Given:
A point (-4,y) lies on the line BC of the triangle.
The objective is to find the value of y.
Explanation:
Consider two points of the straight line BC of the triangle ABC.
[tex]\begin{gathered} (x_1,y_1)=(-3,2) \\ (x_2,y_2)=(0,5) \end{gathered}[/tex]
The slope of the straight line can be calculated as,
[tex]m=\frac{y_2-y_1}{x_2-x_1}\text{ . . . . .(1)}[/tex]
On plugging the coordinates in equation (1),
[tex]\begin{gathered} m=\frac{5-2}{0-(-3)} \\ m=\frac{3}{3} \\ m=1 \end{gathered}[/tex]
To find y:
Now, consider the given coordinate and the point C.
[tex]\begin{gathered} (x_1,y_1)=(-3,2) \\ (x_3,y_3)=(-4,y) \end{gathered}[/tex]
On plugging the obtained values in the equation of slope,
[tex]\begin{gathered} 1=\frac{y-2}{-4-(-3)} \\ 1=\frac{y-2}{-4+3} \\ -4+3=y-2 \\ -1+2=y \\ y=1 \end{gathered}[/tex]
Hence, the value of y is 1.
