Quiz 3

Week 2: Rectangular Coordinate System & Straight Lines

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2026-09-12T00:00:00.000Z
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Cartesian plane, distance formula, section formula, slopes, parallel & perpendicular lines, and line equation forms. # Mathematics I — Week 2: Rectangular Coordinate System & Straight Lines IITM’s Week 2 moves into the **rectangular coordinate system, slope, parallel/perpendicular lines, and different representation...

Mathematics I — Week 2: Rectangular Coordinate System & Straight Lines

IITM’s Week 2 moves into the rectangular coordinate system, slope, parallel/perpendicular lines, and different representations/equations of a line.

TL;DR

Week 2 is basically about this chain:
points→distance/slope→line equations→parallel/perpendicular conditions→geometric interpretation\text{points} \rightarrow \text{distance/slope} \rightarrow \text{line equations} \rightarrow \text{parallel/perpendicular conditions} \rightarrow \text{geometric interpretation}
The highest-value ideas are:
  • coordinates (x,y)(x,y)
  • distance and midpoint
  • slope
  • slope-intercept form y=mx+cy=mx+c
  • point-slope form
  • two-point form
  • general form Ax+By+C=0Ax+By+C=0
  • parallel lines: same slope
  • perpendicular lines: slopes multiply to −1-1
  • vertical/horizontal line edge cases

1. The rectangular coordinate system

The standard Cartesian plane uses two perpendicular axes:
  • horizontal axis: xx-axis
  • vertical axis: yy-axis
Their intersection is the origin:
O=(0,0)O=(0,0)
A point is represented by an ordered pair:
P=(x,y)P=(x,y)
where:
  • xx = horizontal position
  • yy = vertical position
The order matters.
(2,5)≠(5,2)(2,5)\neq(5,2)

2. Why is it called Cartesian?

"Cartesian" comes from René Descartes, whose Latinized name was Cartesius.
The important mathematical idea is more interesting than the history:
Geometry can be translated into algebra.
A geometric object such as a line becomes an equation.
For example:
y=2x+1y=2x+1
is simultaneously:
  • an algebraic equation
  • a set of ordered pairs
  • a geometric straight line
This algebra-geometry bridge is one of the foundational ideas of analytic geometry.

3. Quadrants

The coordinate plane is divided into four regions.
text
          y
          ↑
      II  |  I
          |
----------+----------→ x
          |
     III  |  IV
          |
Signs:
Quadrantxxyy
I++++
II−-++
III−-−-
IV++−-
Mnemonic:
text
I      (+,+)
II     (-,+)
III    (-,-)
IV     (+,-)

4. Distance between two points

Suppose:
P=(x1,y1)P=(x_1,y_1)
and
Q=(x2,y2)Q=(x_2,y_2)
Then:
PQ=(x2−x1)2+(y2−y1)2PQ= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
This is just the Pythagorean theorem.
Horizontal change:
Δx=x2−x1\Delta x=x_2-x_1
Vertical change:
Δy=y2−y1\Delta y=y_2-y_1
So:
PQ=(Δx)2+(Δy)2PQ=\sqrt{(\Delta x)^2+(\Delta y)^2}

5. Example: distance

For:
P=(1,2),Q=(4,6)P=(1,2),\qquad Q=(4,6)
we have:
Δx=4−1=3\Delta x=4-1=3 Δy=6−2=4\Delta y=6-2=4
Therefore:
PQ=32+42PQ=\sqrt{3^2+4^2} =9+16=\sqrt{9+16} =5=5
This is the familiar 33-44-55 triangle.

