Maths Topic 6: Linear Functions Quiz

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

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)

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)

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

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.

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.

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.

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.

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).

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).

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.

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.

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.

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.

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)