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Karnataka Class 11 Commerce Maths Straight Lines

Karnataka Class 11 Commerce Maths Straight Lines

Karnataka Class 11 Commerce Maths Straight Lines

Karnataka Class 11 Commerce Maths Straight Lines : Karnataka Pre-University board is a government body which organizes the higher secondary examination in the state.  The board functions under the Department of Primary & Secondary Education. The board has a total of 1202 government pre-university colleges, 165 unaided Pre-University colleges and about 13 Corporation pre-university colleges.

SLOPE OR GRADIENT OF A LINE

In the co-ordinate plane, a line L which is not parallel to the x axis intersects it such a way that it makes two angles which are supplementary. To be definite we select an angle which is made going anticlockwise direction from the x axis. This angle θ will have values between 00 and 1800 and is called angle of inclination or simply inclination of the line L. all lines parallel to the x axis or coincident with the x-axis have inclination 00. Note that when two lines are parallel they have same inclination. In analytical geometry we associates a number with the inclination of a line called its slope.

Karnataka Class 11 Commerce Maths Straight Lines Karnataka Class 11 Commerce Maths Straight Lines

Definition: The slope m of a line with an inclination θ (not perpendicular to x axis) is defined to be the tangent of the angle made by the line with the x axis in the positive direction i.e. m = tan θ.

The slope of a line perpendicular to the x axis is not defined since the value of tan 900 is undefined.

Following observation follows directly from the definition of the slope of a line
1. Let θ be the angle made by the line with the x axis in the positive direction.
Now
Slope of a line is positive ⇒ Tan θ is positive
⇒ θ is lies between 00 and 900
⇒ θ is acute
Slope of a line is negative ⇒ Tan θ is negative
⇒ θ lies between 900 and 1800
⇒ θ is obtuse

2. From the above observation we have:

a) If the slope of the line is positive, as we move from left to right along the line, then the line will “rise”

b) If the slope of the line is negative, as we move from left to right along the line, then the line will “fall”

3. If a line is parallel to x axis (called as horizontal line) then the angle made by it with the x axis is zero. Hence the slope is zero. In particular, Slope of x axis is zero.

4. The slope of a line parallel to y axis (called as vertical line) is not defined since the value of tan 900 is undefined and here we take the slope as ‘∞’ (infinity).

5. If three points A, B and C are collinear then the slope of AB = slope of BC and conversely. This fact can be used to establish that the given points are
collinear.

SLOPE OF PARALLEL AND PERPENDICULAR LINES:

a) If two lines are parallel then their slopes are equal and conversely.

b) If two lines are perpendicular to each other, then the product of their slopes is -1 and conversely. i.e. If m1 and m2 are the slopes of two lines then we have the condition for parallel lines given by m1 = m2. And if the lines are perpendicular then  m1  m2 = -1.

NOTE:

If the slope of the line is ‘m’ then the slope of any line parallel to it is given by ‘m’. Also the slope of any line perpendicular to it is given -1/m (negative
reciprocal of m)

SLOPE OF THE LINE JOINING TWO POINTS

The slope of the line joining the points (x1 , y1) and (x2 , y2) is given by m = y2 –  y1.

STANDARD FORMS OF EQUATION OF STRAIGHT LINES

Equation of the coordinate axes and the lines parallel to the co-ordinate axes

a) Equation of x axis

We know that the y co-ordinate of every point on the x-axis is zero. This is the common property that every point on the x-axis will satisfy. Thus, the
equation of x axis is y = 0

b) Equation of y axis

The x coordinate of each point on the y axis is zero. Thus the equation of y axis is x = 0.

To Find the equation of the St. line parallel to x axis

Let ‘l’ be a straight line parallel to x axis and at a directed distance ‘h’ from it. Let P (x, y) be a general point on the line ‘l’

∴ y = h

This is the required equation of the line.

Note: If a line is parallel to x axis at a distance of ‘a’ units from the x axis (lying above the x axis) the equation of the line is y = a provided ‘a’ is positive. If ‘a’ is negative, then the line lies below the x axis and so the equation becomes y = -a.

For Example, equation of the line parallel to x axis which is at a distance of 15 units above the x axis is y = 15 or y – 15 = 0

The equation y + 3 = 0 or y = -3 represent the line parallel to x axis, which is 3 units below the x-axis.

