Answer:
The equation of a line which passes through the points (-2,3) and (2,0) is [tex]3 x+4 y=6 \text { or } y=-\frac{3 x}{4}+\frac{3}{2}[/tex]
Solution:
Let us assume that the [tex](x_2, y_2) = (2,0)[/tex] and [tex](x_1, y_1) = (-2,3)[/tex]
The slope of the line m is [tex]=\frac{y_2-y_1}{x_2-x_1}=\frac{0-3}{2-(-2)}=\left(-\frac{3}{4}\right)[/tex]
We know the equation of a line at a given point [tex](x_1, y_1)[/tex] is [tex](y-y_1) = m(x-x_1)[/tex]
Let me take the point (2,0) here,
So the equation of the line is
[tex]\Rightarrow(y-0)=\left(-\frac{3}{4}\right)(x-2)[/tex]
[tex]\Rightarrow y=\left(-\frac{3}{4}\right)(x-2)[/tex]
[tex]\Rightarrow 4 y=(-3) \times(x-2)[/tex]
[tex]\Rightarrow 4 y=-3 x+6[/tex]
[tex]\Rightarrow 3 x+4 y=6 \text[/tex] or [tex]y=-\frac{3 x}{4}+\frac{3}{2}[/tex]
So, the equation is [tex]3 x+4 y=6[/tex]
Or [tex]y=-\frac{3 x}{4}+\frac{3}{2}[/tex]
