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Quiz about Maths Topic 6 Linear Functions
Quiz about Maths Topic 6 Linear Functions

Maths Topic 6: Linear Functions Quiz


This quiz will test you on basic rules and applications of co-ordinate geometry. You will need a pen, paper and calculator for this quiz. Good Luck.

A multiple-choice quiz by dialga483. Estimated time: 4 mins.
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Author
dialga483
Time
4 mins
Type
Multiple Choice
Quiz #
366,154
Updated
Dec 03 21
# Qns
15
Difficulty
Average
Avg Score
9 / 15
Plays
163
-
Question 1 of 15
1. What is the length of the line segment joining the points (1,3) and (-3,0)? Hint


Question 2 of 15
2. What is the midpoint between the points (-1,4) and (5,2)? Hint


Question 3 of 15
3. What is the gradient of the line that passes through the point (2,3) and (-3,4)? Hint


Question 4 of 15
4. What is the gradient of the line that marks an angle of 135° with the x-axis in the positive direction? Hint


Question 5 of 15
5. In degrees and minutes, what is the angle that a straight line makes with the positive direction of the x-axis if the gradient is 0.5? Hint


Question 6 of 15
6. What is a valid equation for the line with a gradient of -4 and passing through the point (-2,3)? Hint


Question 7 of 15
7. What is the equation of the line with an x-intercept of 3 and a y-intercept of 2? Hint


Question 8 of 15
8. What is the equation of the line parallel to the line 2x-y+3=0 and passing through the point (1,-5)? Hint


Question 9 of 15
9. The lines 2x-3y=3 and 5x-2y=13 intersect at which point? Hint


Question 10 of 15
10. If three lines are concurrent, then they are said to what? Hint


Question 11 of 15
11. What is the equation of the line through (-1,2) that passes through the point of intersection of the lines 2x+y-5=0 and x-3y+1=0? Hint


Question 12 of 15
12. What is the perpendicular distance of the point (4,-3) from the line 3x-4y-1=0? Hint


Question 13 of 15
13. What is the size of the acute angle between the lines x+2y=5 and 3y=x+3? Hint


Question 14 of 15
14. If a point P is to divide two points on a line segment externally in a ratio m:n, then what about the given ratio must be changed compared to an internal division? Hint


Question 15 of 15
15. If A is (-2,-1) and B is (1,5), what are the co-ordinates of point P that divides AB externally in the ratio 2:5? Hint



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Quiz Answer Key and Fun Facts
1. What is the length of the line segment joining the points (1,3) and (-3,0)?

Answer: 5 units

To solve this question, we need to use the rule d=sqrt((x2-x1)^2-(y2-y1)^2), where d is the distance between the two points.
So,
d=sqrt((-3-1)^2+(0-3)^2)
d=sqrt(16+9)
d=sqrt(25)
d=5 units
2. What is the midpoint between the points (-1,4) and (5,2)?

Answer: (2,3)

To find the midpoint, we use the rule M=((x1+x2)/2,(y1+y2)/2), where M is the midpoint.
So,
M=((-1+5)/2,(4+2)/2)
M=(4/2,6/2)
M=(2,3)
3. What is the gradient of the line that passes through the point (2,3) and (-3,4)?

Answer: -1/5

To solve this question, we use the rule m=(y2-y1)/(x2-x1), where m is the gradient of the line
So,
m=(4-3)/(-3-2)
m=-(1/5)
4. What is the gradient of the line that marks an angle of 135° with the x-axis in the positive direction?

Answer: -1

To solve this question, we use the rule m=tan(x), where x is the angle made by the line in the positive direction.
So,
m=tan(135°)
m=-1
5. In degrees and minutes, what is the angle that a straight line makes with the positive direction of the x-axis if the gradient is 0.5?

Answer: 26 degrees, 34 minutes

To solve this question, we can once again use the rule m=tan(x)
So, with gradient 0.5
tan(x)=0.5
x=arctan(0.5)
x=26.56505118
Therefore, x=26 degrees, 34 minutes.
6. What is a valid equation for the line with a gradient of -4 and passing through the point (-2,3)?

Answer: 4x+y+5=0

To answer this question, we need to use the rule y-y1=m(x-x1), where m is the gradient.
So,
y-3=-4(x+2)
y-3=-4x-8
4x+y+5=0
Therefore, the equation of the line is 4x+y+5=0.
7. What is the equation of the line with an x-intercept of 3 and a y-intercept of 2?

