Ex 10.3, 8 - Chapter 10 Class 11 Straight Lines (Term 1)
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Ex 10.3, 8 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3. Let equation of line AB be x – 7y + 5 = 0 Let line CD be perpendicular to line AB and having x-intercept 3 Since Line CD has x-intercept 3 So, line CD passes through the point (3, 0) We have to find equation of line CD, Finding slope of line AB x − 7y + 5 = 0 − 7y = −x − 5 − 7y = −(x + 5) 7y = (x + 5) y = 1/7 (x + 5) y = 𝑥/7 + 5/7 The above equation is of the form y = mx + c where m = slope of line Thus, slope of line AB = 1/7 Now, Given that line AB and line CD are perpendicular We know that, product of slope of perpendicular lines is –1 So, (Slope of line AB) × (Slope of line CD) = –1 1/7 × (Slope of line CD) = –1 Slope of line CD = 7 × –1 Slope of line CD = –7 Equation of a line passing through a point (x0, y0) & having slope m is (y – y0)= m(x – x0) Equation of line CD passing through point P(3, 0) & having slope −7 is (y − 0) = –7(x – 3) y = –7 (x – 3) y = –7x – (–7) 3 y = –7x + 21 y + 7x = 21 Thus, the required equation of line is y + 7x = 21
We will learn how to find the equation of a line perpendicular to a line.
Prove that the equation of a line perpendicular to a given line ax + by + c = 0 is bx - ay + λ = 0, where λ is a constant.
Let m\(_{1}\) be the slope of the given line ax + by + c = 0 and m\(_{2}\) be the slope of a line perpendicular to the given line.
Then,
m\(_{1}\) = -\(\frac{a}{b}\) and m\(_{1}\)m\(_{2}\) = -1
⇒ m\(_{2}\) = -\(\frac{1}{m_{1}}\) = \(\frac{b}{a}\)
Let c\(_{2}\) be the y-intercept of the required line. Then its equation is
y = m\(_{2}\)x + c\(_{2}\)
⇒ y = \(\frac{b}{a}\) x + c\(_{2}\)
⇒ bx - ay + ac\(_{2}\) = 0
⇒ bx - ay + λ = 0, where λ = ac\(_{2}\) = constant.
To get it more clear let us assume that ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.
Now convert the ax + by + c = 0 in to slope-intercept form we get,
by = - ax - c
⇒ y = - \(\frac{a}{b}\) x - \(\frac{c}{b}\)
Therefore, the slope of the straight line ax + by + c = 0 is (- \(\frac{a}{b}\)).
Let m be the slope of a line which is perpendicular to the line ax + by + c = 0. Then, we must have,
m × (- \(\frac{a}{b}\)) = - 1
⇒ m = \(\frac{b}{a}\)
Therefore, the equation of a line perpendicular to the line ax + by + c = 0 is
y = mx + c
⇒ y = \(\frac{b}{a}\) x + c
⇒ ay = bx + ac
⇒ bx - ay+ k = 0, where k = ac, is an arbitrary constant.
Algorithm for directly writing the equation of a straight line perpendicular to a given straight line:
To write a straight line perpendicular to a given straight line we proceed as follows:
Step I: Interchange the coefficients of x and y in equation ax + by + c = 0.
Step II: Alter the sign between the terms in x and y of equation i.e., If the coefficient of x and y in the given equation are of the same signs make them of opposite signs and if the coefficient of x and y in the given equation are of the opposite signs make them of the same sign.
Step III: Replace the given constant of equation ax + by + c = 0 by an arbitrary constant.
For example, the equation of a line perpendicular to the line 7x + 2y + 5 = 0 is 2x - 7y + c = 0; again, the equation of a line, perpendicular to the line 9x - 3y = 1 is 3x + 9y + k = 0.
Note:
Assigning different values to k in bx - ay + k = 0 we shall get different straight lines each of which is perpendicular to the line ax + by + c = 0. Thus we can have a family of straight lines perpendicular to a given straight line.
Solved examples to find the equations of straight lines perpendicular to a given straight line
1. Find the equation of a straight line that passes through the point (-2, 3) and perpendicular to the straight line 2x + 4y + 7 = 0.
Solution:
The equation of a line perpendicular to 2x + 4y + 7 = 0 is
4x - 2y + k = 0 …………………… (i) Where k is an arbitrary constant.
According to the problem equation of the perpendicular line 4x - 2y + k = 0 passes through the point (-2, 3)
Then,
4 ∙ (-2) - 2 ∙ (3) + k = 0
⇒ -8 - 6 + k = 0
⇒ - 14 + k = 0
⇒ k = 14
Now putting the value of k = 14in (i) we get, 4x - 2y + 14 = 0
Therefore the required equation is 4x - 2y + 14 = 0.
2. Find the equation of the straight line which passes through the point of intersection of the straight lines x + y + 9 = 0 and 3x - 2y + 2 = 0 and is perpendicular to the line 4x + 5y + 1 = 0.
Solution:
The given two equations are x + y + 9 = 0 …………………… (i) and 3x - 2y + 2 = 0 …………………… (ii)
Multiplying equation (i) by 2 and equation (ii) by 1 we get
2x + 2y + 18 = 0
3x - 2y + 2 = 0
Adding the above two equations we get, 5x = - 20
⇒ x = - 4
Putting x = -4 in (i) we get, y = -5
Therefore, the co-ordinates of the point of intersection of the lines (i) and (ii) are (- 4, - 5).
Since the required straight line is perpendicular to the line 4x + 5y + 1 = 0, hence we assume the equation of the required line as
5x - 4y + λ = 0 …………………… (iii)
Where λ is an arbitrary constant.
By problem, the line (iii) passes through the point (- 4, - 5); hence we must have,
⇒ 5 ∙ (- 4) - 4 ∙ (- 5) + λ = 0
⇒ -20 + 20 + λ = 0
⇒ λ = 0.
Therefore, the equation of the required straight line is 5x - 4y = 0.
● The Straight Line
- Straight Line
- Slope of a Straight Line
- Slope of a Line through Two Given Points
- Collinearity of Three Points
- Equation of a Line Parallel to x-axis
- Equation of a Line Parallel to y-axis
- Slope-intercept Form
- Point-slope Form
- Straight line in Two-point Form
- Straight Line in Intercept Form
- Straight Line in Normal Form
- General Form into Slope-intercept Form
- General Form into Intercept Form
- General Form into Normal Form
- Point of
Intersection of Two Lines
- Concurrency of Three Lines
- Angle between Two Straight Lines
- Condition of Parallelism of Lines
- Equation of a Line Parallel to a Line
- Condition of Perpendicularity of Two Lines
- Equation of a Line Perpendicular to a Line
- Identical Straight Lines
- Position of a Point Relative to a Line
- Distance of a Point from a Straight Line
- Equations of the Bisectors of the Angles between Two Straight Lines
- Bisector of the Angle which Contains the Origin
- Straight Line Formulae
- Problems on Straight Lines
- Word Problems on Straight Lines
- Problems on Slope and Intercept
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