}\right )+a=0\)
Assume that the roots of the lines are \(m_1,m_2\), then
\( m_1=\frac{y}{x}\) and \( m_2=\frac{y}{x}\)
or\( y=m_1x\) and \(y= m_2x\)
These two lines are the straight lines passes through origin.
We know that equation of straight line in slope intercept form is
\( y=mx+c\) .
Therefore, any two lines through origin is written as
\( y=m_x\) and \( y=m_2x\) . (1)
So multiplying both the equation given in (1), we get a pair of lines as
\( (y-m_1x)(y-m_2x)=0\)
The general form of this equation is
\( ax^2+2hxy+by^2=0\) (2)
The equation (2) represents a pair of straight lines.
Simply, it is also known as homogeneous equation of second degree. This equation represents a pair of straight lines passing through the origin.
Angle between two lines
The angle \( \theta\) between two straight lines \( l_1\) and \( l_2\) having slope \( m_1\) and \( m_2\) respectively is
\( \tan \theta = \pm \frac{m_1 -m_2}{1+m_1 m_2 }\)
Proof
Let \( l_1\) and \( l_2\) are two straight lines with equations \( y = m_1 x + c_1\) and \( y = m_2 x + c_2\) respectively. Then, clearly, the slope of \( l_1\) and \( l_2\) are \(m_1\) and \(m_2\) respectively.
Also let \( l_1\) and \( l_2\) make angles \( \theta_1\) and \( \theta_2\) respectively with the positive direction of x-axis.Then,
\( m_1 = \tan \theta_1 \) and \( m_2 = \tan \theta_2\) (1)
Also let \( l_1\) and \( l_2\) intersect at a point P and angle between \( l_1\) and \( l_2\) is \( \theta\)
such that
\( \measuredangle APC = \theta\)
Now, we get,
\( \theta = \theta_2-\theta_1\)
Now taking tangent on both sides, we get,
\( \tan \theta = \tan (\theta_2 - \theta_1)\)
or \( \tan \theta = \frac{\tan\theta_2-tan \theta_1}{1+tan \theta_1 tan\theta_2} \) Using the formula, \( \tan (A +B) =\frac{\tan A-\tan B }{1+\tan A \tan B}\)
or \( \tan \theta = \pm \frac{m_1-m_2}{1+m_1 m_2 }\)
Notes:
- The angle between the lines \( l_1\) and \( l_2\) is acute or obtuse according as the value of \( \frac{m_2 -m_1} {1+m_1 m_2}\) is positive or negative.
- The angle between two intersecting straight lines means the measure of the acute angle between the lines.
- If two lines are parallel then \( m_1=m_2\)
- If two lines are perpendicular then \( m_1m_2=-1\)
Angle between pair of lines
Given the pair of lines are
\( ax^2+2hxy+by^2=0\)
or \( b \left ( \frac{y}{x}\right )^2 +2h\left ( \frac{y}{x}\right )+a=0\)
Assume that the slopes of the lines are \(m_1,m_2\), then
\( m_1+m_2=-\frac{2h}{b}\) and \( m_1m_2=\frac{a}{b}\)
Now, the angle between the lines is given by
\( \tan \theta = \pm \frac{m_1-m_2}{1+m_1m_2} \)
or\( \tan \theta = \pm \frac{ \sqrt{(m_1+m_2)^2-4m_1m_2} }{1+m_1m_2} \)
or\( \tan \theta = \pm \frac{ \sqrt{\left ( \frac{-2h}{b} \right )^2-4 \left ( \frac{a}{b} \right )} }{1+\left ( \frac{a}{b} \right )} \)
or\( \tan \theta = \pm \frac{ 2 \sqrt{h^2-ab}}{a+b} \)
- The angle between pair of lines is acute or obtuse according as the value of \( \frac{ 2 \sqrt{h^2-ab}}{a+b} \) is positive or negative.
- If two lines are coincide/parallel then, \( h^2=ab\)
- If two lines are perpendicularl then, \( a+b=0\)
Example 1
If A (-2, 1), B (2, 3) and C (-2, -4) are three points, fine the angle between the straight lines AB and BC.
Solution
Let the slope of the line AB and BC are \(m_1\) and \(m_2\) respectively. Then
\( m_1 = \frac{1}{2} \) and \( m_2 = \frac{7}{4} \)
Let \( \theta\) be the angle between AB and BC. Then,
\( \tan \theta =\frac{m_2 -m_1}{1+m_1 m_2}=\frac{2}{3}\)
or \( \theta =\tan^{-1}\frac{2}{3}\)
which is the required angle.
