Line VW is to be drawn on the graph such that it is perpendicular to line . If the coordinates of point W are (–1, y), what is the value of y?
Y=

The value of y is 3
Solving linear equation mean calculating the unknown variable from the equation.
Let the linear equation : y = mx + c
If we draw the above equation on Cartesian Coordinates , it will be a straight line with :
m → gradient of the line
( 0 , c ) → y - intercept
Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :
[tex]\large {\boxed {m = \frac{y_2 - y_1}{x_2 - x_1}} }[/tex]
If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :
[tex]\large {\boxed {y - y_1 = m ( x - x_1 )} }[/tex]
Let us tackle the problem.
Firstly , we will calculate the gradient of the line that passes through the point S( -5 , 0 ) and T( 5 , 2 ) .
[tex]m_{ST} = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
[tex]m_{ST} = \frac{2 - 0}{5 - (-5)}[/tex]
[tex]m_{ST} = \frac{2}{10}[/tex]
[tex]\large {\boxed {m_{ST} = \frac{1}{5} } }[/tex]
Next , we will calculate the gradient of the line that passes through the point V( 0 , -2 ) and W( -1 , y ) .
[tex]m_{VW} = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
[tex]m_{VW} = \frac{y - (-2)}{-1 - 0}[/tex]
[tex]\large {\boxed {m_{VW} = \frac{y + 2}{-1} } }[/tex]
The conditions of the line perpendicular to each other will satisfy the following formula.
[tex]m_{ST} \times m_{VW} = -1[/tex]
[tex]\frac{1}{5} \times \frac{y + 2}{-1} = -1[/tex]
[tex]\frac{y + 2}{-5} = -1[/tex]
[tex]y + 2 = -5 \times -1[/tex]
[tex]y + 2 = 5[/tex]
[tex]y = 5 - 2[/tex]
[tex]\large {\boxed {y = 3} }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Linear Equations
Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point