Answer: 2x+3y-6=0

To solve this question we need to use the rule (x/a)+(y/b)=1, where a and b are the x and y intercepts of the equation respectively.
So,
(x/3)+(y/2)=1
2x+3y=6
2x+3y-6=0
Therefore, the equation of the line is 2x+3y-6=0.
8. What is the equation of the line parallel to the line 2x-y+3=0 and passing through the point (1,-5)?

Answer: 2x-y-7=0

For two lines to be parallel, they must share the same gradient. From our given equation, we can re-arrange it to work out that its gradient is 2.
Now using y-y1=m(x-x1)
y+5=2(x-1)
y+5=2x-2
2x-y-7=0
Therefore, the equation of the line is 2x-y-7=0.
9. The lines 2x-3y=3 and 5x-2y=13 intersect at which point?

Answer: (3,1)

To solve this question, we can treat the two equations as simultaneous equations and solve them together.
So,
2x-3y=3 (equation 1)
5x-2y=13 (equation 2)
(equation 1) x 5, (equation 2) x 2
10x-15y=15 (equation 3)
10x-4y=26 (equation 4)
(equation 4)-(equation 3)
11y=11
y=1
Sub into (equation 1)
2x-3=3
2x=6
x=3
Therefore, the lines intersect at (3,1).
10. If three lines are concurrent, then they are said to what?

Answer: Pass through the same point

If three lines are concurrent, then they all pass through the same point. An example of this are the lines 3x-y+1=0, x+2y+12=0 and 4x-3y-7=0, which all intersect at the point (-2,-5).
11. What is the equation of the line through (-1,2) that passes through the point of intersection of the lines 2x+y-5=0 and x-3y+1=0?

Answer: x+3y-5=0

We can solve this question by letting the two given equations be equal to each other, with a constant k be in front of the second equation.
So,
2x+y-5=k(x-3y+1)
Now we can sub in the point (-1,2)
2(-1)+2-5=k((-1)-3(2)+1)
-2+2-5=k(-1-6+1)
-5=-6k
k=(5/6)
Now we can substitute this in
2x+y-5=(5/6)(x-3y+1)
12x+6y-30=5(x-3y+1)
12x+6y-30=5x-15y+5
7x+21y-35=0
7(x+3y-5)=0
Therefore, the equation of the line is x+3y-5=0.
12. What is the perpendicular distance of the point (4,-3) from the line 3x-4y-1=0?

Answer: 23/5 units

To solve this question, we use the formula d=abs((ax+by+c)/sqrt(a^2+b^2)), where d is the perpendicular distance
So, subbing in the given point,
d=abs((3(4)-4(-3)-1)/sqrt((3)^2+(-4)^2))
d=abs((12+12-1)/sqrt(25))
d=23/5
Therefore, the perpendicular distance is 23/5 units.
13. What is the size of the acute angle between the lines x+2y=5 and 3y=x+3?

Answer: 45

To solve this question, we use the rule tan(O)=abs((m1-m2)/(1+m1m2)), where O is the angle between the lines and m1 and m2 are the gradients of the lines.
So for equation 1,
x+2y=5
-2y=x-5
y=(-x/2)+5/2
Therefore, m1=(-1/2)
For equation 2,
3y=x+3
y=(x/3)+1
Therefore, m2=(1/3)
Now we can sub these into the rule. So,
Tan(O)=abs(((-1/2)-(1/3))/(1+(-1/2)(1/3)))
Tan(O)=abs(-1)
Tan(O)=1
Therefore, the angle between the lines is equal to 45.
14. If a point P is to divide two points on a line segment externally in a ratio m:n, then what about the given ratio must be changed compared to an internal division?

Answer: Either m or n must be changed to negative

When one point externally divides another two points in a given ratio, then one of the numbers must be changed to a negative. When doing internal division, both numbers must be kept positive.
15. If A is (-2,-1) and B is (1,5), what are the co-ordinates of point P that divides AB externally in the ratio 2:5?

Answer: (-4,-5)

When dividing an interval externally, we take one of the numbers in the given ratio and make it negative. In this case, we will use 2 and change it to -2.
The rule for dividing an interval is P=((mx2+nx1)/(m+n),(my2+ny1)/(m+n)), where m and n are the numbers from the ratio m:n.
So,
P=((1(-2)+5(-2))/(5-2),(5(-2)+5(-1))/(5-2))
P=((-2-10)/3,(-10-5)/3)
P=((-12/3),(-15/3))
P=(-4,-5)
Source: Author dialga483

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