Assessment 6. Work individually. The coordinates of line [tex]AB[/tex] are [tex]A(5,4)[/tex] and [tex]B(-2, t)[/tex]. The line is perpendicular to a line whose gradient is [tex]-\frac{1}{2}[/tex]. Determine the value of [tex]t[/tex]. The coordinates of line AB are [tex]A (3,6)[/tex] and [tex]B (1,1)[/tex]. Determine the gradient of a line perpendicular to line [tex]AB[/tex]. Without drawing, determine whether each of the following pairs of lines are perpendicular or not.
(a)
[tex]\begin{array}{l}
y=-3 x+4 \
y=\frac{1}{3} x+12
\end{array}[/tex]
(b) [tex]y=\frac{3}{5} x+3[/tex]
[tex]y=5 x-12[/tex]
(c) [tex]y=\frac{1}{7} x-9[/tex]
[tex]y=-\frac{1}{7} x+11[/tex]
Chebet drew line NM with coordinates N(3,5) and M(1,-1). Magera drew line [tex]CD[/tex] with coordinates [tex]C(3,6)[/tex] and [tex]D(4,3)[/tex]. Determine whether line [tex]NM[/tex] is perpendicular to line [tex]CD[/tex] or not.
Answer (2)
Solved for t using the perpendicularity condition: t = − 10 .
Determined the gradient of a perpendicular line: − 5 2 .
Assessed perpendicularity for pairs of lines, finding only pair (a) perpendicular.
Concluded lines NM and C D are not perpendicular.
Explanation
Problem Overview Let's break down this problem step-by-step. We'll tackle each part individually, making sure to clearly explain the concepts and calculations involved.
Finding the value of t First, we need to find the value of t given that line A B with coordinates A ( 5 , 4 ) and B ( − 2 , t ) is perpendicular to a line with a gradient of − 2 1 .
The gradient of line A B is given by: m A B = − 2 − 5 t − 4 = − 7 t − 4
Since line A B is perpendicular to a line with gradient − 2 1 , the product of their gradients is -1: m A B × ( − 2 1 ) = − 1 − 7 t − 4 × ( − 2 1 ) = − 1 14 t − 4 = − 1 t − 4 = − 14 t = − 10
Finding the perpendicular gradient Next, we need to determine the gradient of a line perpendicular to line A B where A ( 3 , 6 ) and B ( 1 , 1 ) .
The gradient of line A B is given by: m A B = 1 − 3 1 − 6 = − 2 − 5 = 2 5
The gradient of a line perpendicular to A B is the negative reciprocal of m A B :
m ⊥ = − m A B 1 = − 2 5 1 = − 5 2
Checking perpendicularity of line pairs Now, let's determine whether the given pairs of lines are perpendicular.
(a) y = − 3 x + 4 and y = 3 1 x + 12 . The gradients are -3 and 3 1 .
Product of gradients: ( − 3 ) × 3 1 = − 1 . Therefore, the lines are perpendicular.
(b) y = 5 3 x + 3 and y = 5 x − 12 . The gradients are 5 3 and 5. Product of gradients: 5 3 × 5 = 3 . Therefore, the lines are not perpendicular.
(c) y = 7 1 x − 9 and y = − 7 1 x + 11 . The gradients are 7 1 and − 7 1 .
Product of gradients: 7 1 × ( − 7 1 ) = − 49 1 . Therefore, the lines are not perpendicular.
Checking perpendicularity of lines NM and CD Finally, let's determine whether line NM with coordinates N ( 3 , 5 ) and M ( 1 , − 1 ) is perpendicular to line C D with coordinates C ( 3 , 6 ) and D ( 4 , 3 ) .
The gradient of line NM is given by: m NM = 1 − 3 − 1 − 5 = − 2 − 6 = 3
The gradient of line C D is given by: m C D = 4 − 3 3 − 6 = 1 − 3 = − 3
Check if the lines are perpendicular: m NM × m C D = 3 × ( − 3 ) = − 9 Since the product is not -1, the lines are not perpendicular.
Final Answers In summary:
The value of t is -10.
The gradient of a line perpendicular to line A B is − 5 2 .
The pairs of lines in (a) are perpendicular, while (b) and (c) are not.
Line NM is not perpendicular to line C D .
Examples
Understanding perpendicular lines is crucial in many real-world applications. For example, architects use this concept to ensure walls are built at right angles for structural stability. Similarly, in navigation, knowing the perpendicular direction to a path helps in making precise turns and avoiding obstacles. In computer graphics and game development, perpendicularity is used to calculate lighting effects and create realistic shadows, enhancing the visual experience.
The value of t is -10, and the gradient of a line perpendicular to line AB is -2/5. The pairs of lines in part (a) are perpendicular, while (b) and (c) are not, and lines NM and CD are also not perpendicular.
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