10.4.1 Gradient of a Straight Line, PT3 Focus Practice


Question 1:
Diagram below shows a straight line RS on a Cartesian plane.
 
Find the gradient of RS.

Solution:
Using gradient formula  y 2 − y 1 x 2 − x 1 Gradient of RS= 3−1 5−( −1 ) = 2 6 = 1 3



Question 2:
In diagram below, PQ is a straight line with gradient  − 1 2 .


Find the x-intercept of the straight line PQ.

Solution:
Using the gradient formula, m=− y-intercept x-intercept − 1 2 =−( −3 x-intercept ) x-intercept=3×( −2 )=−6



Question 3:
Diagram below shows a straight line RS drawn on a Cartesian plane.

 
It is given that the distance of RS is 10 units.
Find the gradient of RS.

Solution:
RS=10 units, OS=6 units OR= 10 2 − ( −6 ) 2 =8 units y-intercept of RS=−6 x-intercept of RS=−8 Using the gradient formula, m=− y-intercept x-intercept ∴ Gradient of RS=−( −6 −8 )=− 3 4


Question 4:
The gradient of the straight line 3x – 4y = 24 is

Solution:
Rearrange the equation in the form y = mx+ c
3x – 4y = 24
4y = 3x – 24
y = 3 4 x − 6

Therefore, gradient of the straight line = 3 4 .


Question 5:
Determine the y-intercept of the straight line 3x + 2y = 5

Solution:
For y-intercept, x = 0
3(0) + 2y = 5
   y= 5 2 ∴y-intercept= 5 2 .



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