Example 2
Find the acute angle between the lines 7x - 4y = 0 and 3x - 11y + 5 = 0.
First we need to find the slope of both the lines.Thus,
Slope of the line 7x - 4y = 0 is
\( \frac{7}{4}\) (1)
Slope of the line 3x - 11y + 5 = 0 is
\( \frac{3}{11}\) (2)
Now, let the angle between the given lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is \( \theta\) , then
\( \tan \theta = \frac{m_2 -m_1}{1+m_1 m_2} = 1\)
or \( \theta = 45^0\)
Condition of Parallelism of Lines
If two lines of slopes \(m_1\) and \(m_2\) are parallel, then the angle \( \theta\) between them is of \( 0^0\) .
Therefore,
\( \tan \theta = \tan 0\)
or \(\tan \theta = 0\)
or \( \frac{m_2 -m_1}{ 1+m_1 m_2 } = 0\)
or \( m_2 -m_1 = 0\)
or \( m_1 = m_2 \)
Thus when two lines are parallel, their slopes are equal.
Example 1
What is the value of k so that the line through (3, k) and (2, 7) is parallel to the line through (-1, 4) and (0, 6)?
Solution
Let A(3, k), B(2, 7), C(-1, 4) and D(0, 6) be the given points. Then,
\( m_1 =\) slope of the line \( AB = \frac{7-k}{2-3} =k -7\)
\( m_2 =\) slope of the line \( CD = \frac{6-4}{0-(-1)} = 2\)
Since, AB and CD are parallel, therefore
\( m_1 = m_2 \)
or k - 7 = 2
or k = 9
Example 2
A quadrilateral has the vertices at the points (-4, 2), (2, 6), (8, 5) and (9, -7). Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram.
Solution
Let A(-4, 2), B(2, 6), C(8, 5) and D(9, -7) be the vertices of the given quadrilateral. Let P,Q, R and S be the mid-points of AB, BC, CD and DA respectively.
Then the coordinates of P, Q, R and S are P(-1, 4), Q (5, 11/2), R(17/2, -1)
and S(5/2, -5/2).
In order to prove that PQRS is a parallelogram, it is sufficient to show that PQ || RS and PQ =RS .
We have,
\( m_1 =\) Slope of the side \( PQ = \frac{1}{4}\)
\( m_2 = \) Slope of the side \( RS = \frac{1}{4}\)
Clearly, \( m_1 = m_2\).
This shows that
PQ || RS. (1)
Now,
\( PQ = \sqrt{(5+1)^2+(112-4)^2}=\sqrt{1532}\)
\( RS = \sqrt{(52-172)^2+(-52+1)^2}=\sqrt{1532}\)
Therefore,
PQ = RS (2)
Using (1) and (2), we claim that
PQRS is a parallelogram.
Condition of Perpendicularity of Two Lines
If two lines \( l_1\) and \( l_2\) of slopes \(m_1\) and \(m_2\) are perpendicular, then the angle between the lines \( \theta\) is of \( 90^0\) .
Therefore,
\( \cot \theta = 0\)
or \( \frac{1+m_1 m_2}{ m_2 -m_1} = 0\)
or \( 1 + m_1 m_2 = 0\)
or \( m_1 m_2 = -1\)
Thus when two lines are perpendicular, the product of their slope is \( -1\) .
NOTE
If m is the slope of a line, then the slope of a line perpendicular to it is \( -1/m\) .
Example 1
If P (6, 4) and Q (2, 12) are two points, then find slope of a line perpendicular to PQ.
Solution
Let m be the slope of PQ. Then
m = -2
Therefore the slope of the line perpendicular to PQ is
\( m_1=\frac{1}{2}\)
Example 2
Without using the Pythagoras theorem, show that P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled triangle.
Solution
In \( \triangle ABC\) , we have:
\( m_1 =\) Slope of the side PQ = -1
\( m_2 =\) Slope of the side PR = 1
Now clearly we see that
\( m_1 m_2 = -1\)
Therefore, the side PQ perpendicular to PR that is \( \measuredangle RPQ = 90^0\) .
Therefore, the given points P (4, 4), Q (3, 5) and R (-1, -1) are the vertices of a right angled
triangle.
Example 3
Find the ortho-centre of the triangle formed by joining the points P (- 2, -3), Q (6, 1) and R (1, 6).
The slope of the side QR of the \( \triangle PQR\) is
-1 ∙
Let PS be the perpendicular from P on QR; hence, if the slope of the line PS be m then,
m=1 .