To Find the equation of the Straight line which is parallel to y axis

Karnataka Class 11 Commerce Maths Straight Lines

Let ‘l’ b a straight line parallel to y axis and at a directed distance ‘k’ from it Let P (x, y) be a general point on the line.

∴ x = k which is the required equation of the line.

Note:

Equation of y axis is x = 0. If k > 0, then the line lies on the right of the y axis and if k < 0 then the line lies on the left of the y axis.

Example 1

Find the equation of the line parallel to y axis and is at a distance of

a) 2 units to the right of it

b) 4/7 units to the left of it

Solution :

Equation of the line parallel to y axis and is at a

a) distance of 2 units to the right of it is given by x = 2

b) Equation of the required line is x = -4/7
i.e. 7x + 4 = 0

Example 2

Find the equation of the line which is

a) Parallel to x axis and passing through (2, -3)

b) Parallel to y axis and passing through (3, -4)

Solution :

Any line parallel to x axis is of the form y = k. By data

a) this should pass through (2, -3). So K = -3. Thus the equation of the line is y = -3 i.e. y + 3 = 0

b) Any line parallel to y axis is of the form x = k. By data.

This should pass through (3, -4). Thus k = 3

Thus equation of the line is x = 3.

Equation of the line in slope point form.

To find the equation of the line whose slope is m and passing through the point A(x1, y1)

Karnataka Class 11 Commerce Maths Straight Lines

Let A = (x1, y1) . Let P (x, y) be any point on the line, Now slope of AP = (y – y1)  / (x – x1)

By data, slope of the line is m = (y – y1)  / (x – x1)

This relation is true for any point on the line.

Thus, the equation of the line whose slope is m and passing through (x1, y1) is given by (y – y1 ) = m (x – x1)

This form of the equation of a line is called slope point form.

Note: Equation of the line passing through the origin and having slope ‘m’ is given by y=mx.

Equation of the line in two point form

Karnataka Class 11 Commerce Maths Straight Lines

Let A = (x1, y1) and B = (x2, y2).

Let P(x,y) be any point on the line joining A and B.Now we observe A, B and P are collinear.

So slope of AP = slope of AB.

(y – y1) / (x – x1) = (y2 – y1) / (x2 – x1)

This relation is true for all points on the line.

Thus the equation of the line joining the points (x1 y1) and (x2 y2) is given by

(y – y1) / (x – x1) = (y2 – y1) / (x2 – x1)

This equation can also be written as

(y – y1) / (y1 – y2) = (x – x1) / (x1 – x2)

This form of equation of a line is called two point form.

Equation of a line in slope intercept form: 

Karnataka Class 11 Commerce Maths Straight Lines

To find the equation of the line whose, slope is ‘m’ which cuts off intercept ‘c’ on y axis.

Solution.:

Given y intercept = OB = c⇒
B = (0, c) By data slope of the line = m

Let P (x, y) be any point on the line Slope of BP = y – c /x – 0 = m (given)
∴  y – c = m (x – 0)

y = mx + c is the equation of the line.

This form is called slope intercept form.

Remarks:

If the equation of the straight line is put in the slope intercept form y = mx + c then the x coefficient gives the slope of the line.

Note:

Equation of the line passing through the origin and having slope m is y = mx.

Equation of the line in the intercept form

Let a line cut the x axis at A and y axis at B. Let OA = a and OB = b. Then the length OA = a is called the x intercept of the line and the length OB = b is
called the y intercept of the line. If the line passes the origin. Then both the x intercept and y intercepts are zero.

Karnataka Class 11 Commerce Maths Straight Lines

Consider the XOY plane. Let the line cut the x axis at A and y axis at B. By data OA = a and OB = b. Thus A = (0, 0) and B = (0, b).

Hence the required line is the line joining A and B, Its equation can be obtained by using two point form of the equation of a line.

Then the equation of AB is 

⇒  y – 0 / x – a = 0 – b / a – 0

⇒ y  / x – a = -b / a

⇒  ay = -bx + ab

⇒ bx + ay = ab

Dividing throughout by ‘ab’ we have

⇒ bx/ab + ay/ab = ab/ab

⇒ x/a + y/b = 1.

This form of equation of the line is called Intercept form.

for more on Karnataka Class 11 Commerce Maths Straight Lines click 1st PU Maths Book

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