Question 1001080
1. Given the linear equation {{{y=-(3/4)x-3}}}, find the y-coordinates of the points
({{{-8}}}, {{{3}}} ), =>{{{y=-(3/cross(4)1)(-cross(8)2)-3}}}=>{{{y=-(3)(-2)-3}}}=>{{{y=6-3}}}=>{{{y=3}}}

({{{-4}}},{{{0}}}  ), =>{{{y=-(3/cross(4)1)(-cross(4)1)-3}}}=>{{{y=-(3)(-1)-3}}}=>{{{y=3-3}}}=>{{{y=0}}}
({{{4}}},{{{-6}}}  )=>{{{y=-(3/cross(4)1)(cross(4)1)-3}}}=>{{{y=-(3)(1)-3}}}=>{{{y=-3-3}}}=>{{{y=-6}}}


Plot those points and graph the linear equation.

 {{{drawing( 600, 600, -10, 10, -10, 10,
circle(-8,3,.12),circle(-4,0,.12),circle(4,-6,.12),
locate(-8,3,p(-8,3)),locate(-4,0,p(-4,0)),locate(4,-6,p(-8,3)),
 graph( 600, 600, -10, 10, -10, 10, -(3/4)x-3)) }}}




2. Given the linear equation {{{y=-(1/3)x+6}}}, find the y-coordinates of the points

({{{-9}}},{{{9}}}  ), =>{{{y=-(1/cross(3)1)(-cross(9)3)+6}}}=>{{{y=-(1)(-3)+6}}}=>{{{y=3+6}}}=>{{{y=9}}}
({{{-3}}},{{{7}}}  ),=>{{{y=-(1/cross(3)1)(-cross(3)1)+6}}}=>{{{y=-(1)(-1)+6}}}=>{{{y=1+6}}}=>{{{y=7}}} 
({{{6}}},{{{4}}}  )=>{{{y=-(1/cross(3)1)(cross(6)2)+6}}}=>{{{y=-(1)(2)+6}}}=>{{{y=-2+6}}}=>{{{y=4}}} 

Plot those points and graph the linear equation.
  {{{drawing( 600, 600, -10, 10, -10, 10,
circle(-9,9,.12),circle(-3,7,.12),circle(6,4,.12),
locate(-9,9,p(-9,9)),locate(-3,7,p(-3,7)),locate(6,4,p(6,4)),
 graph( 600, 600, -10, 10, -10, 10, -(1/3)x+6)) }}}



3. Write the slope-intercept equation for the line that passes through ({{{14}}},{{{ -2}}}) and is perpendicular to {{{7x -10y = 18}}}

the slope-intercept form of the given equation

{{{7x -18 = 10y}}}

{{{y= (7/10)x -9/5 }}}


*[invoke equation_parallel_or_perpendicular "perpendicular", "7/10", "-9/5", 14, -2]




4. Write the slope-intercept equation for the line that passes through ({{{1}}}, {{{-9}}}) and ({{{3}}}, {{{-1}}}). 


*[invoke change_this_name10094 1, -9, 3, -1] 

your equation is:

{{{y=4x -13}}}