Question

A line passes through (2,4) and (-2,2)

Answer 1

Answer:

a) The value of $y$ is 6.

b) This line does not pass through the point of origin.

c) $y = \frac{1}{2}\cdot x + 3$ intersects with $y = -x+32$.

d) Both $y = \frac{1}{2}\cdot x + 3$ and $y = -x+32$ intersect in the first quadrant.

Step-by-step explanation:

a) By Analytical Geometry, we know that an equation of the line can be found by knowing the coordinates of two distinct points on the plane. The equation of the line is defined by this formula:

$y = m\cdot x + b$ (1)

Where:

$x$ - Independient variable.

$y$ - Dependent variable.

$m$ - Slope.

$b$ - x-Intercept.

To determine the slope and x-intercept of the equation of the line, we have to solve the following system of linear equations:

$m\cdot x_{1}+b = y_{1}$ (2)

$m\cdot x_{2}+b = y_{2}$ (3)

If we know that $(x_{1}, y_{1} ) = (2,4)$ and $(x_{2}, y_{2}) = (-2, 2)$, then the solution of the system is:

$m = \frac{1}{2}$, $b = 3$

Then, the equation of the line is $y = \frac{1}{2}\cdot x + 3$. If we know that $x = 6$, then the value of $y$ is:

$y = 6$

The value of $y$ is 6.

b) If this line passes through the point of origin, then the value of $y$ must be zero for $x = 0$. If we know that  $y = \frac{1}{2}\cdot x + 3$ and $x = 0$, then the value of $y$ is:

$y = 3$

The line does not pass through the point of origin since x-intercept is not zero.

c) A system of two linear equations always has an unique solution if and only if slopes and x-intercepts are different to each other. This condition is satisfied by $y = -x+32$, so we conclude that both lines intersect each other.

d) (Note: Question was incorrectly written. Correct form is: If your answer was yes in c, in which quadrant did they intersect?)

First, we solve the following system of linear equations:

$\frac{1}{2}\cdot x -y = -3$ (4)

$-x-y = -32$ (5)

The solution of this system is $(x,y) = (19.333, 12,667)$, which means that both lines intersects each other in the first quadrant.