Given:
A new four way intersection is being constructed in New York Hyde park through point P(-3,-15).
Equation of line of one road : [tex]y=-\dfrac{3}{4}x+7[/tex].
New road that will run perpendicular to the first road
To find:
The equation of line for the new road.
Solution:
The slope intercept form of a line is
[tex]y=mx+b[/tex]
where, m is slope and b is y-intercept.
We have,
[tex]y=-\dfrac{3}{4}x+7[/tex]
Slope of this line is [tex]-\dfrac{3}{4}[/tex] and y-intercept is 7.
Product of slopes of two perpendicular line is -1.
[tex]m_1\times m_2=-1[/tex]
[tex]-\dfrac{3}{4}\times m_2=-1[/tex]
[tex]m_2=\dfrac{4}{3}[/tex]
The point slope form of a line is
[tex]y-y_1=m(x-x_1)[/tex]
where, m is slope.
The slope of new line is [tex]\dfrac{4}{3}[/tex] and it passes through P(-3,-15). So, the equation of line of new road is
[tex]y-(-15)=\dfrac{4}{3}(x-(-3))[/tex]
[tex]y+15=\dfrac{4}{3}(x+3)[/tex]
[tex]y+15=\dfrac{4}{3}x+4[/tex]
Subtract 15 from both sides.
[tex]y=\dfrac{4}{3}x+4-15[/tex]
[tex]y=\dfrac{4}{3}x-11[/tex]
Therefore, the equation of the line representing the new road is [tex]y=\dfrac{4}{3}x-11[/tex].