6. Midpoint

The midpoint between:
(x1,y1)(x_1,y_1)
and
(x2,y2)(x_2,y_2)
is:
(x1+x22,y1+y22)\boxed{ \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) }
You simply average corresponding coordinates.
Example:
P=(2,4),Q=(8,10)P=(2,4),\qquad Q=(8,10)
Then midpoint:
M=(2+82,4+102)M= \left( \frac{2+8}{2}, \frac{4+10}{2} \right) =(5,7)=(5,7)

7. What is slope?

Slope measures how much yy changes as xx changes.
For two points:
(x1,y1),(x2,y2)(x_1,y_1),\qquad(x_2,y_2)
the slope is:
m=y2−y1x2−x1\boxed{ m=\frac{y_2-y_1}{x_2-x_1} }
provided:
x2≠x1x_2\neq x_1
Another common notation:
m=ΔyΔxm=\frac{\Delta y}{\Delta x}
or:
m=riserunm=\frac{\text{rise}}{\text{run}}

8. Intuition behind slope

Slope answers:
If xx increases by 11, approximately how much does yy change?
Suppose:
m=3m=3
Then:
Δx=1⇒Δy=3\Delta x=1 \Rightarrow \Delta y=3
If:
m=12m=\frac12
then:
Δx=2⇒Δy=1\Delta x=2 \Rightarrow \Delta y=1

9. Positive slope

If:
m>0m>0
then the line rises as we move left to right.
text
      /
     /
    /
As xx increases:
y↑y\uparrow

10. Negative slope

If:
m<0m<0
the line falls as we move left to right.
text
\
 \
  \
As xx increases:
y↓y\downarrow

11. Zero slope

A horizontal line has:
m=0m=0
Example:
y=4y=4
For any two points on it:
y2−y1=0y_2-y_1=0
so:
m=0m=0

12. Undefined slope

A vertical line has undefined slope.
Example:
x=5x=5
All points have the same xx coordinate:
x2−x1=0x_2-x_1=0
so:
m=y2−y10m=\frac{y_2-y_1}{0}
which is undefined.
Very common trap:
Vertical slope is not infinity in ordinary elementary algebra.
It is undefined.

13. Equation of a line

The most familiar form is:
y=mx+c\boxed{ y=mx+c }
where:
  • mm = slope
  • cc = yy-intercept
The yy-intercept occurs where:
x=0x=0
Then:
y=cy=c
so the line crosses the yy-axis at:
(0,c)(0,c)

14. Example

y=2x+3y=2x+3
Here:
m=2m=2
and:
c=3c=3
So:
  • slope =2=2
  • yy-intercept =(0,3)=(0,3)

15. Point-slope form

Suppose a line has slope mm and passes through:
(x1,y1)(x_1,y_1)
Then:
y−y1=m(x−x1)\boxed{ y-y_1=m(x-x_1) }
This is one of the most useful line formulas.

16. Example: point + slope

Find the line through:
(2,5)(2,5)
with slope:
m=3m=3
Use:
y−y1=m(x−x1)y-y_1=m(x-x_1)
Therefore:
y−5=3(x−2)y-5=3(x-2)
Expand:
y−5=3x−6y-5=3x-6
so:
y=3x−1y=3x-1

17. Two-point form

Suppose the line passes through:
(x1,y1)(x_1,y_1)
and:
(x2,y2)(x_2,y_2)
First find slope:
m=y2−y1x2−x1m= \frac{y_2-y_1}{x_2-x_1}
Then substitute into point-slope form:
y−y1=y2−y1x2−x1(x−x1)y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
This is sometimes called the two-point form.

18. Example: line through two points

Find the equation through:
(1,2)(1,2)
and:
(3,6)(3,6)
First:
m=6−23−1=42=2m= \frac{6-2}{3-1} = \frac42 = 2
Use point-slope form:
y−2=2(x−1)y-2=2(x-1)
Expand:
y−2=2x−2y-2=2x-2
Thus:
y=2x\boxed{y=2x}

19. General form of a line

Another common representation is:
Ax+By+C=0\boxed{ Ax+By+C=0 }
where AA and BB are not both zero.
Example:
2x+3y−6=02x+3y-6=0
To convert into slope-intercept form:
3y=−2x+63y=-2x+6 y=−23x+2y=-\frac23x+2
Therefore:
m=−23m=-\frac23

20. Slope from general form

For:
Ax+By+C=0Ax+By+C=0
if:
B≠0B\neq0
then:
By=−Ax−CBy=-Ax-C
thus:
y=−ABx−CBy=-\frac ABx-\frac CB
Therefore:
m=−AB\boxed{ m=-\frac AB }

21. The intercepts of a line

For:
Ax+By+C=0Ax+By+C=0

xx-intercept

Set:
y=0y=0
Then:
Ax+C=0Ax+C=0
so:
x=−CAx=-\frac CA
assuming A≠0A\neq0.

yy-intercept

Set:
x=0x=0
Then:
By+C=0By+C=0
so:
y=−CBy=-\frac CB
assuming B≠0B\neq0.

