Verify coordinate incidence and determine a parameter

Line

\(\displaystyle y=2x+3\)

Point

\(\displaystyle P=(4,9)\)

Which check is correct?

Move the point horizontally and vertically. Find positions on the line and positions away from it. Compare the vertical values

Which condition identified every position on the line?

Calculate, then compare

The line rule gives the vertical value for each horizontal coordinate. A candidate point is on the line only when its own vertical coordinate matches that result

1. 1Use the horizontal coordinate Substitute the complete signed input \(\displaystyle y_{\mathrm{rule}}=2(4)+3\)
2. 2Evaluate the rule Multiply before adding the constant \(\displaystyle \begin{aligned}y_{\mathrm{rule}}&=8+3\\&=11\end{aligned}\)
3. 3Compare with the point Different vertical values mean the point is off the line \(\displaystyle 11\ne9\)

Line

\(\displaystyle y=-3x+2\)

Horizontal coordinate

\(\displaystyle x=-4\)

Calculate the line's vertical value for the given horizontal coordinate

Keep the negative input in brackets when you multiply

Line

\(\displaystyle y=3x+14\)

Point

\(\displaystyle Q=(-2,8)\)

Which check correctly decides whether the point lies on the line?

Line

\(\displaystyle y=4x-1\)

Point

\(\displaystyle R=(2,6)\)

What is wrong with this argument?

Testing coordinate incidence

A point is on a line only when both coordinates make the equation true. The residual is zero on the line and changes sign from one side to the other

1. 1Substitute both coordinates Replace x and y in the same equation with the point's coordinates \(\displaystyle r=3(2)-2(5)+4\)
2. 2Use a zero residual A residual of zero means the point lies on the line \(\displaystyle r=3(2)-2(5)+4=0\Rightarrow(2,5)\text{ is on }3x-2y+4=0\)
3. 3Interpret a non-zero residual A non-zero residual means the point is not on the line \(\displaystyle r=3(4)-2(3)+4=10\ne0\Rightarrow(4,3)\text{ is not on the line}\)
4. 4Check every required item Collinearity and concurrency require the remaining point or line to satisfy the same condition \(\displaystyle I=g_1\cap g_2,\;r_{g_3}(I)=0\)

Pipeline

\(\displaystyle 2x-y+3=0\)

Point C

\(\displaystyle (5, 13)\)

Point D

\(\displaystyle (-2, 1)\)

Valve

\(\displaystyle (k, 9)\)

A straight pipeline on a plan follows 2x - y + 3 = 0. Inspection points are marked at C(5, 13) and D(-2, 1), and a valve is to be placed at (k, 9) on the pipeline. Find the residual at D.

Substitute both coordinates of D into 2x - y + 3, keeping each sign

A residual sign needs its coefficient

For Ax+By+C=0, first substitute the point to obtain r. Zero means on the line. A nonzero residual alone does not name the side: multiplying the whole equation by −1 reverses r without moving the line

1. 1Nonvertical line When B is nonzero, divide by B to obtain the point height minus the line height at the same x. Positive means above; negative means below \(\displaystyle y_P-y_{\mathrm{line}}=\frac{r}{B}\)
2. 2Vertical line When B=0, A must be nonzero. Divide by A to obtain the horizontal gap. Positive means right; negative means left. Above/below is not the comparison for a vertical line \(\displaystyle x_P-x_{\mathrm{line}}=\frac{r}{A}\)

Given conditions

\(\displaystyle \begin{gathered}3x-2y+4=0\\P=(2,6)\end{gathered}\)

Where is P relative to this nonvertical line?

Given conditions

\(\displaystyle \begin{gathered}-2x+6=0\\P=(4,-7)\end{gathered}\)

Where is P relative to this vertical line?

A=(2,−5) and B=(2,4). What is the gradient of AB?

The same point (2,3) is supplied twice. Can it determine the gradient of one unique line?

A=(0,1), B=(2,5) and C=(4,8). Do all three points lie on the same straight line?

