The coordinates of line [tex]AB[/tex] are [tex]A(3,6)[/tex] and [tex]B(1,1)[/tex]. Determine the slope of a line perpendicular to line AB. Without drawing, determine whether each of the following pairs of lines are perpendicular or not.
(a) [tex]y=-3 x+4[/tex] and [tex]y=\frac{1}{3} x+12[/tex]
(b) [tex]y=\frac{3}{5} x+3[/tex] and [tex]y=5 x-12[/tex]
(c) [tex]y=\frac{1}{7} x-9[/tex] and [tex]y=-\frac{1}{7} x+11[/tex]
Chebet drew line NM with coordinates N(3,5) and M(1,-1). Magero drew line [tex]CD[/tex] with coordinates [tex]C(3,6)[/tex] and [tex]D(4,3)[/tex]. Determine whether line NM is perpendicular to line [tex]CD[/tex] or not.
Extended activity:
The equation of a straight line is [tex]3 y-6 x+4=0[/tex]. Determine the equation through point [tex](1,2)[/tex] and perpendicular to the given line.
Answer (2)
Calculate the gradient of line AB using the formula m = x 2 − x 1 y 2 − y 1 , resulting in m A B = 2.5 .
Determine the gradient of a line perpendicular to AB by taking the negative reciprocal: m p er p = − 2.5 1 = − 0.4 .
Check the perpendicularity of pairs of lines by multiplying their gradients; lines are perpendicular if the product is -1. Only pair (a) with gradients -3 and 3 1 is perpendicular.
Find the equation of a line through (1,2) perpendicular to 3 y − 6 x + 4 = 0 by finding the perpendicular gradient ( − 2 1 ) and using the point-slope form, resulting in y = − 2 1 x + 2 5 .
y = − 2 1 x + 2 5
Explanation
Gradient of line AB First, let's find the gradient of line AB using the coordinates A(3,6) and B(1,1). The gradient, denoted as m , is calculated as the change in y divided by the change in x :
Calculating gradient of AB m A B = x 2 − x 1 y 2 − y 1 = 1 − 3 1 − 6 = − 2 − 5 = 2.5
Gradient of perpendicular line The gradient of a line perpendicular to AB is the negative reciprocal of the gradient of AB. Therefore:
Calculating gradient of perpendicular line m p er p = − m A B 1 = − 2.5 1 = − 0.4
Checking perpendicularity of lines Now, let's determine whether the given pairs of lines are perpendicular. Two lines are perpendicular if the product of their gradients is -1.
Pair (a) (a) y = − 3 x + 4 and y = 3 1 x + 12 . The gradients are -3 and 3 1 .
m 1 × m 2 = − 3 × 3 1 = − 1 Since the product is -1, these lines are perpendicular.
Pair (b) (b) y = 5 3 x + 3 and y = 5 x − 12 . The gradients are 5 3 and 5. m 1 × m 2 = 5 3 × 5 = 3 Since the product is not -1, these lines are not perpendicular.
Pair (c) (c) y = 7 1 x − 9 and y = − 7 1 x + 11 . The gradients are 7 1 and − 7 1 .
m 1 × m 2 = 7 1 × − 7 1 = − 49 1 Since the product is not -1, these lines are not perpendicular.
Checking perpendicularity of lines NM and CD Next, let's determine whether line NM is perpendicular to line CD. The coordinates of N are (3,5) and M are (1,-1). The coordinates of C are (3,6) and D are (4,3).
Gradient of NM The gradient of NM is: m NM = 1 − 3 − 1 − 5 = − 2 − 6 = 3
Gradient of CD The gradient of CD is: m C D = 4 − 3 3 − 6 = 1 − 3 = − 3
Product of gradients NM and CD The product of the gradients is: m NM × m C D = 3 × − 3 = − 9 Since the product is not -1, line NM is not perpendicular to line CD.
Equation of perpendicular line Finally, let's find the equation of a line through the point (1,2) and perpendicular to the line 3 y − 6 x + 4 = 0 . First, we need to find the gradient of the given line. Rewrite the equation in the form y = m x + c :
Gradient of given line and perpendicular line 3 y = 6 x − 4 y = 2 x − 3 4 The gradient of the given line is 2. The gradient of a line perpendicular to this line is − 2 1 .
Point-slope form Now, use the point-slope form of a line equation, y − y 1 = m ( x − x 1 ) , with the point (1,2) and the gradient − 2 1 :
Simplifying the equation y − 2 = − 2 1 ( x − 1 ) y = − 2 1 x + 2 1 + 2 y = − 2 1 x + 2 5
Examples
Understanding perpendicular lines is crucial in many real-world applications. For example, architects use this concept to ensure walls are perfectly vertical to the ground, providing structural integrity to buildings. Similarly, in navigation, knowing the perpendicular direction to a path helps in determining the shortest route to avoid obstacles. These principles ensure safety and efficiency in design and planning.
The slope of line AB is 2.5, and a perpendicular line has a slope of -0.4. The pairs of lines (a) are perpendicular, while (b) and (c) are not, and line NM is not perpendicular to line CD. The perpendicular line through (1,2) is represented by the equation y = -\frac{1}{2}x + \frac{5}{2}.
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