22. Intercept form

If a line intersects the axes at:
(a,0)(a,0)
and:
(0,b)(0,b)
its equation can be written:
xa+yb=1\boxed{ \frac xa+\frac yb=1 }
This is called intercept form.

23. Deriving intercept form

The slope between:
(a,0)(a,0)
and:
(0,b)(0,b)
is:
m=b−00−a=−bam= \frac{b-0}{0-a} = -\frac ba
Using:
y=mx+cy=mx+c
and c=bc=b:
y=−bax+by=-\frac ba x+b
Divide through by bb:
yb=−xa+1\frac yb=-\frac xa+1
Therefore:
xa+yb=1\frac xa+\frac yb=1

24. Parallel lines

Two nonvertical lines are parallel when they have equal slopes:
m1=m2\boxed{ m_1=m_2 }
Example:
y=2x+1y=2x+1
and:
y=2x−7y=2x-7
Both have slope:
22
Therefore they are parallel.
Their intercepts differ, so they are distinct lines.

25. Coincident lines

If two equations represent the same line, they are coincident.
Example:
2x+4y=62x+4y=6
and:
x+2y=3x+2y=3
The first equation is simply twice the second.
Therefore these are not two separate parallel lines.
They are the same line.

26. Perpendicular lines

For two nonvertical/nonhorizontal lines:
m1m2=−1\boxed{ m_1m_2=-1 }
Equivalently:
m2=−1m1m_2=-\frac1{m_1}
This is called the negative reciprocal relationship.

27. Example

Suppose:
m1=2m_1=2
Then the slope of a perpendicular line is:
m2=−12m_2=-\frac12
because:
2(−12)=−12\left(-\frac12\right)=-1

28. Important perpendicular edge case

A horizontal line:
y=cy=c
has slope:
00
A vertical line:
x=kx=k
has undefined slope.
They are perpendicular.
So the formula:
m1m2=−1m_1m_2=-1
cannot literally be used here because one slope is undefined.
Geometric reasoning handles this case.

29. Parallel condition in general form

Suppose:
A1x+B1y+C1=0A_1x+B_1y+C_1=0
and:
A2x+B2y+C2=0A_2x+B_2y+C_2=0
Their slopes are:
−A1B1-\frac{A_1}{B_1}
and:
−A2B2-\frac{A_2}{B_2}
For parallel lines:
−A1B1=−A2B2-\frac{A_1}{B_1} = -\frac{A_2}{B_2}
Therefore:
A1B2=A2B1\boxed{ A_1B_2=A_2B_1 }
assuming the normal nonvertical situation.

30. Perpendicular condition in general form

Using:
m1m2=−1m_1m_2=-1
we get:
(−A1B1)(−A2B2)=−1\left(-\frac{A_1}{B_1}\right) \left(-\frac{A_2}{B_2}\right) =-1
Thus:
A1A2B1B2=−1\frac{A_1A_2}{B_1B_2}=-1
so:
A1A2+B1B2=0\boxed{ A_1A_2+B_1B_2=0 }
This formula is useful because it avoids explicitly computing slopes.

31. Angle of inclination

The angle a line makes with the positive xx-axis is called its angle of inclination, usually θ\theta.
Its slope satisfies:
m=tan⁡θ\boxed{ m=\tan\theta }
where:
0∘≤θ<180∘0^\circ\leq\theta<180^\circ
for ordinary line orientation.