Therefore, the equation of the straight line PS is
y + 3 = 1 (x + 2)
or x - y = 1 (1)
Again, the slope of the side RP of the \( \triangle PQR \) is 3 ∙
Let QT be the perpendicular
from Q on RP; hence, if the slope of the line QT be m_1 then,
\( m_1=-\frac{1}{3}\)
Therefore, the equation of the straight line QT is
\( y-1 =-\frac{1}{3}(x-6) \)
or x + 3y = 9 (2)
Now, solving equations (1) and (2) we get
x = 3, y = 2 .
Therefore, the co-ordinates of the ortho-centre of the \( \triangle PQR\) is (3, 2).
Example 4
What is the single equation of straight lines passing through the origin and perpendicular to the lines represented by ax^2-2hxy+by^2=0?
Solution
Given the pair of lines are
\( ax^2+2hxy+by^2=0\) (A)
or \( b \left ( \frac{y}{x}\right )^2 +2h\left ( \frac{y}{x}\right )+a=0\)
Assume that the slopes of the lines are \(m_1,m_2\), then
\( m_1+m_2=-\frac{2h}{b}\) and \( m_1m_2=\frac{a}{b}\)
Now, the slopes of the lines perpendicular to (A) are \(\frac{-1}{m_1}\) and \(\frac{-1}{m_2}\), then
The required equation is given by
\( \left ( \frac{y}{x}\right )^2 +\left ( \frac{-1}{m_1}+\frac{-1}{m_2} \right )+ \left ( \frac{-1}{m_1}.\frac{-1}{m_2} \right ) =0\)
or \( bx^2+2hxy+ay^2=0\)
Angle Bisector
Bisector of the angles between two lines (or Angle bisector of two lines) are the lines which bisects the angle between the two given lines. These Angle bisector of two lines are the locus of a point which is equidistant from the two lines. In other words, an angle bisector has equal perpendicular distance from the two lines.Equation of Angle Bisector of Two Lines
Let us consider a pair of straight lines given by
\(l_1 : a_1x + b_1y + c_1= 0\) and \(l_2 : a_2x + b_2y + c_2= 0\)
Also let, P(x, y) be a point lies on the angle bisectors, then length of perpendicular from the point P to both the lines \(l_1,l_2\) are equal.
Thus
\( \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2+b_2^2}}\)(1)
Solving (1), we will get two bisectors \(b_1,b_2\) as required.
Equation of Angle Bisector of the pair of lines
-
Equation of the angle bisectors of the pair of lines \( ax^2+2hxy+by^2=0\)
Given the pair of lines are
\( ax^2+2hxy+by^2=0\)
or \( b \left ( \frac{y}{x}\right )^2 +2h\left ( \frac{y}{x}\right )+a=0\)
Assume that the slopes of the lines \(l_1\) and \(l_2\) are \(m_1,m_2\), then
\( m_1+m_2=-\frac{2h}{b}\) and \( m_1m_2=\frac{a}{b}\)Now, angle bisector \(l\) of two lines \(l_1\) and \(l_2\) is the locus of a point which is equidistant from the two lines \(l_1\) and \(l_2\) . In other words, it has equal perpendicular distance from the two lines.Thus, the euqtion of angle bisector is
\( \frac{y-m_1x}{\sqrt{1+m_1^2}} = \pm \frac{y-m_2x}{\sqrt{1+m_2^2}} \)
or \( \frac{(y-m_1x)^2(1+m_2^2)-(y-m_2x)^2(1+m_1^2) }{(1+m_1^2)(1+m_2^2)}=0\)
or \( y^2(m_2^2-m_1^2)-x^2(m_2^2-m_1^2)+2xy(m_2-m_1)-2xym_1m_2(m_2-m_1)=0\)
or \( y^2(m_2+m_1)-x^2(m_2+m_1)+2xy-2xym_1m_2=0\)
or \( y^2(m_2+m_1)-x^2(m_2+m_1)+2xy(1-m_1m_2)=0\)
or \( \frac{x^2-y^2}{a-b}=\pm \frac{xy}{h}\) -
Equation of the angle bisectors of the pair of lines \( ax^2+2hxy+by^2=0\)
Given the pair of lines are
\( ax^2+2hxy+by^2=0\)
or \( b \left ( \frac{y}{x}\right )^2 +2h\left ( \frac{y}{x}\right )+a=0\)
Assume that the slopes of the lines \(l_1\) and \(l_2\) are \(m_1,m_2\), then
\( m_1+m_2=-\frac{2h}{b}\) and \( m_1m_2=\frac{a}{b}\)Now, angle bisector \(l\) of two lines \(l_1\) and \(l_2\) is the locus of a point which bisects the angle \( (\theta _2 - \theta _1 )\) , where \( (\theta _2 - \theta _1 )\) is the angle made by two lines \(l_1\) and \(l_2\) . In other words, the angle bisector makes an angle \( \frac{\theta _2 - \theta _1}{2}+ \theta _1 \)
Thus, the angle bisector makes an angle \( \frac{\theta _1 + \theta _2}{2} \)
Let us suppose that,
\( \frac{\theta _1 + \theta _2}{2} = \phi\)
Then,
\( \tan 2 \phi= \theta _1 + \theta _2\)
or \( \frac{2 \tan \phi}{1-\tan^2 \phi}= \frac{\tan \theta _1+ \tan \theta _2}{1-\tan \theta _1 \tan \theta _2} \)
or \( \frac{2 \frac{y}{x} }{1-\frac{y^2}{x^2} }= \frac{m_1+ m_2}{1-m_1 m_2} \)
or \( \frac{2 \frac{y}{x} }{1-\frac{y^2}{x^2} }= \frac{\frac{-2h}{b}}{1-\frac{a}{b}} \)
or \( \frac{x^2-y^2}{a-b}=\frac{xy}{h}\)
Solved Examples
- Find the combined equation of the straight lines whose separate equations are \(x-2y-3 = 0\) and \(x + y+5 = 0\).
