Mathematics, 29.12.2019 09:31 jackiecroce1

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?

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Line vw is to be drawn on the graph such that it is perpendicular to line . if the coordinates of po

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Answer from: infoneetusinghoyg22o

The value of y is 3.

Since the lines are to be perpendicular, that means that their slopes must be negative reciprocals of each other.

First we find the slope of line ST. The formula for slope is:
$m=\frac{y_2-y_1}{x_2-x_1}$

Substituting our coordinates (using point T as point 2 and point S as point 1), we have:
$m=\frac{2-0}{5--5}=\frac{2-0}{5+5}=\frac{2}{10}=\frac{1}{5}$

Since the slope of line VW must be the negative reciprocal of 1/5, it would be -5/1=-5.

The coordinates of V are (-2, 0). Using V as point 1 and W as point 2, in our slope formula for this line we have:
$-5=\frac{-2-y}{0--1}=\frac{-2-y}{0+1}=\frac{-2-y}{1}=-2-y
\\
\\-5=-2-y$

Add 2 to both sides:
-5+2=-2-y+2
-3=-y

Divide both sides by -1:
-3/-1=-y/-1
3=y

Answer from: doggosbepis

The value of y is 3

Further explanation

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 :

$\large {\boxed {m = \frac{y_2 - y_1}{x_2 - x_1}} }$

If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :

$\large {\boxed {y - y_1 = m ( x - x_1 )} }$

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 ) .

$m_{ST} = \frac{y_2 - y_1}{x_2 - x_1}$

$m_{ST} = \frac{2 - 0}{5 - (-5)}$

$m_{ST} = \frac{2}{10}$

$\large {\boxed {m_{ST} = \frac{1}{5} } }$

Next , we will calculate the gradient of the line that passes through the point V( 0 , -2 ) and W( -1 , y ) .

$m_{VW} = \frac{y_2 - y_1}{x_2 - x_1}$

$m_{VW} = \frac{y - (-2)}{-1 - 0}$

$\large {\boxed {m_{VW} = \frac{y + 2}{-1} } }$

The conditions of the line perpendicular to each other will satisfy the following formula.

$m_{ST} \times m_{VW} = -1$

$\frac{1}{5} \times \frac{y + 2}{-1} = -1$

$\frac{y + 2}{-5} = -1$

$y + 2 = -5 \times -1$

y + 2 = 5

y = 5 - 2

$\large {\boxed {y = 3} }$

Learn moreInfinite Number of Solutions : System of Equations : System of Linear equations : Answer details

Grade: High School

Subject: Mathematics

Chapter: Linear Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point

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