Collection of Maths Problems

Line through two points

Task number: 2350

• Variant 1

  \(A=(1{,}2)\) and \(B=(3{,}4)\)

• Hint

  For \((x_0,y_0)\) choose the point \(A\), then find \((p,q)\), s.t. \(\pi\) goes through \(B\) for \(t=1\).

  For general form combine two equations into a single (eliminate \(t\)).

  Derive the division form from the general by division by \(-c\).

  The slope-intercept form derive from the general by extracting \(y\) (when \(b\ne 0\)).

• Resolution

  We deduce the parametric form from equation \((3{,}4)=(1{,}2)+(p,q)\), i.e. \((x,y)=(1{,}2)+t(2{,}2)\). This form could be simplified by reducing \(p\) and \(q\) by their greatest common divisor and by subtracting \(\frac{x_0}{p}(p,q)\) from \((x_0,y_0)\). This results into parametric form \((x,y)=(0{,}1)+t(1{,}1)\).

  The general form fould be obtained by setting \(t=x\) into \(y=1+t\), i.e. \(x-y+1=0\). As a general method the first equation could be multiplied by \(q\), the second \(-p\), and the both added. This results into \(a=q,\ b=-p,\ c= py_0-qx_0\).

  The division form yields coefficients \(g=-1\), \(h=1\), i.e. \(\frac{x}{-1} + \frac{y}{1} = -x+y=1\)

  The slope-intercept form is \(y=x+1\).

  Ad uniqueness: \((x_0,y_0)\) could be any point on \(\pi\), the coefficents \((p,q)\) could be multiplied by any nonzero number as well as the coefficients \(a,b\) and \(c\) of the general form.

  The slope-intercept and division forms are unique (if exist).

• Variant 2

  \(A=(-3{,}0)\) and \(B=(2{,}3)\)

• Result

  Parametric form: \((x,y)=(-3{,}0)+t(5{,}3)\), general: \(3x-5y+9=0\), division: \(\frac{x}{-3}+\frac{y}{9/5}=1\), and slope-intercept: \(y=\frac{3}{5}x+\frac{9}{5}\).

• Variant 3

  \(A=(1,-1)\) and \(B=(2{,}2)\)

• Result

  \((x,y)=(1,-1)+t(1{,}3)\), \(3x-y-4=0\), \(\frac{x}{4/3}+\frac{y}{-4}=1\), \(y=3x-4\).

• Variant 4

  \(A=(2,-1)\) and \(B=(2{,}4)\)

• Result

  \((x,y)=(2,-1)+t(0{,}1)\), \(x-2=0\), division and slope-intercept forms do not exist, since \(\pi\) is parallel with the \(y\)-axis.

• Variant 5

  \(A=(0{,}1)\) and \(B=(4{,}3)\)

• Result

  \((x,y)=(0{,}1)+t(2{,}1)\), \(x-2y+2=0\), \(\frac{x}{-2}+\frac{y}{1}=1\), \(y=\frac{1}{2}x+1\).

• Variant 6

  \(A=(2{,}1)\) and \(B=(4{,}2)\)

• Result

  \((x,y)=(0{,}0)+t(2{,}1)\), \(x-2y=0\), division form does not exists, since \(\pi\) goes through the origin, \(y=\frac{1}{2}x\).

• Variant 7

  \(A=(0{,}3)\) and \(B=(-2{,}3)\)

• Result

  \((x,y)=(0{,}3)+t(1{,}0)\), \(y-3=0\), division form does not exist, since \(\pi\) is parallel with the \(x\)-axis, \(y=3\).