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```Equation of a Straight Line in Different Forms

Straight lines constitute an important topic of the three dimensional geometry. The topic is a bit tricky but with a little hard work can fetch you some direct questions. A straight line can be represented in different forms and questions are often framed in the exam on these forms. We shall first discuss some of the forms of representing the straight line and then proceed to some related questions.

What Exactly is a Straight Line?

In three dimensional geometry, a straight line is defined as the intersection of two planes. So general equation of straight line is stated as the equations of both plane together i.e. general equation of straight line is a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0              ……(1)

So, equation (1) represents straight line which is obtained by intersection of two planes.

View the video on straight lines

Equation of Straight Line in Different Forms

Symmetrical Form:

Equation of straight line passing through point P (x1, y1, z1) and whose direction cosines are l, m, n is

(x–x1)/l = (y – y1)/m = (z – z1) / n.

Equation of straight line passing through two points P (x1, y1, z1) and Q (x2, y2, z2) is

x–x1 / x2– x1 = y–y1 / y2 – y1 = z – y1 / z2 – z1

Section formula: If P(x, y) divides the line joining A(x1, y1) and B(x2, y2) in the ratio of m:n then,

x = (mx2 + nx1)/ (m+n) and y = (my2 + ny1)/ (m+n)

Intercept form: If a straight line makes an intercept of say ‘a’ and ‘b’ on x and y axis respectively, then the equation of the straight line is given as

x/a + y/b = 1

The general coordinates of a point on a line are given by (x1 + lr, y1 + mr, z1 + nr) where r is the distance between point (x1, y1, z1) and the point whose coordinates are to be written.

Illustration: Find the equations of the straight lines through the point (a, b, c) which are

(a) parallel to z-axis                              (b) perpendicular to z-axis

Solution: (i) Equation of straight lines parallel to z-axis have

α = 900, β = 900, γ = 00

=> l = 0, m = 0, n = 1.

Therefore, the equation of straight line which is parallel to z-axis and passing through (a, b, c) is

(x – a) / 0 = (y – b) / 0 = (z – c) / 1

(ii) Equation of straight line perpendicular to z-axis

Let us assume that it makes an angle of α and β with x and y axes respectively.

Then the equation of straight line perpendicular to z axis and passing through (a, b, c) is

(x–a) / cos α = (y – b) / sin α = (z – c) / 0

=> (x–a) / l = (y – b) / m = (z – c) / 0.

Illustration: Find the coordinates of the point where the line joining the points (2, –3, 1) and (3, –4, –5) cuts the plane 2x + y + z = 7.

Solution: The direction ratios of the line are 3 – 2, –4 – (–3), –5 – 1 i.e. 1, –1, –6

Hence equation of the line joining the given points is

(x–2) / 1 = (y + 3) / –1 = (z – 1) / – 6 = r (say)

Coordinates of any point on this line are (r + 2, –r – 3, –6r + 1).

If this point lies on the given plane 2x + y + z = 7, then

2(r + 2) + (–r – 3) + (–6r + 1) = 7 => r = –1

Coordinates of the point are (–1 + 2, –(–1) – 3, –6(–1) + 1) i.e. (1, –2, 7).

Remark: If the equation of a straight line is given in general form, it can be changed into symmetrical form. The method is described in following illustration.

Illustration: Find in symmetrical form the equations of the line

3x + 2y – z – 4 = 0 = 4x + y – 2z + 3.

Solution: The equations of the line in general form are

3x + 2y – z – 4 = 0, 4x + y – 2z + 3 = 0                  ……(1)

Let l, m, n be the direction cosines of the line. Since the line is common to both the planes, it is perpendicular to the normal to both the planes.

Hence 3l + 2m – n = 0, 4l + m – 2n = 0

Solving these we get,

L /–4+1 = m–4+6 = n/3–8

i.e. 1/–3 = m/2 = n/–5 = 1/√(–3)2 + 22 + (–5)2 = 1/√38

So, the direction cosines of the line are –3/√38, 2/√38, –5/√38.

Now to find the coordinates of a point on a line: Let us find out the point where it meets the plane z = 0. Putting z = 0 in the equation given by (1), we have

3x + 2y – 4 = 0, 4x + y + 3 = 0

Solving these, we get x = –2, y = 5

So, one point of the line is (–2, 5, 0).

∴ Equation of the line in symmetrical form is

Illustration: If P = (1, 0), Q = (-1, 0) and R = (2, 0) are three given points then locus of the points satisfying the relation SQ2 + SR2 = 2SP2, is ….? (1988)

Solution: Let the coordinate of S be (x, y).

Then, according to given condition SQ2 + SR2 = 2SP2

Since the coordinates are given, we substitute them in the condition and obtain

(x+1)2 + y2 + (x-2)2 + y2 = 2[(x-1)2 + y2]

This gives, x2 + 2x + 1 + y2 + x2 - 4x + 4 + y2 = 2 (x2 + 1 - 2x + y2)

This yields 2x + 3 = 0.

Hence, x = -3/2.

Hence, the line satisfying the given condition is a straight line parallel to y-axis.

Ilustration: Let a and b be non-zero real numbers then what does the equation

(ax2 + by2 + c)(x2 - 5xy + 6y2) represent? (2008)

Solution: let a and b be two non-zero real numbers.

Hence, the equation (ax2 + by2 + c)(x2 - 5xy + 6y2) implies that either

(x2 - 5xy + 6y2) = 0

This gives, (x-2y)(x-3y) = 0

Or x = 2y and x = 3y

These represent two straight lines passing through origin or

(ax2 + by2 + c) = 0 when c = 0 and a and b are of same signs then

(ax2 + by2+c) = 0 and x= 0 and y = 0

This is actually a point specified at the origin.

When a = b and c is of opposite sign to that of a, (ax2 + by2+c) represents a circle. Hence the given equation may represent two straight lines and a circle.

askIITians is an online portal which acts as a platform for the JEE aspirants where they can ask any kind of questions on 3D like the equation of y-axis in 3D, demonstration of the general form of straight line or equation of z-axis in 3D.

Related Resources

Look into the Past Year Papers with Solutions to get a hint of the kinds of questions asked in the exam.

You can get the knowledge of Useful Books of Mathematics.

To read more, Buy study materials of 3D Geometry comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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