Answer:
y = 3x + 1
Step-by-step explanation:
Coordinates of the points given on line BC,
A(0, 1), B(-3, 2) and C(3, 0)
Since, slope of a line passing through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is,
m = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
Therefore, slope of line BC,
[tex]m_1=\frac{2-0}{-3-3}[/tex]
[tex]=-\frac{1}{3}[/tex]
Let the slope of a line perpendicular to BC = [tex]m_2[/tex]
By the property of perpendicular lines,
[tex]m_1\times m_2=-1[/tex]
[tex](-\frac{1}{3})\tims m_2=-1[/tex]
[tex]m_2=3[/tex]
Since, equation of a line passing through point [tex](x_1,y_1)[/tex] and slope m is given by,
[tex]y-y_1=m(x-x_1)[/tex]
Therefore, equation of a line passing through A(0, 1) and slope = 3,
y - 1 = 3(x -0)
y - 1 = 3x
y = 3x + 1
Equation of the perpendicular line on BC and passing through point A is,
y = 3x + 1