Solution :
Combined equation of straight lines :
(x-2y-3)(x + y + 5)
or \(x^2 + xy + 5x-2xy - 2y^2 - 10y-3x-3y-15\)
or \(x^2-xy - 2y^2 + 2x - 13-15\) -
Show that \(4x^2 + 4xy + y^2-6x-3y-4 = 0 \)represents a pair of parallel lines.
Solution :
Given pair of lines is
\( 4x^2 + 4xy + y^2-6x-3y-4 = 0\)
By comparing the given equation with the general equation of pair of straight lines
\( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\)
We get,
a = 4, b = 1, 2h = 4 ==> h = 2
Now, we can see that
\(h^2=ab\)
or\(2^2=4.1\)
Therefore, the two lines represented by \(4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0 \) are parallel - Show that \(2x^2 + 3xy − 2y^2 + 3x + y + 1 = 0 \)represents a pair of perpendicular lines.
Solution :
\(2x^2 + 3xy − 2y^2 + 3x + y + 1 = 0\)
By comparing the given equation with the general equation of pair of straight lines
\( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\)
We get,
a = 2, b = -2, 2h = 3 ==> h = 3/2
If two lines are perpendicular then
a + b = 0
or 2 + (-2) = 0
Hence the given pair of straight line is perpendicular -
Prove that the equation \(6x^2+13xy+6y^2+8x+7y+2=0\)
represents a pair of straight lines.
Solution:
To show that this equation represents a pair of straight lines, we use the determinant condition
\( \begin{pmatrix} a & h & g \\ h &b&f\\g&f&c \end{pmatrix}\)
or \( \begin{pmatrix} 6 &13/2&4 \\13/2&6&7/2\\4&7/2&2 \end{pmatrix} =0\)
which confirms the stated assertion
Exercise
- Find the combined equation of the straight lines whose separate equations are \(x - 2y -3 = 0\) and \(x + y+5 = 0\)
- Show that \( 4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0 \) represents a pair of parallel lines
- Show that \( 2x^2 + 3xy - 2y^2 + 3x + y + 1 = 0\) represents a pair of perpendicular lines.
- Show that the equation \( 2x^2 -xy-3y^2 -6x + 19y - 20 = 0 \) represents a pair of intersecting lines. Show further that the angle between them is \( \tan^{-1}(5) \).
- Prove that the equation to the straight lines through the origin, each of which makes an angle \( \alpha \) with the straight line y = x is \( x^2 - 2xy \sec 2 \alpha + y^2 = 0 \)
- Find the equation of the pair of straight lines passing through the point (1, 3) and perpendicular to the lines \(2x - 3y+1 = 0\) and \(5x + y - 3 = 0\)
- Find the separate equation of the following pair of straight lines
- \( 3x^2 + 2xy - y^2 = 0 \)
- \(6(x - 1)^2 + 5(x - 1)(y - 2) - 4(y - 2)^2 = 0\)
- \(2x^2 - xy - 3y^2 - 6x + 19y - 20 = 0 \)
- The slope of one of the straight lines \(ax^2 + 2hxy + by^2 = 0 \) is twice that of the other, show that \( 8h^2 = 9ab \) .