32. Examples of slope-angle relation

If:
θ=0∘\theta=0^\circ
then:
m=tan⁡0∘=0m=\tan0^\circ=0
Horizontal line.
If:
θ=45∘\theta=45^\circ
then:
m=tan⁡45∘=1m=\tan45^\circ=1
If:
θ=135∘\theta=135^\circ
then:
m=tan⁡135∘=−1m=\tan135^\circ=-1
A vertical line corresponds to:
θ=90∘\theta=90^\circ
where:
tan⁡90∘\tan90^\circ
is undefined.
Exactly consistent with vertical slope being undefined.

33. Angle between two lines

For slopes m1m_1 and m2m_2, the angle θ\theta between the lines satisfies:
tan⁡θ=∣m2−m11+m1m2∣\boxed{ \tan\theta= \left| \frac{m_2-m_1}{1+m_1m_2} \right| }
If:
1+m1m2=01+m_1m_2=0
then:
m1m2=−1m_1m_2=-1
which means the lines are perpendicular.

34. Collinearity

Three points are collinear if they lie on the same straight line.
For:
A=(x1,y1)A=(x_1,y_1) B=(x2,y2)B=(x_2,y_2) C=(x3,y3)C=(x_3,y_3)
one method is to compare slopes:
mAB=mBCm_{AB}=m_{BC}
provided denominators are valid.

35. Safer collinearity test

Slope formulas can become awkward for vertical lines.
A safer determinant-style condition is:
x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0\boxed{ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) =0 }
If this equals zero, the three points are collinear.

36. Why that test works

Three points form a triangle.
If they are collinear, the triangle has area:
00
The area formula is:
Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac12 \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
So collinearity means:
Area=0\text{Area}=0

37. Equation of horizontal and vertical lines

These are worth treating separately.

Horizontal

Through:
(a,b)(a,b)
the horizontal line is:
y=b\boxed{y=b}
Example:
Through:
(4,7)(4,7)
the horizontal line is:
y=7y=7

Vertical

Through:
(a,b)(a,b)
the vertical line is:
x=a\boxed{x=a}
Example:
Through:
(4,7)(4,7)
the vertical line is:
x=4x=4

38. A line as a set of solutions

This is an important conceptual upgrade.
Consider:
2x+y=52x+y=5
The line is not merely "the drawing."
It is the entire set:
L={(x,y)∈R2:2x+y=5}L= \{(x,y)\in\mathbb R^2:2x+y=5\}
Every point satisfying the equation belongs to the line.
Examples:
(0,5)(0,5)
because:
2(0)+5=52(0)+5=5
and:
(2,1)(2,1)
because:
2(2)+1=52(2)+1=5
But:
(1,1)(1,1)
does not belong because:
2(1)+1=3≠52(1)+1=3\neq5
This way of thinking becomes useful throughout mathematics.

39. Normal vector intuition

For:
Ax+By+C=0Ax+By+C=0
the vector:
(A,B)(A,B)
is perpendicular to the line.
It is called a normal vector.
This explains why:
A1A2+B1B2=0A_1A_2+B_1B_2=0
appears in some line relationships: it is related to the dot product of normal vectors.
You do not need to over-focus on this if IITM has not yet introduced vectors, but the intuition is useful.

40. Distance from a point to a line

For point:
(x0,y0)(x_0,y_0)
and line:
Ax+By+C=0Ax+By+C=0
the perpendicular distance is:
d=∣Ax0+By0+C∣A2+B2\boxed{ d= \frac{|Ax_0+By_0+C|} {\sqrt{A^2+B^2}} }
Example:
Point:
(1,2)(1,2)
Line:
3x+4y−10=03x+4y-10=0
Then:
d=∣3(1)+4(2)−10∣32+42d= \frac{|3(1)+4(2)-10|} {\sqrt{3^2+4^2}} =∣3+8−10∣5= \frac{|3+8-10|}{5} =15= \frac15

41. Intersection of two lines

Consider:
y=2x+1y=2x+1
and:
y=−x+7y=-x+7
At their intersection, both equations have the same xx and yy.
Therefore:
2x+1=−x+72x+1=-x+7
so:
3x=63x=6 x=2x=2
Then:
y=2(2)+1=5y=2(2)+1=5
Intersection:
(2,5)\boxed{(2,5)}
Geometrically, solving simultaneous linear equations means:
Find the common point of the lines.