Check a gradient calculation

Use signed changes in a consistent point order. A gradient is a ratio of changes, not a midpoint or a distance

1. 1Reverse both differences The numerator and denominator both change sign, leaving the quotient unchanged
2. 2Check a third point Matching finite gradients from one fixed point establish that distinct points lie on the same nonvertical straight line. For a vertical line, check that all horizontal coordinates match
3. 3Context and units The gradient has vertical-axis units per horizontal-axis unit. Later rate problems use this same calculation

One line must contain every point

Use two DISTINCT points to define a candidate line, then check each remaining point. For a nonvertical line you may compare gradients from one fixed point. A general equation also handles vertical lines without division by a zero run

1. 1Define the candidate P and Q lie on the line below. Their distinct coordinates determine one line \(\displaystyle \begin{gathered}P=(1,2),\quad Q=(3,6)\\2x-y=0\end{gathered}\)
2. 2Check the remaining point R=(4,8) also lies on that line, so all three are collinear. If R were (4,9), its residual would be −1 and the claim would fail \(\displaystyle 2(4)-8=0\)
3. 3Vertical case Distinct points with the same x-coordinate share a vertical line. Do not divide by zero or call its gradient zero

Concurrent means the SAME intersection

First solve two nonparallel lines as a simultaneous pair. Their intersection is only a candidate: substitute that same pair into every remaining line

1. 1Find the candidate Adding the first two equations gives 2x=6, then y=2. \(\displaystyle \begin{gathered}g_1:x+y=5\\g_2:x-y=1\\I=(3,2)\end{gathered}\)
2. 2Check the third This third line contains I, so all three are concurrent \(\displaystyle g_3:2x+y=8,\quad2(3)+2=8\)
3. 3Do not count pairwise crossings If the third line were 2x+y=9, all pairs would still intersect, but not at the same point. If the first two lines are parallel or coincident, choose another suitable pair or analyse that case separately; this practice bank supplies a nonparallel first pair

Given conditions

\(\displaystyle \begin{gathered}I=(3,2)\\g_3:2x+y-9=0\end{gathered}\)

The first two lines meet at I=(3,2). Calculate the residual in the THIRD line shown

The graph shows three pairwise crossings. What does I establish?

Determine a parameter from incidence

Treat k as an unknown coefficient or coordinate. Substitute the entire given point, collect the k terms, solve the resulting equation, then check the result in the original line

1. 1Parameter in a coordinate For (k,9) on 2x−y+3=0, substitution gives the equation below \(\displaystyle 2k-9+3=0\Rightarrow k=3\)
2. 2Parameter in a coefficient For P=(2,1) on (2k−1)x+y=7, substitute before solving \(\displaystyle \begin{gathered}2(2k-1)+1=7\\4k=8\Rightarrow k=2\end{gathered}\)
3. 3Check the original At k=2 the second example is 3x+y=7, and P gives 3(2)+1=7. A coefficient, constant or coordinate is not itself an answer without the stated condition

Pipeline

\(\displaystyle 2x-y+3=0\)

Point C

\(\displaystyle (5, 13)\)

Point D

\(\displaystyle (-2, 1)\)

Valve

\(\displaystyle (k, 9)\)

Find the value of k that puts the valve at (k, 9) on the pipeline

Substitute x = k and y = 9, then solve 2k - 9 + 3 = 0.

Given conditions

\(\displaystyle \begin{gathered}(2k-1)x+y=7\\P=(2,1)\end{gathered}\)

Find k so that P lies on the parameterised line

Parallelism uses direction, not height

For nonvertical lines, compare gradients. More generally, nonzero normal vectors (A,B) are parallel when their cross-product is zero. Check the constants too if you need distinct parallel lines rather than the same line

1. 1Direction condition This avoids dividing by a zero coefficient and handles vertical lines \(\displaystyle A_1B_2-A_2B_1=0\)
2. 2Example The direction condition gives −2(k+1)+6=0, so k=2. At that value the constants differ while the direction coefficients agree, so these are distinct parallel lines \(\displaystyle \begin{gathered}(k+1)x-2y+3=0\\3x-2y+8=0\end{gathered}\)
3. 3A genuine line is required A and B cannot both be zero. A condition with zero coefficient of k may yield no k or many k values, rather than one unique answer. The ordinary bank deliberately avoids those degeneracies

Given conditions

\(\displaystyle \begin{gathered}(k+1)x-2y+3=0\\3x-2y+8=0\end{gathered}\)

Find k so that these lines are parallel

Given conditions

\(\displaystyle \begin{gathered}x+(2k-3)y=4\\x=7\end{gathered}\)

Find k so that the parameterised line is vertical and parallel to x=7.

Given conditions

\(\displaystyle \begin{gathered}(k+1)x-2y+3=0\\3x-2y+8=0\end{gathered}\)

At k=1, are the two displayed lines parallel?

The point (0,0) is substituted into kx=0. Does this determine one unique k for a genuine line?

What does a correct supplied verdict demonstrate?