- The slope of one of the straight lines \(ax^2 + 2hxy + by^2 = 0\) is three times the other, show that \( 3h^2 = 4ab . \)
- A \( \triangle OPQ\) is formed by the pair of straight lines \(x^2 -4xy +y^2 = 0\) and the line PQ . The equation of PQ is x + y - 2 = 0 . Find the equation of the median of the triangle \( \triangle OPQ\) drawn from the origin O .
- Find p and q , if the following equation represents a pair of perpendicular lines \( 6x^2 + 5xy - py^2 + 7x + qy - 5 = 0 \)
- Find the value of k , if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting, \( 12x^2 + 7xy - 12y^2 - x + 7y + k = 0 \)
- For what value of k does the equation \( 12x^2 +2kxy+2y^2 +11x-5y+2 = 0 \) represent two straight lines.
Exercise 9.1 [BCB pg 244]
- Find the lengths of perpendiculars drawn from
- (0,0) to the line \(3x+y+1=0\)
- (-3,0) to the line \(3x+4y+7=0\)
- (2,3) to the line \(8x+15y+24=0\)
- If p is the length of the perpendicular dropped from the origin on the line \(\frac{x}{a}+\frac{y}{b}=1\) prove that \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}\)
- Find the distance between the parallel lines
- \(3x+5y=11\) and \(3x+5y=-23\)
- \(2x-5y=6\) and \(6x-15y+11=0\)
- Find the equation of bidectors of the angles between the lines
- \(3x-4y+2=0\) and \(5x+12y+5=0\)
- \(3x-4y+6=0\) and \(5x+12y+10=0\)
- \(x-2y=0\) and \(2y-11x=6\)
- Find the equation of the bisectors of the angles between the lines containing the origin in each of the following cases
- \(4x-3y+1=0\) and \(12x-5y+7=0\)
- \(7x-y+11=0\) and \(x+y-15=0\) Prove that the bisectors of angles are at right angles to each other.
- Find the equation of the bisectors of acute angles between each of the following pair of lines
- \(y=x\) and \(y=7x+4\)
- \(x+2y=5\) and \(4x+2y+9=0\) Prove that the bisectors of angles are at right angles to each other.
- Find the equation of the bisectors of the angles between the lines containing the origin in each of the following cases
- Are the points (1,2) and (-5,6) on the same side or on opposite side of the line \(3x+5y-8=0\)?
- On what side of the line \(5x-4y+6=0\) do the points (0,0) and (-1,3) lie?
- Show that two of the three points (0,0),(2,3) and (3,4) lie on one side and the remaining on the other side of the line \(x-3y+3=0\)
- The length of the perpendicular drawn from the point (a,3) on the line 3x+4y+5=0 is 4. Find the value of a.
- What are the points on the axis of X whose perpendicular distance from the striaght line \(\frac{x}{a}+\frac{y}{b}=1\) is a ?
- Determine the equation and length of the altitude drawn from the vertex A to the opposite side of the triangle A(1,0), B(1,3), C(4,-2).
- Find the equation of two straight lines each of which is parallel to and at a distance of \(\sqrt{5}\) from the line x+2y-7=0
- Find the equations of the two straight lines drawn through the point (0,a) on which the perpendicular drawn from the point (2a,2a) are each of length a
- Find the equation of line which is at right angles to \(3x+4y=12\) such that its perpendicular distance from the origin is equal to the length of the perpendicular from (3,2) on the given line.
- The equation of the diagonal of a parallelogram is \(3y=5x+k\). The two opposite vertices of a parallelogram are the points (1,-2) and (-2,1). Find the value of k.
- If p and p' be the length of the perpendiculars from the origin upon the straight line whose equations are \(\sec \theta + y \csc \theta = a \) and \(x \cos \theta - y \sin \theta = a \cos 2 \theta \), prove that \(4p^2 + p'^2 = a^2\)
- Show that the product of the perpendiculars drawn from the two points \( ( \pm \sqrt{a^2-b^2},0) \) upon the line \(\frac{x}{a}\cos \theta +\frac{y}{b} \sin \theta =1\) is \(b^2\)
- The origin is a corner of a square and two of its sides are \(y + 2x = 0\) and \(y + 2x = 3\). Find the equation of the other two sides.
-
- A triangle is fomed by lines \(x+ y= 6, 7x - y + 10 = 0\) and \(3x + 4y + 9 = 0\). Find the equation of the internal bisector of the angles between the first two sides.
- Find the equations of the internal bisectors of the angles of the triangle whose sides are \(4x - 3y + 2 = 0, 3x - 4y + 12 = 0\) and \(3x + 4y - 12 = 0\). Also find the incentre of the triangle.