42. Three possibilities for two lines

For two lines in the plane:

One intersection

Different slopes:
m1≠m2m_1\neq m_2
One unique solution.

No intersection

Parallel distinct lines:
m1=m2m_1=m_2
but different intercepts.
No solution.

Infinitely many intersections

Same line.
Infinitely many solutions.
This connects geometry directly to systems of equations.

43. Common question pattern: line through point parallel to another

Given:
2x−3y+7=02x-3y+7=0
Find a parallel line through:
(1,4)(1,4)
Original line:
−3y=−2x−7-3y=-2x-7 y=23x+73y=\frac23x+\frac73
So:
m=23m=\frac23
Parallel line also has:
m=23m=\frac23
Use point-slope:
y−4=23(x−1)y-4= \frac23(x-1)
Done.

44. Faster general-form method

If a line parallel to:
Ax+By+C=0Ax+By+C=0
is required, keep the same AA and BB:
Ax+By+K=0Ax+By+K=0
Then determine KK using the given point.
Example:
Parallel to:
2x−3y+7=02x-3y+7=0
through:
(1,4)(1,4)
write:
2x−3y+K=02x-3y+K=0
Substitute:
2(1)−3(4)+K=02(1)-3(4)+K=0 2−12+K=02-12+K=0 K=10K=10
Therefore:
2x−3y+10=0\boxed{ 2x-3y+10=0 }
Much faster.

45. Line through a point perpendicular to another

Suppose original slope:
m=3m=3
Then perpendicular slope:
m⊥=−13m_\perp=-\frac13
If the desired line passes through:
(6,2)(6,2)
then:
y−2=−13(x−6)y-2=-\frac13(x-6)
Common mistake: reciprocal without negative sign
If:
m=4m=4
the perpendicular slope is:
−14-\frac14
not:
14\frac14
and not:
−4-4
Rule:
m⊥=−1mm_\perp=-\frac1m
Common mistake: line slope from general equation
Given:
3x−2y+5=03x-2y+5=0
Do not say:
m=32m=\frac32
without checking signs.
Rearrange:
−2y=−3x−5-2y=-3x-5 y=32x+52y=\frac32x+\frac52
Thus:
m=32m=\frac32
Here it happens to be positive.
General shortcut:
m=−ABm=-\frac AB
For:
A=3,B=−2A=3,\qquad B=-2
we get:
m=−3−2=32m=-\frac{3}{-2} =\frac32
Common mistake: confusing intercepts
For:
y=3x+5y=3x+5
55 is the yy-intercept, not the xx-intercept.
To find the xx-intercept:
y=0y=0
so:
0=3x+50=3x+5 x=−53x=-\frac53
Thus xx-intercept:
(−53,0)\left(-\frac53,0\right)
Common mistake: treating ordered pairs as unordered
For coordinates:
(2,7)≠(7,2)(2,7)\neq(7,2)
Unlike sets:
{2,7}={7,2}\{2,7\}=\{7,2\}
The entire coordinate system depends on order.
This connects directly to Week 1's Cartesian products, where ordered pairs also appeared.

50. Week 1 → Week 2 connection

Week 1 introduced:
A×BA\times B
and ordered pairs:
(x,y)(x,y)
Week 2 specializes this to:
R×R=R2\mathbb R\times\mathbb R = \mathbb R^2
The coordinate plane is essentially:
R2\boxed{\mathbb R^2}
Every point is an ordered pair of real numbers.
So the syllabus is more connected than it initially looks.

51. One very useful mental model

A straight line equation is a constraint on two variables.
For:
y=2x+1y=2x+1
xx and yy cannot vary independently.
Once xx is chosen:
yy
is forced.
For example:
x=0⇒y=1x=0\Rightarrow y=1 x=1⇒y=3x=1\Rightarrow y=3 x=2⇒y=5x=2\Rightarrow y=5
This is also why a nonvertical line can be interpreted as a function:
f(x)=mx+cf(x)=mx+c

52. Vertical line and function connection

Consider:
x=3x=3
At:
x=3x=3
there are infinitely many yy values.
Therefore it cannot be written as:
y=f(x)y=f(x)
as an ordinary real-valued function of xx.
This foreshadows the vertical-line test that appears later in the course.

