Question 129738
The x-intercept of a linear equation in two variables is that point where the line crosses the x-axis, i.e. the point that has a y-coordinate of 0.  So, substitute 0 for y and solve for x.


{{{5x+4(0)-20=0}}}
{{{5x=20}}}
{{{x=4}}}


So the x-intercept is (4,0).


Two lines are perpendicular if and only if their slopes are negative reciprocals of each other, that is {{{m[1]=-1/m[2]}}}.


So, we need to find the slope of {{{4x+3y-12=0}}}.  Put it into slope-intercept form by solving for y:


{{{4x+3y-12=0}}}
{{{3y=-4x+12}}}
{{{y=-4x/3+4}}}


That gives us a slope of {{{-4/3}}}.   Any line perpendicular to this line must have a slope that is the negative reciprocal of {{{-4/3}}}, namely {{{3/4}}}.


Now that we have identified the slope of the desired line and a point through which it must pass, we can use the point-slope form of the line to derive the required equation.  The point-slope form of the line is {{{y-y[1]=m(x-x[1])}}} where {{{x[1]}}} and {{{y[1]}}} are the coordinates of the given point and {{{m}}} is the slope.


{{{y-0=(3/4)(x-4)}}}
{{{y=3x/4-3}}}


Or, in standard form:
{{{3x-4y=12}}}


Check:
The red line is the derived equation, the green one is the perpendicular and the blue one has the common x-intercept.



{{{drawing(400,400,-5,5,-5,5,
grid(1),
graph(400,400,-5,5,-5,5,3x/4-3,-4x/3+4,-5x/4+5))}}}