*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 0 ^{0} and 180^{0} 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 0*

^{0}. 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 90 ^{0} 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 m _{1} = m_{2}. And if *

*the lines are perpendicular then m*

_{1}m_{2}= -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 (x _{1} , y_{1}) and (x_{2} , y_{2}) is given by *

*m = y*_{2 – }_{ }y_{1.}*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 – y_{1)}*

_{ /}

_{ (x – x1)}*By data, slope **of the line is m = (y – y_{1)} _{ / (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 – y*_{1 )}= m (x – x_{1})*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 = (x _{1}, y_{1}) and B = (x_{2}, y_{2}). *

*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 – y _{1}) / (x – x_{1}) = (y_{2} – y_{1}) / (x_{2} – x_{1})**

*This relation is true for all points on the line.*

*Thus the equation of the line joining the points (x _{1} y_{1}) and (x_{2} y_{2}) is given by*

**(y – y _{1}) / (x – x_{1}) = (y_{2} – y_{1}) / (x_{2} – x_{1})**

*This equation can also be written as*

**(y – y _{1}) / (y_{1 }– y_{2}) = (x – x_{1}) / (x_{1} – x_{2})**

*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.*

*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.*

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