53. Formula sheet

Distance

d=(x2−x1)2+(y2−y1)2\boxed{ d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

Midpoint

M=(x1+x22,y1+y22)\boxed{ M= \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) }

Slope

m=y2−y1x2−x1\boxed{ m= \frac{y_2-y_1}{x_2-x_1} }

Slope-intercept form

y=mx+c\boxed{ y=mx+c }

Point-slope form

y−y1=m(x−x1)\boxed{ y-y_1=m(x-x_1) }

General form

Ax+By+C=0\boxed{ Ax+By+C=0 }

Slope from general form

m=−AB\boxed{ m=-\frac AB }

Parallel

m1=m2\boxed{ m_1=m_2 }

Perpendicular

m1m2=−1\boxed{ m_1m_2=-1 }

Point-to-line distance

d=∣Ax0+By0+C∣A2+B2\boxed{ d= \frac{|Ax_0+By_0+C|} {\sqrt{A^2+B^2}} }

54. What you should actually master

Don't judge Week 2 by whether you have read all of this.
You should be able to do these without much hesitation:
  1. Plot/read (x,y)(x,y) coordinates.
  2. Identify quadrants.
  3. Find distance between two points.
  4. Find midpoint.
  5. Calculate slope.
  6. Recognize positive, negative, zero and undefined slopes.
  7. Construct a line from a point and slope.
  8. Construct a line from two points.
  9. Move between: y=mx+cy=mx+c and Ax+By+C=0.Ax+By+C=0.
  10. Find xx- and yy-intercepts.
  11. Test whether lines are parallel.
  12. Test whether lines are perpendicular.
  13. Construct parallel/perpendicular lines through a given point.
  14. Determine whether points are collinear.
  15. Solve for the intersection of two lines.
If those are comfortable, Week 2 is basically done.

55. Week 2 conceptual map

text
Coordinate plane
│
├── point (x,y)
├── distance
└── midpoint
     │
     ▼
Slope
│
├── positive
├── negative
├── zero
└── undefined
     │
     ▼
Straight line
│
├── y = mx + c
├── point-slope
├── two-point
├── intercept form
└── Ax + By + C = 0
     │
     ▼
Relationships
│
├── parallel
├── perpendicular
├── coincident
└── intersecting

What comes next

Week 3 = Quadratic Functions: quadratic functions, their graphs, minima/maxima, vertex, slope-related interpretation, and quadratic equations.
That is where Maths 1 starts becoming more function-centric rather than purely coordinate geometry.

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:
Question :: Answer.
  • Distance between (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) :: (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • Slope of line through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) :: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}
  • Condition for parallel lines with slopes m1,m2m_1, m_2 :: m1=m2m_1 = m_2
  • Condition for perpendicular lines with slopes m1,m2m_1, m_2 :: m1m2=−1  ⟹  m2=−1m1m_1 m_2 = -1 \implies m_2 = -\frac{1}{m_1}
  • Slope of general line ax+by+c=0ax + by + c = 0 :: m=−abm = -\frac{a}{b}
  • Perpendicular distance from (x0,y0)(x_0, y_0) to ax+by+c=0ax + by + c = 0 :: ∣ax0+by0+c∣a2+b2\frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}
  • Midpoint of segment between (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) :: (x1+x22,y1+y22)(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.
1

Question 1

What is the slope of a line that is perpendicular to the line 3x−6y+7=03x - 6y + 7 = 0?
MCQ
3 Unit Assessment
A. 1/2
B. -2
C. 2
D. -1/2
2

Question 2

What is the perpendicular distance from the point (1,2)(1, 2) to the line 3x+4y−1=03x + 4y - 1 = 0?
NAT
3 Unit Assessment

Active Recall Session

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