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 y2−y1x2−x1Gradient of RS=3−15−(−1)=26=13Using gradient formula y2−y1x2−x1Gradient of RS=3−15−(−1)=26=13



Question 2:
In diagram below, PQ is a straight line with gradient  −12−12 .


Find the x-intercept of the straight line PQ.

Solution:
Using the gradient formula, m=−y-interceptx-intercept−12=−(−3x-intercept)x-intercept=3×(−2)=−6Using the gradient formula, m=−y-interceptx-intercept−12=−(−3x-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 unitsOR=√102−(−6)2=8 unitsy-intercept of RS=−6x-intercept of RS=−8Using the gradient formula, m=−y-interceptx-intercept∴ Gradient of RS=−(−6−8)=−34RS=10 units, OS=6 unitsOR=√102−(−6)2=8 unitsy-intercept of RS=−6x-intercept of RS=−8Using the gradient formula, m=−y-interceptx-intercept∴ Gradient of RS=−(−6−8)=−34


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=34x−6y=34x−6

Therefore, gradient of the straight line = 34.34.


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

Solution:
For y-intercept, x = 0
3(0) + 2y = 5
   y=52∴y-intercept=52.   y=52∴y-intercept=52.



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